Imagine there are 1,000 doors and you pick 1. All other doors except 1 are opened and you're given the offer: keep the door you picked, or pick this other door. What are the chances you picked the right door (vs. this other door)?
People seem to intuitively understand that having only one door unopened is a massive "hint" to where the prize is.
For someone who doesn't get it, the problem is then whether the correct extension is all the doors being opened. With 3 doors it's the same.
For me the most sensible explanation requires you to know that a dud door is always opened, thus the probability from the 2/3 is the one you are switching to.
In the 1,000 doors problem, my odds of being right initially were something like 1/1000 and then it changes to something like 998/1000 or 999/1000 for switching, I can’t intuitively grasp exactly what the odds become of winning if I switch, I just know it’s high. Bringing it down to 3 doors doesn’t help me much — it’s still something like 1/2 or 1/3.
You don't need to know the exact odds to understand that it's higher. I think that's the main takeaway of making it a 1,000 door problem. It makes it intuitive that the correct solution is to switch. The exact probability doesn't matter.
Try thinking of it this way. It may or may not be any more useful. I've seen different people come to understand the problem from different examples.
1. Observe that 3/3 = 1. Pedantic, yes, but good for frame of mind here.
2. Pick one of three doors. (1/3 odds)
3. Gain information that one of the three doors is a loser.
4. Note your odds on choosing the original door correctly are still 1/3.
5. Note that if you change doors, there are still 2/3 doors there to choose.
5. Note you're not going to switch to the known loser door, so if you change doors you know 100% which of the other 2/3 of doors to choose.
The intuition usually is that you're down to two doors after the loser door is opened, but that's not the case. There are still three doors. The host has just told you that if you trade doors, you know which door to trade for. So trade for it.
Note there's a newer version of "Let's Make a Deal", hosted by Wayne Brady, but there is no option to switch after a losing door has been shown in that version.
Because it isn't different really.
It is always a goat door that is opened, so you don't gain any information about your door by the opening of the goat door.
I'm thinking of a number between 1 and 10, guess it. If I now tell you a number I promise is not the one I was thinking and not your number, you have no more information about if you were correct.
Because it really centers on the initial premise: Monty will always open a goat door after your choice, no matter what.
So, you make a totally random choice. That choice must be 1/3 right, right? Now the thing that you already knew would definitely happen happens: Monty opens a goat door. How can your odds suddenly jump to 1/2?
Are you saying every single time you play the game, you always have a 1/2 chance of getting it right first time?
You choosing a door in the first place gave you 1-in-3 chance. But your 1-in-3 choice is deducted from the 3-of-3 choices that Monty could have had, so Monty only had a 2-in-3 choice: there's a 2/3 chance that the car is in the pool of doors from which Monty could select. Monty has a 100% chance of choosing a door without a car. Therefore, you inherit the 2-in-3 chances if you change your selection.
The first door will have the car 1/3 of the time. The second door's chances had been expanded to the remaining 2/3 percent thanks to Monty always choosing the last 1/3 door which is guaranteed to not have a car.
What you knew about the door you initially picked hasn't changed at all. What you now know about the other doors has. By giving you information about 1/2 of the other 2/3 doors, Monty has given you an extra 1/3 chance if you pick among those.
IMO what helps is to imagine slightly changing the order.
First step is still that you pick a door. There's a 1/3 chance it has the car. Now you can either keep that single door (with a 1/3 chance of a car), or switch and get both of the other two doors (each with a 1/3 chance of the car, for a total of 2/3 chance). After you pick, I'll reveal all the goats.
Assume there are seven billion people on the planet. One of them knows the location of a specific hidden treasure. I know who it is and I ask you to guess who it is, but you have no possible way of knowing or even getting a hint about it.
You pick some random person. I then bring in another stranger and tell you that the person who knows where the hidden treasure is is either the random person you chose or the one I brought in.
At this point, there are only two possibilities:
1. You happened to randomly choose the right person on Earth and in my surprise, I had to pick some other random stranger to pretend they knew the secret.
2. You chose a total rando who has no idea what's going on and the person I brought in is in fact the one who knows where the treasure is
That explanation never worked for me, because you can turn it around to the situation where Monty does not know where the car is. Say there are 1000 doors, and you pick door 429. On his way to open door 429, Monty stumbles, falls, and accidentally knocks open every door except door 128. If by some coincidence all opened doors happened to contain goats, you will have nothing to gain from switching. Very counter-intuitive, but just as true as the original problem.
A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round".
Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.
You can't really turn it around, because Monty knowing and using his knowledge of where the car is to reveal only goats is what makes switching advantageous.
In the case of the clumsy Monty of your example, it goes like this:
1. There is a 1/1000 chance door 429 has the car.
2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000.
2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds.
3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch.
3a. Your door has the car, which happens 1/1000 games.
3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games.
In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching.
In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this:
1. There is a 1/1000 chance your door, 429, has the car.
2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours.
2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car.
3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door.
3a. You get there via 2a in 1/1000 games.
3b. You get there via 2b in 999/1000 games.
If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.
In the original problem, Monty always opens a door to reveal a goat. This strongly implies that he knows which doors have goats.
Your alternative is different. If the doors opened are truly selected randomly, then the odds are high that he would reveal a car, especially in the 1000-door version. What are the odds that Monty could choose 998 doors randomly out of 1000 and NOT pick the one with the car?
If Monty doesn't know where the car is, then you're arguing with a completely different version of the game that could result in him opening the door with the car, so the strategy is going to be different.
Changing the scenario is the exactly the point of why I am unsure increasing the number of doors gives true intuition. In the original problem with the original scenario, people intuitively think switching doesn't matter, when it provably does. With 1000 doors and the Monty Fall scenario, again the intuition is wrong. So are we gaining true intuition for the Monty Hall scenario by expanding the setup, or is it just some subconscious Bayesian thinking accidentally steering us toward the right answer.
I was lucky enough to get this explanation from my high school physics teacher, who first presented the classic Monty Hall problem and then illustrated the changing of the odds by substituting all of the lockers in our school for the three doors. Switching gives a clear advantage.
The rest of this post is an anecdote from the same class that this brought to mind, and is unrelated to the topic. Maybe we can say it shows how good teachers engage their students or something, but really it’s just a good yarn.
We were learning about inelastic vs elastic collisions, and how an elastic collision has 2x the energy of an inelastic one. The teacher asked for a volunteer, and a bright-eyed student rose to the occasion. The teacher gave him some safety glasses and told him to lie down on the floor.
The teacher took the inelastic ball and said, “Okay, I’m gonna drop this on your forehead now, ready?” PLONK. “Ow.”
“Remember that feeling! This is the elastic one, and it has the same mass, so it should hurt twice as much.” PLONK. “Ow.”
The teacher asked, “So, did the second one hurt more than the first?” The rest of us anticipated the experimental confirmation of what we’d just learned about.
“...I couldn’t really tell the difference,” said the student.
“Yeah,” said the teacher, “I knew you wouldn’t. I just wanted to see if you’d let me do it.”
That's the explanation that helped me, and led me to realize something about this problem. The problem asks what is the best strategy. It doesn't ask what should you do at that moment. At that moment implies you should process the information available to you right then and there. The information available to you at that time - two doors, ignores prior information, which I think is the counterintuitive aspect that trips people up.
When I first heard about the problem, I struggled with the seemingly 50/50 chances. either you picked the right door and now switch and lose, or you picked a wrong door and switch and win. Switching seemed a zero sum game.
The explanation that works best for me is that you were more likely to have picked a wrong door in the first place, so while the impacts are opposite equals, the likelihoods are not equivalent.
Wow. I "understood" the reasoning behind the original and knew that was the right answer, but your anecdote just made it totally click. Of course if I choose one random locker there is a 1/1000 chance of getting a prize, and if Monty opens 998 other empty lockers, of course I should switch. That would make switching be the correct choice in 999/1000 times.
This is the most intuitive explanation by far. I'd even say 1 million doors to really drive the point home.
You choose a door with only 1 in a million odds of it being the door with a prize. Monty Hall know where the prize is and will only open the remaining doors he KNOWS doesn't have the prize. If he then opens up 999,998 doors without a prize behind them and asks if you want to keep your original door or switch, you'd obviously know that Monty's last remaining door must be the one with the prize.
The opening of the door seems irrelevant and only serves to confuse. There will always be a goat door to open, who cares if it gets opened prior to the choice? The choice you are being given is "keep one door" or "choose both of the other doors." That is functionally the choice because Monty always opens a goat door. The chance of the other two doors containing a car is not affected.
I like this phraseology. How to make the answer intuitive is the objective, and underlining that you're choosing two doors, one of which is going to be wrong, really helps.
Yes I finally understood it. At the beginning you are choosing one door vs "the others". If you could you would already chose "the others" because they have a 67% win probability but you are only allowed to chose one single door. Then he opens all the wrong doors of "the other doors". The "other doors" still have 67% probability and now only one door is remaining in the "other doors". Obviously that last remaining door now has 67%.
That's what I dont understand about all the other explanations trying to "simplify" the situation. Just ask people, would you prefer 1 door or 2 doors. I've never understood how expanding it to 100 or 1000 doors is simplier than asking "which is better odds 1/3rds or 2/3rds."
I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random. The intent was for him to have prior knowledge or impeccable luck, and the wordsmithing of the question came later as, in my opinion, a failed rebuttal to the simplicity of the question. The question might have been worded to not be immediately obvious, but it was not intended to have different correct outcomes depending on interpretation.
Understanding the probability tree is fairly interesting though.
If he opens a door and tells you he knows it's a goat, you double your likelihood of winning by switching to the remaining door.
If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
>If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching.
At best, that was intended as a red herring to make the correct solution less obvious. It was not intended as an alternate correct answer. It's not how the game ever worked.
The initial version of the question said he opens a goat door. Whether he knew it would be a goat door or not is slightly irrelevant, because your odds of being right the first time were 1/3. As far as the premise of the thought experiment, a goat door always gets opened, either through peaking, premonition, or consistent luck.
>Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
I disagree, and if you don't see it as that simple, you are falling for the trap that makes the question fun.
I don't have a particularly strong opinion on this, but if we accept that its interesting as a thought experiment and not a useful strat should one find themselves time traveled into the monte hall game show, I think it's worth exploring the very closely related variants of the thought experiment.
> If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
What do you mean "opens a door randomly"? Is he picking from all three doors? Yours, and the two others?
In that cases you get interesting but trivially-obvious-what-move-to-make scenarios like "he chose your door and showed you you were right" "he chose your door and showed you you were wrong" "he chose a different door which had the car"...
Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?
>Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because
Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.
so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times.
there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too.
this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.
Assume you choose the car (1/3). He will always choose a goat. If you switch, you have a 100% chance of losing. (1/3) * (1/1) = 1/3 chance of losing if you switch.
Assume you choose a goat (2/3). There is a 50% chance that he opens a goat if he chooses randomly. If this happens, there's a 100% chance that you win if you switch. (2/3) * (1/2) * 100% = 1/3 chance that you win if you switch.
The remaining 1/3 chance is voided because we've conditioned the game on him randomly choosing a goat, and not a car. So the probability of winning if you switch is (1/3) / ((1/3) + (1/3)) = (1/2)
exactly 50/50.
In the traditional monty hall, switching gives a 2/3 chance of winning. This is still true in the random variant but only if you're allowed to choose the car door in monty opens that one by chance.
> Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?
He chooses randomly from the doors you did not choose, and draws a goat. When offered the choice to switch to the third door, it offers you no benefit. Both are equally likely to contain the car.
> I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random.
That’s extremely obvious! The entire reason to phrase the problem as a game show with a game show host is to make it abundantly clear that the host “knows the answers.”
No, if Monty sometimes opens the door with the prize, and you ignore those games, then the result is different, and you don't benefit from switching doors.
Monty’s knowledge is irrelevant. The original problem can be modified with no change in probabilities to:
Pick one of three doors. You can have whatever is behind that door, or you can change your pick to both of the other doors and win whatever is behind both of them. Should you switch? Obviously.
Step 1: Take three cards face down, one of which is an ace.
Step 2: Divide it into a pair of cards, and a single card.
Step 3: Reveal one of the pairs of cards. If you reveal and ace, return to step 1 since this history is eliminated from the story we've been told: we know this wasn't a possible path to our endgame. Otherwise continue to step 4.
Step 4: Which of the remaining two cards is more likely to be an ace?
Step 5: Realize that they're just as likely as each other to be the ace. Step 2 didn't magically imbue the card remaining in the pair with extra probability juice.
Long story short, you definitely need Monty to make an intelligent selection, so Monty's knowledge is far from irrelevant. It matters whether he revealed a goat by luck or by knowledge, because he's 2x as likely to get "lucky" in the case where there are 2 goats behind the doors you didn't choose in your original guess.
I mean, step 2 did "imbue the card remaining in the pair with extra probability juice." It went from 1/3 to 1/2. But so did the card you picked initially.
When he shows (randomly from the doors you didn’t pick) the car, switching to open door (where the car is) and the other door (which hides a goat) you increase your chances from 0% to 100%.
When he shows a goat, switching to the open door (where there is a goat) and the other door (which may or may not hide the car) your chances stay at 50%.
> It’s important that Monty looked behind the doors before choosing which to open. This is where people’s intuition usually fails. If he had chosen a door at random — in a way that he risked possibly exposing a car, then the situation would be different. (In that case, there’s no advantage or harm in switching.)
This one took a second to rationalize. The reason it works is you have the same chance of winning overall (assuming you can safely choose the car if he opens the car by chance), it's just that the value of winning from switching vs. staying has been shifted into the probability of winning by default. The usual mental trick is to extend to 1,000,000 doors.
If you pick one door, then are told that all of the alternatives except one are the correct answer, you should obviously reason that the door you didn't choose is the correct answer, unless you got the 1/1,000,000 guess. Odds of winning if you switch are 999,999 / 1,000,000
If the host instead opens 999,998 doors randomly, you have a 999,998 / 1,000,000 chance of winning by default. The remaining two doors have equal chance of winning, giving you the same total odds of 999,999 / 1,000,000 no matter which you choose.
Which makes sense, because both situations are, more or less, being given 999,999 chances to guess the lucky door.
Most of the subtlety of the problem lies in this small sentence in the Side Notes:
"He deliberately chooses to show you goats."
This is not made as explicit in the standard formulation
"the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat."
Which could be read as "which happens to have a goat".
It's the ambiguity in Monte's door opening strategy that leads to different answers.
One of the realities we face here is that different people have different brains which work differently. Thus different phraseology is going to help. Adding variant phrases such as "He will never show you a car", will probably help for some.
It's also possible that Monty only gives you the option to switch if you chose the car. Or only if you chose door No 1. All of which affect the answer. Therefore, it must be made fully explicit what Monty can and cannot do.
Whether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does.
The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck.
The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with.
tldr: it doesnt matter how he opened a goat door, all that matters is that he did.
How does it matter how he opened a goat door? If you allow the possibility of Monty opening a car door, you change the problem entirely, and you before stating whether it changes or doesn't change the odds, you need to tell us your new interpretation of the problem.
1) picks randomly and opens the door, or
2) picks randomly and then, if that door has a car, switches.
His application of knowledge is converting those cases that would have been thrown away in case 1 into victories for "switch".
The counts bear this out. If you simulate it, 1/3 of all games go to 'stick' regardless. Either 1/3 or 2/3 go to 'switch', and either 1/3 or 0/3 (respectively) end up with Monty revealing the car.
Problem-1:
"You pick a door. Then Monty opens a door he knows having the goat. Calculate probability of winning if staying."
Problem-2:
"You pick a door. Monty opens a door at random, which just happens to be a goat. Calculate probability of winning if staying."
Both are conditional probability P(A|B) (that is, probability of A happening, under the assumption B has happened).
- A is probability of picking the correct door at the first try (or switching if you prefer)
- B is the probability that Monty picked a door with a goat
P(A|B) is defined as: P(A|B) = P(A and B) / P(B)
In both problems P(A) is 1/3.
In Problem-1, P(B) is 1 because Monty knows and it's not a random event and P(A and B) = P(A), so P(A|B) = P(A)*P(B)/P(B) = P(A) = 1/3.
In Problem-2, P(A) is 1/3, P(B) is 2/3, P(A and B) is 1/3 (if you pick a car, Monty is guaranteed to pick a goat), so P(A|B) = 1/3 / 2/3 = 1/2.
> Specifically, whether there was a chance that he might not have.
That is false. Him opening the car is outside the constraints of the premise of the thought experiment. He can't not open a goat. The chance of him not opening a goat is irrelevant, and the probability is 0%.
So in a new, different, and entirely unrelated thought experiment:
You pick a door. Monty picks another door at random - he has clearly committed to doing so, flips a coin in front of you or whatever. God only knows what would have happened if he'd revealed a car, but this time he didn't. Having found yourself in that situation, what are your odds if you switch doors?
If you simulate this procedure, you'll find that the number of outcomes where "switch" wins is about 1/3 of the total games; "stick" wins in about 1/3 of the total games; and 1/3 of the total games Monty revealed the car. When Monty does pick the goat reliably, as in the Monty Hall Problem as it was intended to be understood, Monty is doing the work to convert those "revealed" games into wins for "switch".
I'm sorry, I understand the difference you're implying, but can't figure out how it matters in practice. How can I even differentiate if Monty opens the door to reveal a goat by knowledge, or by chance?
I mean that literally - how can I write this into a simulation to test that there's a difference?
It depends on the context. If it's a riddle, you could see if it's made clear by the wording, interrogate the person asking the riddle, or state the assumption you're making in your answer.
If you find yourself actually in the situation, you need to make your best guess from what information is available.
As for simulation, write a function that handles a single play through of the game and returns the outcome if you switch. Make Monty's strategy a parameter. Remember that "Monty revealed the car" is a separate outcome from "win" or "lose". Run that function a whole bunch of times, counting the outcomes. Compare the ratio between wins and losses, at various choices of strategy.
To really drive home the point that strategy can matter, consider another possibility: maybe this season the studio executives have decided that cutting costs is more important than having an interesting game show, and Monty's strategy is now "reveal the car if you haven't already chosen it". Monty has revealed a goat. Is there still a 2/3 chance that you didn't pick the car?
What door Monty opens is irrelevant. He shows you a door with a goat, you’re better off switching. He shows you a door with a car, you’re still better off switching. What matters is that when you first choose a door, you have most likely chosen a goat.
First caveat: If he shows you a door with a car, there is literally no point switching. There's a goat behind both doors you're allowed to choose from, which are the remaining two.
(Below I use non-chosen to mean non-chosen by the contestants initial choice.)
Second point: If we are in the subset of all possible histories where Monty picked randomly revealed a goat, then we will have 50% of histories where both non-chosen doors contain 1 goat selected by our history subset, and 100% of histories where both non-chosen doors contain 2 goats selected by our history subset.
Since there are twice as many possible histories where the non-chosen doors contains 1 goat vs 2 goats, after selection, we have an equal number of histories in our sample where we have 1 goat or 2 goats behind the non-chosen doors. Or equivalently, we have a 50% chance that the non-chosen doors contain a car.
Therefore it is irrelevant whether you switch.
Monty needs to make an intelligent selection to change the game.
I interpreted the original game as follows: you can freely choose any of the three doors. Monty then opens one of the doors that you didn't pick, and then you may again freely choose any of the three doors. (In the standard problem, there is no reason why you would choose Monty's door). The question is: does the strategy "choose a new door" have better odds of winning?
We can work out odds for this "always switch away from original door" strategy: suppose you initially chose a door with the car (1/3 probability). Then choosing a new door makes you surely lose regardless. On the other hand, suppose you initially chose a door with a goat (2/3 probability). Then regardless of which of the two doors Monty opens, you can choose the car (if he revealed the car, choose that. If he revealed the goat, choose the other door). So our odds of winning with this strategy are still 2/3.
So it's up to the interpretation of the modified game I guess.
Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to it and won. But he didn't.[0]
Now you have a choice to make.
His probability of not revealing a car in the case that both doors you didn't pick had goats behind them, is 100%.
His probability of not revealing a car in the case that one door you didn't pick had a car behind it, is 50%.
We can reason then, that of the three possible scenarios for the two non picked doors {g,c}, {c,g}, {g,g}, that it is equally likely now that since he didn't randomly reveal a car, {g,g} must be weighted twice as much as the other two.
Therefore, we must be as likely to have a goat behind the remaining door as we didn't.
In conclusion, Monty only changes the problem if he selects with information.
[0] Notice that at this point, we have to throw away from the decision tree 1/3 of the histories, and in all of them switching was beneficial. In what remains then, switching is less beneficial than it was before this trimming.
> So our odds of winning with this strategy are still 2/3.
That’s true before he opens a door. Then either
A) he shows a car and the odds of winning with this strategy are 100%
or
B) he shows a goat and the odds of winning with this strategy are between 1/2 and 2/3 depending on how the choice of door was made
If the choice is random, he shows a car with probability 1/3. The probability of winning is 1/3 x 1 + 2/3 x 1/2 = 2/3
But that analysis is for the unconditional problem and what we’re asked is what to do in case B, after a goat has been unveiled.
If he shows always a goat, the probability of winning is 0 x 1 + 1 x 2/3 = 2/3. Here there is no difference between the unconditional and the conditional problems because A never happens.
I got out three playing cards and did the experiment myself over and over. That was many years ago and now whenever I see something on the Monty Hall problem it seems so obvious as if I can no longer even see why people think its unintuitive. The reason its unintuitive is that people don't really have experience with anything that works this way. I'm not sure that the linked explanation will be that helpful to people because of that, although it does make it plainly obvious that Monty Hall knows which door has the prize and he's telegraphing to you which door it might be by only opening doors which don't have it in there. I suppose if that's the missing part of the puzzle for you, then that will help.
I've always thought of it like the 'denominator' of probability for the second stage became 2/3 (vs. the usual/first stage 1).
So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3).
But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others.
Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick!
(But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')
> I suppose if that's the missing part of the puzzle for you, then that will help.
Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge.
The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge.
> I got out three playing cards and did the experiment myself over and over.
How did you do that when you didn't know which cards were 'goats'?
I've had issues understanding the Monty Hall problem for years until now. What made it click was that, at the end of the article, it's explained that the doors are not opened by Monty at random. My previous reads about the problem did not disclose this, so I had made the assumption that the door opening was random. I've also never seen the show, for what it's worth.
The problem explicitly says that he opens the door to reveal a goat, though. If he was opening the door at random, then he would have a 1/3 chance to reveal the car. Since he always reveals the goat, to me that implies that it's not random.
An accurate and precise description of the problem is essential to understanding. Without explicitly stating that the door choice is non-random, I did not make the assumption.
> It’s important that Monty looked behind the doors before choosing which to open.
This is often not explicitly stated when the problem is given. It is even not a 100% clear from the statement above. Monty always chooses a door with a goat. So:
1. You choose a door.
2. Prob that there is a car behind it: 1/3
3. Prob that the car is behind the two other doors: 2/3
4. If the car was behind the two other doors (which, remember, has p=2/3), Monty will choose the door without a car for you, and the door with a car will remain closed. In this case you are guaranteed to have the car if you switched.
So with switching, the overall probability is 2/3. Without, its the original 1/3.
If you did not understand that Monty always chooses a goat door, but the person giving you the problem does, or vice versa, then what usually happens is that both of you try to explain why your intuition is correct. Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.
>> It’s important that Monty looked behind the doors before choosing which to open.
> This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand.
If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.
I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)
What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply.
I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.
Your first paragraph looks very plausible, though I am not aware of any data saying it is a prevalent one. Are you suggesting that people still look at this problem as if the choice was random, even when they know it is not? (e.g., if they think the 'random' and 'chosen' cases are equivalent.) That is also plausible, but if it is common, that would imply that the phrasing of the problem is not the root cause of their misunderstanding of the optimal strategy.
Personally, IIRC, my first reaction was to assume the second box opening was not random (perhaps only because the question is not phrased as being conditional on this act revealing a goat) but did not see how this gave any useful information.
Here's another possible way of getting it wrong, regardless of the phrasing of the problem: assuming that, after the reveal (and whether one thinks of it as random or not), one is, as it were, starting over, except with a choice between two boxes rather than three, and no other information.
I definitely don't think it is the only point of confusion, but I definitely think it is a big one. We discussed the problem in my discrete math class and the professor had stated the problem without the constraint. Many in the class, being far too young to even know who Monty Hall is, were confused, and several had their 'aha!' moment once that constraint was mentioned.
> According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.
Interestingly, coding up my own simulation took me from understanding the problem to grokking it. I'd definitely suggest anyone who has programming ability but doesn't grok the problem to write up a simulation.
If anyone reading this uses this example in a class, or is thinking of doing so, perhaps it could be set up as an experiment to see how much the phrasing influences the result?
I agree that "Monty knowing the answer" isn't the "whole reason" MH is confusing.
IMO, people are confusing MH with a case where the door is opened randomly before your choice.
Let's say Monty shows you three doors, then knowingly opens a wrong door (B) before you choose. Do you want A or C? It's a coin toss, of course.
But suppose you pick door A first and THEN he knowingly opens a wrong door you didn't choose (B). Now the odds of winning are 2/3rds if you switch. But it feels the same as in the previous case. ("How could choosing a door first affect the odds?! It couldn't possibly change anything!!")
The answer is that your choice of a door constrains Monty.
Suppose you choose a door secretly before Monty opens a door. Suppose you choose A, and Monty opens door A, revealing a goat. Well, uh, you can't pick that any more, now, can you? You're forced to choose between B and C, a toss up.
But if you publicly pick door A then Monty can't/won't open A; he has to open one of the other doors. Since you probably chose the wrong door to begin with, and he always opens a wrong door, the other remaining closed door probably has the car.
Especially if you walk through what that would look like. You choose door #14. Monty then goes from door to door, opening every single one except door #67. What's special about door #67? Either you picked the door with the car behind it at 1/100 probability, and Monty picked it at random, leaving it closed. Or, with 99/100 probability, Monty is skipping opening that door because it's got the car behind it...
I've tried that with people with limited success, though it could be that I've explained it badly.
When you put the games in terms of "you and a friend playing the game simultaneously", most people get it.
You and a friend are playing the game simultaneously; you both decide ahead of time to:
a. always pick the same initial door
b. no matter what Monty does, you ALWAYS switch, your friend NEVER does.
The possible outcomes before the reveal are:
* Monty wins, because he shows you the car. We agree ahead of time that this never happens.
Now, since you know one of you is going to win, since Monty didn't, who has the better chance? Your friend started with a 1/3 chance, and he hasn't switched; does his chance suddenly change because Monty opened a door? Most people (not all!) will agree that it has not.
Since one of you MUST win, and friend has 1/3 chance, what's left?
It doesn't matter at all if Monty knows which doors are winners and losers.
If Monty doesn't know, then sometimes the game will be ruined because he will expose the grand prize and then the game is moot. But if is simply lucky by showing the goat door vs he picked it with foreknowledge doesn't change the odds in any way.
It does matter. If you are looking at a single instance where Monty got lucky and chose a goat door, then the probability that the door you have chosen has the car is 1/2, and switching doesn't change anything. This can easily be tested: just write a simulation that runs the experiment with Monty choosing a random door, and discard the instances where the game was "ruined" because he picked the car. In the remaining instances, both strategies will perform the same.
It matters because in the case where Monty does know to avoid the door, those games end up as wins for the person who selected to switch, as opposed to broken games.
You are taking an empty result and replacing it with a strict win.
I think lqet's point is that if it's just recounted as being thrown into a situation -- you've picked a door, and Monty opens a door, and it's a goat -- you don't know if you're in a world where Monty always picks a goat or one where he just happened to be lucky. In the second scenario, it doesn't matter if you switch.
You’re forgetting about the possibility of being in a universe where Monty only opens a door if you chose the correct door originally (otherwise he does something else like ending the game). If you know nothing about the rule set of the game show, Monty opening the door tells you nothing and you have no reason to either switch or not switch.
Indeed. It's like the argument that you may as well believe in God, because if he doesn't exist it doesn't hurt you.
That said, though, if you're trying to convince someone of the correctness of the Monty Hall Problem, and they're stuck on that misconception, saying "well you may as well switch in case I'm right" isn't a great argument.
Sure, here's the simulation. There are only six distinct cases, so we can "simulate" it exhaustively, and a Monte Carlo will approximate this.
Let "A", "B", and "C" represent the prizes; A is valuable, B and C are goats. They're shuffled behind doors, so the player and Monty choose "1" or "2" or "3", and after the fact we'll map ABC to 123. As such, we can assume without loss of generality that the player always chooses "1" and Monty always chooses "2".
What happens?
123
ABC - player was right at first, so they lose if they switch
ACB - player was right at first, so they lose if they switch
BAC - player was wrong, Monty chooses 2 which is the car, so the game is ruined
BCA - player was wrong, Monty finds a goat, player wins if they switch.
CAB - player was wrong, Monty chooses 2 which is the car, so the game is ruined
CBA - player was wrong, Monty finds a goat, player wins if they switch.
Huh. Of the 4 non-ruined games, half of them are improved by switching, half are worsened.
Exactly and if Monty knows of the car location (as in the original formulation), he will basically save BAC and CAB from being ruined and both of these will become "player wins if they switch" outcomes. So instead of 2 out of 4, we now have 4 out of 6 cases where switching is better.
I think where I differ (also expressed on the parent comment) is that when Monty ruins the game, you are tossing out many cases where the "I'm sticking with my original choice" strategy failed, while I'm counting them as a failure in the strategy.
You are free to count them as a failure of the "I'm sticking with my original choice" strategy, but only if you also count them as a failure of the "I'm switching" strategy, because when Monty ruins the game there is no way to win the car by switching to the remaining closed door.
If Monty chooses randomly both strategies fail 2/3 of the time.
Precisely. Either way switch or no switch have the same success rate if Monty ruins the game. You can count it as 1/2 or 2/3 but either way both choices are the same. So it’s a very different scenario from when Monty knows the door is a goat.
Whether Monty knows for sure or is just picking at random only affects whether games are ruined, not the probability that the door the contestant picked is a winner.
If there are N doors and the player picks door 1, there is a 1/N chance he will be correct. If he sticks to that door, there is nothing that will ever change that 1/N odds that it was the winner.
Downthread someone enumerated the cases, but again, they discard the 1/3 of the cases where Monty ruined it. What that does is throw out half of the cases where contestant picked the wrong door and had the wrong strategy. Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.
>Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.
By that logic, so does the remaining door. 1/3 vs 1/3.
Because that is a fundamental part of the "Monty Hall" problem. You are correct in the case where Monty will randomly pick doors, and sometimes he will pick the winning door and ruin the game, but that is a fundamentally different problem than what is described as the "Monty Hall problem", and therefore irrelevant.
Its like saying soccer players could score more if they picked up the ball and ran with it. While true, you are no longer describing the game of soccer.
It's weirdly great that you had enough misplaced conviction to motivate you to write a simulation which ended up proving you wrong.
It's actually an interesting, possibly general story, that kind of makes me think. Perhaps for certain personality types, misplaced conviction will put you on a higher velocity trajectory toward truth than honest confusion.
For sure. It's a small fraction of the population whose first instinct to prove themselves right is to do which also has the possibility of also proving themselves wrong.
It's the story of science! Unlike lawyers or debaters, scientists are delighted when their hypotheses are clearly disproven; this is at the crux of the scientific method, and is essential for discovery of the truth. Assert something falsifiable, then experiment. The truth wins.
In terms of the logic problem, I can't see any distinction between not counting moot games where he reveals a grand prize vs assuming he never reveals the grand prize. It's just two different ways to say that the only games under consideration are the ones where Monty reveals a goat.
So I'm not getting the point you're making, unless you're saying that would be a more grokable way to state the problem?
It makes a difference how you got there, assuming in the second one that he reliably/intentionally never reveals the grand prize.
Imagine you have two identical looking bags of marbles, one contains two red marbles, and one contains one red and one black. You reach into a random one and draw a marble. It is red. If you picked the marble randomly, you now know you're more likely to be holding the bag that contained two red marbles than the mixed bag. The red marble you randomly drew provides bayesian evidence for which bag you're holding.
Marble 1 2
Bag
2R R R
BR B R
Out of the three possible equally likely worlds (given that you know you didn't roll into BR-1), two of them have you holding bag 2R, aka a 2/3rds chance.
If, on the other hand, you use a red-marble-picking robot, you get:
Marble 1 2
Bag
2R R R
BR R R
Now you have four possible worlds and are equally likely to be holding either bag, ie 1/2.
So, if our friend Monty opens a random door and you only look at simulations where it reveals a goat... There is a 1/3rd chance you initially get the car, but if you do he's guaranteed to reveal a goat. So 1/3rd of games, you have a car. There's a 2/3rd chance you initially get a goat, but then he has only a 50/50 chance of revealing the goat. So half of the 2/3rds of games, the game is discard as invalid, and the other half of 2/3rds (aka 1/3) you have a goat. As such, the potential outcomes are split 1:1:1 between car:goat:invalid. Removing the invalid cases, it's 1:1 car:goat, despite the only 1/3rd chance of you choosing the car initially.
tl;dr: If he's randomly opening doors and you discard any where it's the car, those cases are entirely ones where you didn't initially choose the car, which introduces some bias. If he's opening doors deliberately, no cases are thrown out since it's always possible to reveal a goat, maintaining the original 1:2 probability ratio.
OK, so I pick a door, Monty randomly reveals a goat. Your claim is that this information should make me increase the odds that my initial pick was a car. That sounds right. In the normal Monty problem, Monty doesn't give me any information about my original pick, just about the remaining door.
Alright just working through your logic here. Let's say I pick the first door always. I'll write out all the possible configurationss with both possibilities for Monty.
CGG > Monty reveals mid (Goat)
CGG > Monty reveals right (Goat)
----
GCG > Monty reveals mid (Car)
GCG > Monty reveals right (Goat)
---
GGC > Monty reveals mid (Goat)
GGC > Monty reveals right (Car)
So there's 4 worlds where Monty reveals a goat. In two of them, I picked a car initially. In the other two, I get a car by switching.
By the time you have the decision, yeah, that's true.
There are other situations where the game is over before you get a decision.
Both GCG and GGC are eliminated half the time before you get to make a decision. But CGG is never eliminated before you get to make a decision.
Once you're making a decision and the game isn't already over, you can use this information -- that the game didn't end -- to assign a 50% chance of being in CGG, and 25% of each of the other two.
You only remove games as moot when Monty reveals a car, which can only happen when you haven't initially selected the car.
Therefore, no games are removed as moot when you have initially selected a car, and 50% of games are removed as moot when you have initially selected a goat.
So it ends in a coin toss whether you switch or not.
The 2/3 to 1/3 split only happens when Monty removes a goat using privileged information.
Is it important the host looked behind the doors?
If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.
This is gonna have anthropic principle vibes but here we go. Either the host looks behind the door, or they don't and we just don't talk about games where they accidentally show you the car. The probabilities are the same. We're already conditioned on being in the "they show you a goat" universe.
If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge.
Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
I thought as you think. I wrote a simulation to show I was right. I was wrong.
I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
I think you’ve simply misread what they are saying. They are saying that the situation in which Monty opens a door and it reveals the grand prize, are the set of cases that we do not care about because we have already lost, and can thus discard them.
This is basically just a different way of saying that Monty looks behind the door to be sure to only reveal goats.
Yes, the probability of wins will be different in these two scenarios, but it doesn’t affect the conclusion: when Monty reveals a goat, you should always switch.
Edit: if you believe I am the confused one after this comment, I will go make the simulation as you suggest.
That simulates a different scenario than what’s being described here; it simulates a situation where we count the times when Monty picks the car. But those situations are irrelevant because they have no bearing on the fundamental question of whether or not to switch doors when Monty shows you a goat. Effectively, we discard all outcomes when Monty shows you the car, making it an identical “game” as when Monty simply does not ever choose the car.
As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.
Thank you (and everyone else in this thread) for sticking with it with me. It was a struggle to read that single line of code on a cell-phone screen, and that was compounded by the fact that it turns out that I don't understand the Monty Hall problem when I thought that I did. I'm going to sit with this one for awhile.
So let's say you and I sit down, and do the following:
1) I roll a 3-sided die (or a 6 sided, wrapping) and keep it covered.
2) You pick a number, 1-3
3) I flip a coin. If it's heads, I pick the lower available number; if it's tails I pick the higher.
4) I peak at the die. If it's my number, we reveal and start over.
5) I (always, at this point) offer you a wager: if the die shows your number (so you would have lost if you switch), you pay me $7; if the die doesn't show your number, I pay you $5.
Assuming you believe the die and coin are fair, etc, would you agree to play that game 1000 times? In those games, is there a reason you would turn down the wager?
Is it the same game if I flip the coin secretly, peek at the die, announce my number (picked algorithmically in the obvious way), and then pay out according to whether switching would win (as above)?
Because (assuming I've explained these games as I intend... it's getting late) I would play the former with you not the latter (but I would play the latter if we switched the payments around - I chose 5 and 7 because 7/12 is halfway between 1/2 and 2/3).
It will be different. The question comes down to P(winning by switching | monty reveals a goat). The key difference is whether P(mony reveals a goat | you chose a goat door the first time) is 50% or 100% (if you choose the car, there's a 100% chance he reveals a goat in either case). Since 'winning by switching' is the same as 'choosing a goat door first', you can apply bayes theorem to see how the results change.
To put it another way, if Monty is choosing randomly, half the time where you would win by switching, instead the game just ends/isn't counted, but the same is not true of the case where you win by not switching. From a bayesian point of view, Monty randomly revealing a goat should increase your belief that you picked the car the first time.
That said, switching is not worse than not switching unless Monty is biased towards revealing the car instead.
They're not the same if the host chooses randomly. Check it out:
WLOG assume the car is in door A (The rest will be the same by symmetry.)
You pick A, B, or C.
The host picks one of the other two doors to show you.
The universes can be described by your choice, followed by the host's random choice. There are 6 possibilities: AB, AC, BA, BC, CA, CB.
Since we're not in a universe where the host chose a car, we eliminate BA and CA.
We are left with four possibilities: AB, AC, BC, CB.
AB = You chose the correct door. (probability 1/4)
AC = You chose the correct door. (probability 1/4)
BC = You chose the wrong door. (probability 1/4)
CB = You chose the wrong door. (probability 1/4)
Since the host's choice was random, it's 50/50 as to whether the car is behind the door you chose or the remaining door.
This same reasoning breaks down when the host knows what's behind the doors because you're no longer eliminating a couple of the universes and it's a 2/6 vs 4/6 situation.
Imagine there are two contestants and each one selects a door. The host opens the remaining one and it happens to show goat. Do you think they will both increase their odds of winning by switching?
Can both select the same door? If the answer is no, you've changed the parameters of the game, because the person who chooses second can only pick from two doors.
The subset of games where they close different doors is smaller than all possible games.
If participants are allowed to pick the same door, even if they don't in practice, then yes: after Monty shows a goat behind the third goat then both participants should switch. This is because they are essentially playing two simultaneous but independent games, with no interference between each other.
As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.
No, your intuition is misleading you: it's not a choice between two doors. If the probabilities don't convince you, you can easily simulate this by using three cards and considering one is the car and the other two goats. Pick one at random, then pick the remaining "goat" as if you were Monty; lastly, switch your initial choice as a player. Repeat the experiment 10 times and count how many times you win; remember you must always switch. You will win more often than lose! I tried this, that's how I know this will convince you.
Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable.
The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)
So when they don’t switch each one has probability 1/3 of getting the car? Where does the car go if no one gets it?
And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing?
Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again.
Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
> So when they don’t switch each one has probability 1/3 of getting the car?
Yes.
> Where does the car go if no one gets it?
I don't understand your question. What does this have to do with probabilities?
> And if they switch both have probability 2/3 of getting the car?
Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously.
They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's.
> Is this some form of car sharing?
That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch.
> Maybe you should try to play with three cards and one ace.
There are only 3 doors, so you should only use 3 cards.
> Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".
Player #1 had a 1/3 probability of picking the right door. Therefore, he has a 2/3 probability of winning if he switches.
Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches.
In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).
No. Because the experiments are completely independent (i.e. player #1 & player #2 don't interact at all; their choices don't interfere), the events are effectively "choosing door A and switching to door B" and "choosing door A' and switching to door B'". You have doors A, B, A' & B'. This of course assumes all doors are revealed simultaenously; i.e. no player gets to see the result of the other's choice beforehand.
You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent.
These are two independent games, so for every player:
a) The probability of winning if staying: 1/3
b) The probability of winning if switching: 2/3
c) The probability of the car being behind door the player chose is the same as a): 1/3
d) The probability of the car being behind the door Monty didn't open is 2/3
e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat!
Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.
There is one single game. It's my version and I make the rules!
Rules2: "One player picks a door, the second picks a door (maybe the same), the host opens a door that has not been picked (if there are two he picks at random). Then the players can either keep their choice or switch to the other closed door."
Scenario2: "Player #1 picked one door, player #2 picked another door, the third door was opened and there was a goat behind it."
In that scenario, what are the probabilities?
I agree it's different from the original problem, that's the point!
Rules0: "The player picks a door, the host opens a door that has not been picked with full knowledge that there will be a goat behind it. Then the player can either keep his choice or switch to the other closed door."
Scenario0: "The player picked one door, the host opened another door and there was a goat behind it."
But the game I proposed is equivalent to the version of the problem that some people, including you apparently [0], insist in this thread that is equivalent to the original problem.
Rules1: "The player picks a door, the host opens another door selected at random. Then the player can either keep his choice or switch to the other closed door."
Scenario1: "The player picked one door, the host opened another door and there was a goat behind it."
The latter is not equivalent to the original problem. The situation is similar but the rules are different. The solution is different.
However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors.
They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat.
Rules1': "The player picks a door, and opens another door selected at random. Then he can either keep his choice or switch to the other closed door."
Scenario1': "The player picked one door, then opened another door and there was a goat behind it."
Hang on, I think you've missed something. They are not completely independent, because if they pick different doors, that forces the host to pick the last door. Removing the host's freedom to choose a non-winning door (and effectively help the player) changes the math. Having two players with the ability to choose different doors (but still 3 doors and the host must choose the third) makes it a different game with different dynamics. It doesn't work the same because it's not the same game.
Or, because they are two independent games, the host can flip different goats for different people. You would have to add an element of "no screen peaking" where the contestants cant see which door opened for the other person.
So you have Player A who chose door 1. The host flips the goat in door 3.
You have Player B who chooses door 2. For them the host flips open the goat in door 1.
Player A should switch, and wins the car. Player B should switch, but had the car and loses it. The right move in both their cases was to switch.
Really, its no different than playing two separate games. Unless, like you said, you somehow change the rules and limit someones choices, in which case Im not sure why its a part of the conversation, other than being introduced through confusion.
If the players pick different doors, the host will open the third door. That prevents him from being sure to get a goat, in the same way that opening a door at random in the single-player case prevents him from being sure to get a goat.
Well, you're right: in fact, in some cases Monty cannot open any door! That is, if players A & B both chose goats, then the remaining door is the car, and Monty cannot open it without ruining the game.
For this reason, both players should be playing "blind", without seeing the other's choices or the door Monty opens for the other player. Anything else doesn't make sense for the game.
> Removing the host's freedom to choose a non-winning door (and effectively help the player) changes the math.
As it does removing the host's ability to choose a non-winning door in the variant where "the host had just picked at random from the two doors, and it happened to show a goat". That doesn't work the same either because it's not the same game.
In your example, the host didn't get to choose their door, which kept them from choosing goat. As I follow this discussion, the key is the knowledge that given a choice, the host will pick a non-winning door. If the host can't choose, you lose the benefit of that fact.
The host has no choice in this case so his action has no effect. The chance to win is 1/3 for all. And after the host gets a goat one of the contestant has to wins so its 1/2 for both whether they switch or not.
That’s indeed what happens when the host picks at random (the hypothesis in this subthread). He has a 1/3 chance of showing the car. If he doesn’t, the car is behind any of the other doors with probability 1/2.
> Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.
If the host picked a door at random, and it happened to show the prize, then the game would be ruined. That would make some terrible television. Therefore, he can't be picking a door at random.
It's also not always emphasized that the host never picks the same door you did. The host's behavior is not done in isolation. It is a response to your choice.
What kind of game shows are people watching where the host opens a door and says, hey, there's the prize but too bad, now you can't win it? Of course he always picks a door with a goat.
Deal or No Deal does this - the contestant selects a briefcase, then opens up briefcases randomly. It's very similar to the Monty Hall problem, with some twists iirc.
Why not? Reveal all the choices the contestant didn't make, one at a time, slowly increasing the odds of a prize. This adds to the suspense. It's used all the time.
I get what you are saying, but if I look at it differently I still intuitively feel like it should be p=1/2 if they switch instead of 2/3. For example:
- the bottom line is Monty will always eliminate one wrong choice
- therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2)
(edit: thanks i see, no need for any more answers)
It’s not regardless of what came before. That’s like saying, if there were two doors and Monty simply tells you which one the car is behind, there’s still a 50/50 chance of each door being right. No, there’s a 100% chance the one he told you is right. You learn real new information when Monty opens a door just like when he tells you something.
> therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2)
I get why you intuit this - I had a lot of trouble with it at first as well. Common sense is lying to you, however, because you are still dealing with the original odds (1/3 vs 2/3). Monty's choice is always to choose a Goat (probability 1.0), therefore his effect on the probabilities will always be with a weighting of 1.0. You end up with a probably spread of (1/3 * 1.0 vs 2/3 * 1.0).
Therefore it is always in your interests to choose Monty's remaining door.
I think if you really read over the article, you'll see this is wrong. If not, tell us on which step you disagree with the author.
A more common re-telling of the story:
Monty has 1 million doors, with only one car behind one. When you pick a door, Monty will always open 999,998 other doors, all of which show goats, leaving one other door conspicuously closed.
Do you still think you have 50% chance of being right by sticking with your choice?
This got me closer to understanding. But, after 999,998 doors are opened, although the one other door is conspicuously closed, my door is also closed. That seems conspicuous, too. In that moment, there are 2 doors to pick from and so 50/50 makes sense.
As some other comments have mentioned, I think this isn't a fully satisfying explanation because it's hard to reason about the difference between a general strategy vs. an in-the-moment choice; why does it matter that I've watched Monty narrow the choices down to 2 doors, vs. seeing 2 doors from the start?
I think the 1 million doors explanation does a very poor job of explaining why switching is the correct choice and only gives a more intuitive answer, not more intuitive understanding.
This is how I understand the Monty Hall problem (Using 3 doors):
1) You are asked to choose a door from A, B or C. You choose A.
Probabilities:
p(A = Car) = 1/3
p(not A = Car) = p(B or C = Car) = 2/3
2) Monty opens a door and shows you a goat. Let's say he opens door B and gives you the choice to switch from door A.
Probabilities with B eliminated:
p(A = Car) = 1/3
p(not A = Car) = p(C = Car) = 2/3
3) The probability of the car being behind door A remains unchanged at 1/3, therefore the probability of the car being behind not A is also still 2/3. Since door B has been eliminated, this means the probability that the car is behind door C _must_ be 2/3! Therefore, we choose to switch from door A to door C.
Monty never opens your door until the end of the game. So Monty opening 999,998 other doors doesn’t give you any information about your door, just the other one that remains closed.
If you made a one in a million guess the first time, your door has a car. If you didn’t, the other remaining door must have the car. Which do you choose?
Let's say Monty doesn't open any door, but after you've chosen one, he offers the chance to switch to both of the other doors. Pretty clear then that switching gives you a 2/3 chance of getting the car, and that is essentially the choice Monty is offering.
The problem is confusing because it appears that the choices are independent when they really are dependent. That's what confuses people and what needs to be adequately explained (and thus explicitly acknowledged).
> So with switching, the overall probability is 2/3. Without, its the original 1/3.
Thank you for spelling this out so clearly. If explained in terms of strategies (always switch v. always stick), the probabilities should be crystal clear.
What does the switcher have to do to win? Pick the wrong door first (2/3). What does the sticker have to do to win? Pick the right door first (1/3).
This carries over to a general problem with wordy riddles and puzzles. They rest on a slew of assumptions and tacit knowledge. If the riddler fails to pluck out the essential parts, then there's a good chance the interaction will fail.
Yes, this is the single most critical piece of information, and when people struggle to understand the Monty Hall problem, its almost always because this wasn't made clear. When it is, the problem becomes much more intuitive.
I think the confusion comes from the oracle-like knowledge of Monty Hall - he knows what is behind each door and decides which door to open based on that knowledge. If Monty Hall wasn't an oracle, and just opened a random door after the guess, people's intuitions would be correct.
Your comment would be better if you could explain why you think it is wrong, rather than merely making an empty assertion. I think you are just misinterpreting the analysis. It says "Option B still gets you the car 90% of the time", and Option B entails switching. Do you read it differently?
If there are infinite number of doors, then does switching guarantee us the car?
Also, Game 5:
There are 2 doors. A car is randomly placed behind one, and goats behind the others.
You pick one door.
Monty looks behind the other doors. He chooses 0 of them with goats behind them, and opens them.
You get two options:
Option A: You get whatever is behind the door you picked.
Option B: You get whatever is behind the other closed door.
Should you switch?
It's counterintuitive, as magically the car appears behind the door that Monty chose. But tbf chances of it being behind the door you've chosen were 0 to begin with.
Q: In the monty hall problem, how does opening the second door skew the probability in favor of the initially unchosen door?
A: The Monty Hall problem is generally poorly described, in order to make the conclusion seem more surprising then it is.
The actual Monty Hall game — as imagined by the people who are asking the question— is set up like this:
In front of you are 3 doors, there is a goat behind two
of them, and a car behind the other one.
In *round 1* of the game, you select a *pair* of doors,
from which *one* “goat containing door” will be
*automatically eliminated from*, leaving only *one* door
of the selected pair of doors in play, (and only *two* of
initial *three* doors in play).
In round 2 of the game, you guess which of the two
remaining doors in play has the car.
The choice is this: should you choose the remaining door from the pair selected in round 1, or should you choose the door which was not part of the selected pair in round 1?
When phrased like this, the answer is fairly obvious: the pair of doors contains a car 2/3 of the time, whereas the non-paired door contains a car 1/3 of the time.
The Monty Hall problem— as normally described— messes this all up by introducing a game show host. This is an agent who— seemingly by their own whim— changes the game you thought you were playing, and introduces round 1 of the game once you have guessed the door you initially think the car is behind. The rules this game show host agent are following are almost never described to a sufficient degree to ensure the game is equivalent to the game laid out above. And yet, the people asking this problem pretend that it is exactly equivalent when they ask you for an answer.
It’s generally a poorly described problem, whose answer depends entirely on what kind of agent the game show host is.
Don’t worry if it doesn’t make sense to you as it’s usually described. If you can understand why— in the two round game I describe above— it’s better to pick the door from the pair of doors, rather than the single door, you understand probability just fine.
The assumption is that you don’t know how he picks the door when there are two goats. It’s a reasonable assumption when the problem statement doesn’t say anything about it.
It could be random, it could be the highest, it could be the lowest, it could depend on the position of the stars. The point is that you don’t know so the answer cannot depend on it.
Of course you stick -- Monte obviously opened the other doors just to trick you into switching because he knows you got the right answer. Of course he can do this; you already know that 999 doors have goats, so whether or not you picked correctly he can open 998 doors to reveal goats.
If you had picked a goat door, he would have just opened the door you picked or opened the door that has the money to show you that you got the wrong answer.
The assumption that Monty Hall will always offer you a chance to switch is what is broken in the problem statement and the reason why so many people think that the correct answer is unintuitive.
As has been explained in the other comments, Door A has a 1/3 chance of being the right choice and Door C (the unopened door) has a 2/3 chance of being the right choice.
All true... but if you toss a coin to choose whether to switch or not, the odds are 50:50...
Choosing between any two outcomes with a coin flip where one outcome is good X% of the time and the other is good 1-X% (the rest) of the time will always give you a 50/50 good outcome...
If you were correct with your initial choice, switching is a loss. If you were wrong on your initial choice, switching is a win. Your initial choice is only correct 1/3 times, but your initial choice is wrong 2/3 times. It's pretty simple.
This worked for me. Avoiding the "and now you have 2 doors..." moment in your explanation I think made it easier to stay away from seeing it as a 50/50
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[ 1.7 ms ] story [ 258 ms ] threadImagine there are 1,000 doors and you pick 1. All other doors except 1 are opened and you're given the offer: keep the door you picked, or pick this other door. What are the chances you picked the right door (vs. this other door)?
People seem to intuitively understand that having only one door unopened is a massive "hint" to where the prize is.
(I learned this idea from Better Explained: https://betterexplained.com/articles/understanding-the-monty...)
For me the most sensible explanation requires you to know that a dud door is always opened, thus the probability from the 2/3 is the one you are switching to.
In the 1,000 doors problem, my odds of being right initially were something like 1/1000 and then it changes to something like 998/1000 or 999/1000 for switching, I can’t intuitively grasp exactly what the odds become of winning if I switch, I just know it’s high. Bringing it down to 3 doors doesn’t help me much — it’s still something like 1/2 or 1/3.
probability of winning if you don't is 1 / n.
By switching, you are simply betting that your original guess of 1/n was wrong.
Note there's a newer version of "Let's Make a Deal", hosted by Wayne Brady, but there is no option to switch after a losing door has been shown in that version.
I'm thinking of a number between 1 and 10, guess it. If I now tell you a number I promise is not the one I was thinking and not your number, you have no more information about if you were correct.
So, you make a totally random choice. That choice must be 1/3 right, right? Now the thing that you already knew would definitely happen happens: Monty opens a goat door. How can your odds suddenly jump to 1/2?
Are you saying every single time you play the game, you always have a 1/2 chance of getting it right first time?
The first door will have the car 1/3 of the time. The second door's chances had been expanded to the remaining 2/3 percent thanks to Monty always choosing the last 1/3 door which is guaranteed to not have a car.
First step is still that you pick a door. There's a 1/3 chance it has the car. Now you can either keep that single door (with a 1/3 chance of a car), or switch and get both of the other two doors (each with a 1/3 chance of the car, for a total of 2/3 chance). After you pick, I'll reveal all the goats.
You pick some random person. I then bring in another stranger and tell you that the person who knows where the hidden treasure is is either the random person you chose or the one I brought in.
At this point, there are only two possibilities:
1. You happened to randomly choose the right person on Earth and in my surprise, I had to pick some other random stranger to pretend they knew the secret.
2. You chose a total rando who has no idea what's going on and the person I brought in is in fact the one who knows where the treasure is
https://youtu.be/GPoPSNxV1D4?t=365
I've timestamped the relevant bit - but you should watch the full thing from the start, it's very entertaining :)
A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round".
Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.
In the case of the clumsy Monty of your example, it goes like this:
1. There is a 1/1000 chance door 429 has the car.
2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000.
2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds.
3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch.
3a. Your door has the car, which happens 1/1000 games.
3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games.
In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching.
In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this:
1. There is a 1/1000 chance your door, 429, has the car.
2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours.
2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car.
3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door.
3a. You get there via 2a in 1/1000 games.
3b. You get there via 2b in 999/1000 games.
If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.
In the original problem, Monty always opens a door to reveal a goat. This strongly implies that he knows which doors have goats.
Your alternative is different. If the doors opened are truly selected randomly, then the odds are high that he would reveal a car, especially in the 1000-door version. What are the odds that Monty could choose 998 doors randomly out of 1000 and NOT pick the one with the car?
If Monty doesn't know where the car is, then you're arguing with a completely different version of the game that could result in him opening the door with the car, so the strategy is going to be different.
The rest of this post is an anecdote from the same class that this brought to mind, and is unrelated to the topic. Maybe we can say it shows how good teachers engage their students or something, but really it’s just a good yarn.
We were learning about inelastic vs elastic collisions, and how an elastic collision has 2x the energy of an inelastic one. The teacher asked for a volunteer, and a bright-eyed student rose to the occasion. The teacher gave him some safety glasses and told him to lie down on the floor.
The teacher took the inelastic ball and said, “Okay, I’m gonna drop this on your forehead now, ready?” PLONK. “Ow.”
“Remember that feeling! This is the elastic one, and it has the same mass, so it should hurt twice as much.” PLONK. “Ow.”
The teacher asked, “So, did the second one hurt more than the first?” The rest of us anticipated the experimental confirmation of what we’d just learned about.
“...I couldn’t really tell the difference,” said the student.
“Yeah,” said the teacher, “I knew you wouldn’t. I just wanted to see if you’d let me do it.”
The explanation that works best for me is that you were more likely to have picked a wrong door in the first place, so while the impacts are opposite equals, the likelihoods are not equivalent.
You choose a door with only 1 in a million odds of it being the door with a prize. Monty Hall know where the prize is and will only open the remaining doors he KNOWS doesn't have the prize. If he then opens up 999,998 doors without a prize behind them and asks if you want to keep your original door or switch, you'd obviously know that Monty's last remaining door must be the one with the prize.
I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random. The intent was for him to have prior knowledge or impeccable luck, and the wordsmithing of the question came later as, in my opinion, a failed rebuttal to the simplicity of the question. The question might have been worded to not be immediately obvious, but it was not intended to have different correct outcomes depending on interpretation.
If he opens a door and tells you he knows it's a goat, you double your likelihood of winning by switching to the remaining door.
If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
At best, that was intended as a red herring to make the correct solution less obvious. It was not intended as an alternate correct answer. It's not how the game ever worked.
The initial version of the question said he opens a goat door. Whether he knew it would be a goat door or not is slightly irrelevant, because your odds of being right the first time were 1/3. As far as the premise of the thought experiment, a goat door always gets opened, either through peaking, premonition, or consistent luck.
>Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
I disagree, and if you don't see it as that simple, you are falling for the trap that makes the question fun.
What do you mean "opens a door randomly"? Is he picking from all three doors? Yours, and the two others?
In that cases you get interesting but trivially-obvious-what-move-to-make scenarios like "he chose your door and showed you you were right" "he chose your door and showed you you were wrong" "he chose a different door which had the car"...
Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?
Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.
- you choose door 1, goat
-- host chooses door 1, goat
--- switch: 1/2 chance of being right
--- stay: you're wrong
-- host chooses door 2, goat
--- switch: you are right
--- stay: you are wrong
-- host chooses door 3, car
--- switch: you are right
--- stay: you are wrong
- you choose door 2, goat
-- host chooses door 1, goat
--- switch: you are right
--- stay: you're wrong
-- host chooses door 2, goat
--- switch: 1/2 chance of being right
--- stay: you are wrong
-- host chooses door 3, car
--- switch: you are right
--- stay: you are wrong
- you choose door 3, car
-- host chooses door 1, goat
--- switch: you are wrong
--- stay: you are right
-- host chooses door 2, goat
--- switch: you are wrong
--- stay: you are right
-- host chooses door 3, car
--- switch: you are wrong
--- stay: you are right
so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times.
there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too.
this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.
No. He picks goat 100% of the time. Choosing not goat is outside the bounds/parameters of the initial constraints.
Assume you choose the car (1/3). He will always choose a goat. If you switch, you have a 100% chance of losing. (1/3) * (1/1) = 1/3 chance of losing if you switch.
Assume you choose a goat (2/3). There is a 50% chance that he opens a goat if he chooses randomly. If this happens, there's a 100% chance that you win if you switch. (2/3) * (1/2) * 100% = 1/3 chance that you win if you switch.
The remaining 1/3 chance is voided because we've conditioned the game on him randomly choosing a goat, and not a car. So the probability of winning if you switch is (1/3) / ((1/3) + (1/3)) = (1/2)
exactly 50/50.
In the traditional monty hall, switching gives a 2/3 chance of winning. This is still true in the random variant but only if you're allowed to choose the car door in monty opens that one by chance.
He chooses randomly from the doors you did not choose, and draws a goat. When offered the choice to switch to the third door, it offers you no benefit. Both are equally likely to contain the car.
Hey lets put a car behind three doors, have people choose, open one of the other two doors, then ask them to switch.
Sounds good Bob, but wait, won't the show end prematurely 1/3 of the time because you will randomly open a car?
Huh, good point. Um, let's sneak a peek before opening so we never open a car?
Sounds good Bob.
That’s extremely obvious! The entire reason to phrase the problem as a game show with a game show host is to make it abundantly clear that the host “knows the answers.”
Reordering the sentence makes it clearer. "In all games where Monty chooses goat...". The premise creates a subset.
Monty’s knowledge is irrelevant. The original problem can be modified with no change in probabilities to:
Pick one of three doors. You can have whatever is behind that door, or you can change your pick to both of the other doors and win whatever is behind both of them. Should you switch? Obviously.
Step 2: Divide it into a pair of cards, and a single card.
Step 3: Reveal one of the pairs of cards. If you reveal and ace, return to step 1 since this history is eliminated from the story we've been told: we know this wasn't a possible path to our endgame. Otherwise continue to step 4.
Step 4: Which of the remaining two cards is more likely to be an ace?
Step 5: Realize that they're just as likely as each other to be the ace. Step 2 didn't magically imbue the card remaining in the pair with extra probability juice.
Long story short, you definitely need Monty to make an intelligent selection, so Monty's knowledge is far from irrelevant. It matters whether he revealed a goat by luck or by knowledge, because he's 2x as likely to get "lucky" in the case where there are 2 goats behind the doors you didn't choose in your original guess.
https://news.ycombinator.com/item?id=24713352
When he shows (randomly from the doors you didn’t pick) the car, switching to open door (where the car is) and the other door (which hides a goat) you increase your chances from 0% to 100%.
When he shows a goat, switching to the open door (where there is a goat) and the other door (which may or may not hide the car) your chances stay at 50%.
Switching is the best ex-ante strategy, not necessarily so after a door has been opened.
Try visualizing how you’d pseudocode this game - it literally didn’t click for me until right now, and now it seems much more intuitive.
This one took a second to rationalize. The reason it works is you have the same chance of winning overall (assuming you can safely choose the car if he opens the car by chance), it's just that the value of winning from switching vs. staying has been shifted into the probability of winning by default. The usual mental trick is to extend to 1,000,000 doors.
If you pick one door, then are told that all of the alternatives except one are the correct answer, you should obviously reason that the door you didn't choose is the correct answer, unless you got the 1/1,000,000 guess. Odds of winning if you switch are 999,999 / 1,000,000
If the host instead opens 999,998 doors randomly, you have a 999,998 / 1,000,000 chance of winning by default. The remaining two doors have equal chance of winning, giving you the same total odds of 999,999 / 1,000,000 no matter which you choose.
Which makes sense, because both situations are, more or less, being given 999,999 chances to guess the lucky door.
This is not made as explicit in the standard formulation "the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat."
Which could be read as "which happens to have a goat".
It's the ambiguity in Monte's door opening strategy that leads to different answers.
The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck.
The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with.
tldr: it doesnt matter how he opened a goat door, all that matters is that he did.
But the standard formulation just says Monte opened a door and it had a goat behind it. No explicit mention of intention.
It matters not that he happened to do. It matters that he will.
It's a Conditional Probability of non independent events.
I originally thought it couldn't matter. I wrote a simulation to demonstrate how right I was. I was wrong. I encourage duplicating the experience.
The counts bear this out. If you simulate it, 1/3 of all games go to 'stick' regardless. Either 1/3 or 2/3 go to 'switch', and either 1/3 or 0/3 (respectively) end up with Monty revealing the car.
Problem-2: "You pick a door. Monty opens a door at random, which just happens to be a goat. Calculate probability of winning if staying."
Both are conditional probability P(A|B) (that is, probability of A happening, under the assumption B has happened).
P(A|B) is defined as: P(A|B) = P(A and B) / P(B)In both problems P(A) is 1/3.
In Problem-1, P(B) is 1 because Monty knows and it's not a random event and P(A and B) = P(A), so P(A|B) = P(A)*P(B)/P(B) = P(A) = 1/3.
In Problem-2, P(A) is 1/3, P(B) is 2/3, P(A and B) is 1/3 (if you pick a car, Monty is guaranteed to pick a goat), so P(A|B) = 1/3 / 2/3 = 1/2.
The complete cases for the choices is:
Problem-1:
Problem-2: We know Monty didn't pick a car, that reduces it to:Problem-1:
Problem-2: Or, 1/3 for Problem-1, 2/4 (=1/2) for Problem-2That is false. Him opening the car is outside the constraints of the premise of the thought experiment. He can't not open a goat. The chance of him not opening a goat is irrelevant, and the probability is 0%.
You pick a door. Monty picks another door at random - he has clearly committed to doing so, flips a coin in front of you or whatever. God only knows what would have happened if he'd revealed a car, but this time he didn't. Having found yourself in that situation, what are your odds if you switch doors?
If you simulate this procedure, you'll find that the number of outcomes where "switch" wins is about 1/3 of the total games; "stick" wins in about 1/3 of the total games; and 1/3 of the total games Monty revealed the car. When Monty does pick the goat reliably, as in the Monty Hall Problem as it was intended to be understood, Monty is doing the work to convert those "revealed" games into wins for "switch".
If you disagree, I'll happily take your money.
I mean that literally - how can I write this into a simulation to test that there's a difference?
If you find yourself actually in the situation, you need to make your best guess from what information is available.
As for simulation, write a function that handles a single play through of the game and returns the outcome if you switch. Make Monty's strategy a parameter. Remember that "Monty revealed the car" is a separate outcome from "win" or "lose". Run that function a whole bunch of times, counting the outcomes. Compare the ratio between wins and losses, at various choices of strategy.
To really drive home the point that strategy can matter, consider another possibility: maybe this season the studio executives have decided that cutting costs is more important than having an interesting game show, and Monty's strategy is now "reveal the car if you haven't already chosen it". Monty has revealed a goat. Is there still a 2/3 chance that you didn't pick the car?
(Below I use non-chosen to mean non-chosen by the contestants initial choice.)
Second point: If we are in the subset of all possible histories where Monty picked randomly revealed a goat, then we will have 50% of histories where both non-chosen doors contain 1 goat selected by our history subset, and 100% of histories where both non-chosen doors contain 2 goats selected by our history subset.
Since there are twice as many possible histories where the non-chosen doors contains 1 goat vs 2 goats, after selection, we have an equal number of histories in our sample where we have 1 goat or 2 goats behind the non-chosen doors. Or equivalently, we have a 50% chance that the non-chosen doors contain a car.
Therefore it is irrelevant whether you switch.
Monty needs to make an intelligent selection to change the game.
We can work out odds for this "always switch away from original door" strategy: suppose you initially chose a door with the car (1/3 probability). Then choosing a new door makes you surely lose regardless. On the other hand, suppose you initially chose a door with a goat (2/3 probability). Then regardless of which of the two doors Monty opens, you can choose the car (if he revealed the car, choose that. If he revealed the goat, choose the other door). So our odds of winning with this strategy are still 2/3.
So it's up to the interpretation of the modified game I guess.
Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to it and won. But he didn't.[0] Now you have a choice to make.
His probability of not revealing a car in the case that both doors you didn't pick had goats behind them, is 100%.
His probability of not revealing a car in the case that one door you didn't pick had a car behind it, is 50%.
We can reason then, that of the three possible scenarios for the two non picked doors {g,c}, {c,g}, {g,g}, that it is equally likely now that since he didn't randomly reveal a car, {g,g} must be weighted twice as much as the other two.
Therefore, we must be as likely to have a goat behind the remaining door as we didn't.
In conclusion, Monty only changes the problem if he selects with information.
[0] Notice that at this point, we have to throw away from the decision tree 1/3 of the histories, and in all of them switching was beneficial. In what remains then, switching is less beneficial than it was before this trimming.
That’s true before he opens a door. Then either
A) he shows a car and the odds of winning with this strategy are 100%
or
B) he shows a goat and the odds of winning with this strategy are between 1/2 and 2/3 depending on how the choice of door was made
If the choice is random, he shows a car with probability 1/3. The probability of winning is 1/3 x 1 + 2/3 x 1/2 = 2/3
But that analysis is for the unconditional problem and what we’re asked is what to do in case B, after a goat has been unveiled.
If he shows always a goat, the probability of winning is 0 x 1 + 1 x 2/3 = 2/3. Here there is no difference between the unconditional and the conditional problems because A never happens.
So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3).
But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others.
Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick!
(But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')
Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge.
The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge.
> I got out three playing cards and did the experiment myself over and over.
How did you do that when you didn't know which cards were 'goats'?
This is often not explicitly stated when the problem is given. It is even not a 100% clear from the statement above. Monty always chooses a door with a goat. So:
1. You choose a door.
2. Prob that there is a car behind it: 1/3
3. Prob that the car is behind the two other doors: 2/3
4. If the car was behind the two other doors (which, remember, has p=2/3), Monty will choose the door without a car for you, and the door with a car will remain closed. In this case you are guaranteed to have the car if you switched.
So with switching, the overall probability is 2/3. Without, its the original 1/3.
If you did not understand that Monty always chooses a goat door, but the person giving you the problem does, or vice versa, then what usually happens is that both of you try to explain why your intuition is correct. Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.
> This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand.
If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.
I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)
I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.
Personally, IIRC, my first reaction was to assume the second box opening was not random (perhaps only because the question is not phrased as being conditional on this act revealing a goat) but did not see how this gave any useful information.
Here's another possible way of getting it wrong, regardless of the phrasing of the problem: assuming that, after the reveal (and whether one thinks of it as random or not), one is, as it were, starting over, except with a choice between two boxes rather than three, and no other information.
> According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.
Interestingly, coding up my own simulation took me from understanding the problem to grokking it. I'd definitely suggest anyone who has programming ability but doesn't grok the problem to write up a simulation.
IMO, people are confusing MH with a case where the door is opened randomly before your choice.
Let's say Monty shows you three doors, then knowingly opens a wrong door (B) before you choose. Do you want A or C? It's a coin toss, of course.
But suppose you pick door A first and THEN he knowingly opens a wrong door you didn't choose (B). Now the odds of winning are 2/3rds if you switch. But it feels the same as in the previous case. ("How could choosing a door first affect the odds?! It couldn't possibly change anything!!")
The answer is that your choice of a door constrains Monty.
Suppose you choose a door secretly before Monty opens a door. Suppose you choose A, and Monty opens door A, revealing a goat. Well, uh, you can't pick that any more, now, can you? You're forced to choose between B and C, a toss up.
But if you publicly pick door A then Monty can't/won't open A; he has to open one of the other doors. Since you probably chose the wrong door to begin with, and he always opens a wrong door, the other remaining closed door probably has the car.
When you put the games in terms of "you and a friend playing the game simultaneously", most people get it.
You and a friend are playing the game simultaneously; you both decide ahead of time to: a. always pick the same initial door b. no matter what Monty does, you ALWAYS switch, your friend NEVER does.
The possible outcomes before the reveal are: * Monty wins, because he shows you the car. We agree ahead of time that this never happens.
Now, since you know one of you is going to win, since Monty didn't, who has the better chance? Your friend started with a 1/3 chance, and he hasn't switched; does his chance suddenly change because Monty opened a door? Most people (not all!) will agree that it has not.
Since one of you MUST win, and friend has 1/3 chance, what's left?
If Monty doesn't know, then sometimes the game will be ruined because he will expose the grand prize and then the game is moot. But if is simply lucky by showing the goat door vs he picked it with foreknowledge doesn't change the odds in any way.
You are taking an empty result and replacing it with a strict win.
That said, though, if you're trying to convince someone of the correctness of the Monty Hall Problem, and they're stuck on that misconception, saying "well you may as well switch in case I'm right" isn't a great argument.
Let "A", "B", and "C" represent the prizes; A is valuable, B and C are goats. They're shuffled behind doors, so the player and Monty choose "1" or "2" or "3", and after the fact we'll map ABC to 123. As such, we can assume without loss of generality that the player always chooses "1" and Monty always chooses "2".
What happens?
Huh. Of the 4 non-ruined games, half of them are improved by switching, half are worsened.I believed you were wrong, but now I believe you.
1/3: you picked car, switch fails
1/3: you picked goat, monty ruins. Not included
1/3: you picked goat, monty open goat. Switch wins.
Of the two non ruined scenarios, half win
We can also demonstrate by using the 100 door example. Monty opens 98 of the 99 doors you didn’t choose. 98 times out of 99, monty ruins the game.
So random choice dramatically changes things.
If Monty chooses randomly both strategies fail 2/3 of the time.
Whether Monty knows for sure or is just picking at random only affects whether games are ruined, not the probability that the door the contestant picked is a winner.
If there are N doors and the player picks door 1, there is a 1/N chance he will be correct. If he sticks to that door, there is nothing that will ever change that 1/N odds that it was the winner.
Downthread someone enumerated the cases, but again, they discard the 1/3 of the cases where Monty ruined it. What that does is throw out half of the cases where contestant picked the wrong door and had the wrong strategy. Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.
By that logic, so does the remaining door. 1/3 vs 1/3.
Its like saying soccer players could score more if they picked up the ball and ran with it. While true, you are no longer describing the game of soccer.
I wrote a simulation to demonstrate how right I was.
I was wrong and you are too.
It's actually an interesting, possibly general story, that kind of makes me think. Perhaps for certain personality types, misplaced conviction will put you on a higher velocity trajectory toward truth than honest confusion.
The key part of this sentence that makes it true, is the "for certain personality types".
I would wager the majority of personality types however, it would put you on a higher velocity trajectory to willful ignorance.
So I'm not getting the point you're making, unless you're saying that would be a more grokable way to state the problem?
Imagine you have two identical looking bags of marbles, one contains two red marbles, and one contains one red and one black. You reach into a random one and draw a marble. It is red. If you picked the marble randomly, you now know you're more likely to be holding the bag that contained two red marbles than the mixed bag. The red marble you randomly drew provides bayesian evidence for which bag you're holding.
Out of the three possible equally likely worlds (given that you know you didn't roll into BR-1), two of them have you holding bag 2R, aka a 2/3rds chance.If, on the other hand, you use a red-marble-picking robot, you get:
Now you have four possible worlds and are equally likely to be holding either bag, ie 1/2.So, if our friend Monty opens a random door and you only look at simulations where it reveals a goat... There is a 1/3rd chance you initially get the car, but if you do he's guaranteed to reveal a goat. So 1/3rd of games, you have a car. There's a 2/3rd chance you initially get a goat, but then he has only a 50/50 chance of revealing the goat. So half of the 2/3rds of games, the game is discard as invalid, and the other half of 2/3rds (aka 1/3) you have a goat. As such, the potential outcomes are split 1:1:1 between car:goat:invalid. Removing the invalid cases, it's 1:1 car:goat, despite the only 1/3rd chance of you choosing the car initially.
tl;dr: If he's randomly opening doors and you discard any where it's the car, those cases are entirely ones where you didn't initially choose the car, which introduces some bias. If he's opening doors deliberately, no cases are thrown out since it's always possible to reveal a goat, maintaining the original 1:2 probability ratio.
EDIT: formatting.
Alright just working through your logic here. Let's say I pick the first door always. I'll write out all the possible configurationss with both possibilities for Monty.
CGG > Monty reveals mid (Goat)
CGG > Monty reveals right (Goat)
----
GCG > Monty reveals mid (Car)
GCG > Monty reveals right (Goat)
---
GGC > Monty reveals mid (Goat)
GGC > Monty reveals right (Car)
So there's 4 worlds where Monty reveals a goat. In two of them, I picked a car initially. In the other two, I get a car by switching.
My prior post is mistaken -- thank you!
(C)GG - switch and lose
(G)CG - switch and win
(G)GC - switch and win
There are other situations where the game is over before you get a decision.
Both GCG and GGC are eliminated half the time before you get to make a decision. But CGG is never eliminated before you get to make a decision.
Once you're making a decision and the game isn't already over, you can use this information -- that the game didn't end -- to assign a 50% chance of being in CGG, and 25% of each of the other two.
Therefore, no games are removed as moot when you have initially selected a car, and 50% of games are removed as moot when you have initially selected a goat.
So it ends in a coin toss whether you switch or not.
The 2/3 to 1/3 split only happens when Monty removes a goat using privileged information.
The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.
If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge.
Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
I thought as you think. I wrote a simulation to show I was right. I was wrong.
I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
This is basically just a different way of saying that Monty looks behind the door to be sure to only reveal goats.
Yes, the probability of wins will be different in these two scenarios, but it doesn’t affect the conclusion: when Monty reveals a goat, you should always switch.
Edit: if you believe I am the confused one after this comment, I will go make the simulation as you suggest.
As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.
No, it does not.
>> if (montys_choice === car_door) { i -= 1; continue; }
This discards the game where Monty chose a car.
>> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door);
This makes Monty never choose a door with a car.
Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.
Is it the same game if I flip the coin secretly, peek at the die, announce my number (picked algorithmically in the obvious way), and then pay out according to whether switching would win (as above)?
Because (assuming I've explained these games as I intend... it's getting late) I would play the former with you not the latter (but I would play the latter if we switched the payments around - I chose 5 and 7 because 7/12 is halfway between 1/2 and 2/3).
To put it another way, if Monty is choosing randomly, half the time where you would win by switching, instead the game just ends/isn't counted, but the same is not true of the case where you win by not switching. From a bayesian point of view, Monty randomly revealing a goat should increase your belief that you picked the car the first time.
That said, switching is not worse than not switching unless Monty is biased towards revealing the car instead.
In one, Monty reliably reveals a goat. In the other, Monty picks randomly between the other two doors.
We are, in parallel universes, playing both of those games. We know which. The host has just revealed a goat.
I read fouronnes3 as saying that these two situations are the same. They are not.
WLOG assume the car is in door A (The rest will be the same by symmetry.)
You pick A, B, or C.
The host picks one of the other two doors to show you.
The universes can be described by your choice, followed by the host's random choice. There are 6 possibilities: AB, AC, BA, BC, CA, CB.
Since we're not in a universe where the host chose a car, we eliminate BA and CA.
We are left with four possibilities: AB, AC, BC, CB.
AB = You chose the correct door. (probability 1/4)
AC = You chose the correct door. (probability 1/4)
BC = You chose the wrong door. (probability 1/4)
CB = You chose the wrong door. (probability 1/4)
Since the host's choice was random, it's 50/50 as to whether the car is behind the door you chose or the remaining door.
This same reasoning breaks down when the host knows what's behind the doors because you're no longer eliminating a couple of the universes and it's a 2/6 vs 4/6 situation.
The subset of games where they close different doors is smaller than all possible games.
The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)
As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.
Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable.
The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)
And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing?
Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again.
Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
Yes.
> Where does the car go if no one gets it?
I don't understand your question. What does this have to do with probabilities?
> And if they switch both have probability 2/3 of getting the car?
Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously.
They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's.
> Is this some form of car sharing?
That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch.
> Maybe you should try to play with three cards and one ace.
There are only 3 doors, so you should only use 3 cards.
> Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".
Player #1 has picked door A
Player #2 has picked door B
The host has opened door C and there was a goat
At this point:
What are the chances of winning of player #1 if he keeps his pick of door A? And if he switches to door B?
What are the chances of winning of player #2?
Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches.
In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).
The probability of player #1 getting the car if he keeps his door, which is equal to the probability that the car is behind door A, is 1/3
The probability of player #1 getting the car if he switches, which is equal to the probability that the car is behind door B, is 2/3
The probability of player #2 if he keeps his door, which is equal to the probability that the car is behind door B, is 1/3
The probability of player #2 getting the car if he switches, which is equal to the probability that the car is behind door A, is 2/3
In summary, the car is:
Behind door C with probability 0
Behind door A with probability 1/3
Behind door B with probability 2/3
Behind door B with probability 1/3
Behind door A with probability 2/3
You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent.
But don't trust me: try it!
After player #1 has picked door A, player #2 has picked door B and door C has been opened revealing a goat:
a) if player #1 stays with door A his probability of winning is ___
b) if player #1 switches to door B his probability of winning is ___
c) if player #2 stays with door B his probability of winning is ___
d) if player #2 switches to door A his probability of winning is ___
e) the probabilty that the car is behind door A is ___
f) the probabilty that the car is behind door B is ___
g) the probabilty that the car is behind door C is ___
a) The probability of winning if staying: 1/3
b) The probability of winning if switching: 2/3
c) The probability of the car being behind door the player chose is the same as a): 1/3
d) The probability of the car being behind the door Monty didn't open is 2/3
e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat!
Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.
I agree it's different from the original problem, that's the point!
But the game I proposed is equivalent to the version of the problem that some people, including you apparently [0], insist in this thread that is equivalent to the original problem. The latter is not equivalent to the original problem. The situation is similar but the rules are different. The solution is different.However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors.
They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat.
[0] The problem I proposed at https://news.ycombinator.com/item?id=24707305 is also equivalent:
So you have Player A who chose door 1. The host flips the goat in door 3.
You have Player B who chooses door 2. For them the host flips open the goat in door 1.
Player A should switch, and wins the car. Player B should switch, but had the car and loses it. The right move in both their cases was to switch.
Really, its no different than playing two separate games. Unless, like you said, you somehow change the rules and limit someones choices, in which case Im not sure why its a part of the conversation, other than being introduced through confusion.
https://news.ycombinator.com/item?id=24713352
If the players pick different doors, the host will open the third door. That prevents him from being sure to get a goat, in the same way that opening a door at random in the single-player case prevents him from being sure to get a goat.
For this reason, both players should be playing "blind", without seeing the other's choices or the door Monty opens for the other player. Anything else doesn't make sense for the game.
As it does removing the host's ability to choose a non-winning door in the variant where "the host had just picked at random from the two doors, and it happened to show a goat". That doesn't work the same either because it's not the same game.
If the host picked a door at random, and it happened to show the prize, then the game would be ruined. That would make some terrible television. Therefore, he can't be picking a door at random.
I was always told Monty reveals new info and that changes probabilities. Never made sense. Your 2/3 vs 1/3 explanation makes it obvious.
Then there is the ambiguity as whether the host will always offer you to switch or just when he hasn't revealed the car.
- the bottom line is Monty will always eliminate one wrong choice
- therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2)
(edit: thanks i see, no need for any more answers)
I get why you intuit this - I had a lot of trouble with it at first as well. Common sense is lying to you, however, because you are still dealing with the original odds (1/3 vs 2/3). Monty's choice is always to choose a Goat (probability 1.0), therefore his effect on the probabilities will always be with a weighting of 1.0. You end up with a probably spread of (1/3 * 1.0 vs 2/3 * 1.0).
Therefore it is always in your interests to choose Monty's remaining door.
A more common re-telling of the story:
Monty has 1 million doors, with only one car behind one. When you pick a door, Monty will always open 999,998 other doors, all of which show goats, leaving one other door conspicuously closed.
Do you still think you have 50% chance of being right by sticking with your choice?
As some other comments have mentioned, I think this isn't a fully satisfying explanation because it's hard to reason about the difference between a general strategy vs. an in-the-moment choice; why does it matter that I've watched Monty narrow the choices down to 2 doors, vs. seeing 2 doors from the start?
This is how I understand the Monty Hall problem (Using 3 doors):
1) You are asked to choose a door from A, B or C. You choose A.
2) Monty opens a door and shows you a goat. Let's say he opens door B and gives you the choice to switch from door A. 3) The probability of the car being behind door A remains unchanged at 1/3, therefore the probability of the car being behind not A is also still 2/3. Since door B has been eliminated, this means the probability that the car is behind door C _must_ be 2/3! Therefore, we choose to switch from door A to door C.If you made a one in a million guess the first time, your door has a car. If you didn’t, the other remaining door must have the car. Which do you choose?
Thank you for spelling this out so clearly. If explained in terms of strategies (always switch v. always stick), the probabilities should be crystal clear.
What does the switcher have to do to win? Pick the wrong door first (2/3). What does the sticker have to do to win? Pick the right door first (1/3).
Yes, this is the single most critical piece of information, and when people struggle to understand the Monty Hall problem, its almost always because this wasn't made clear. When it is, the problem becomes much more intuitive.
Also, Game 5: There are 2 doors. A car is randomly placed behind one, and goats behind the others. You pick one door. Monty looks behind the other doors. He chooses 0 of them with goats behind them, and opens them. You get two options: Option A: You get whatever is behind the door you picked. Option B: You get whatever is behind the other closed door. Should you switch?
1. https://en.wikipedia.org/wiki/Almost_surely
Simulated many more variations at https://simonduff.net/monty_hall/
https://www.quora.com/In-the-monty-hall-problem-how-does-ope...
Reprinted:
Q: In the monty hall problem, how does opening the second door skew the probability in favor of the initially unchosen door?
A: The Monty Hall problem is generally poorly described, in order to make the conclusion seem more surprising then it is.
The actual Monty Hall game — as imagined by the people who are asking the question— is set up like this:
The choice is this: should you choose the remaining door from the pair selected in round 1, or should you choose the door which was not part of the selected pair in round 1?When phrased like this, the answer is fairly obvious: the pair of doors contains a car 2/3 of the time, whereas the non-paired door contains a car 1/3 of the time.
The Monty Hall problem— as normally described— messes this all up by introducing a game show host. This is an agent who— seemingly by their own whim— changes the game you thought you were playing, and introduces round 1 of the game once you have guessed the door you initially think the car is behind. The rules this game show host agent are following are almost never described to a sufficient degree to ensure the game is equivalent to the game laid out above. And yet, the people asking this problem pretend that it is exactly equivalent when they ask you for an answer.
It’s generally a poorly described problem, whose answer depends entirely on what kind of agent the game show host is.
Don’t worry if it doesn’t make sense to you as it’s usually described. If you can understand why— in the two round game I describe above— it’s better to pick the door from the pair of doors, rather than the single door, you understand probability just fine.
What happens if Monty does not ever choose at random. Say that Monty always opens the the highest possible door.
If you choose door 1 and Monty reveals door 2, then switching (to 3) is 100% win.
If you choose door 1 and Monty reveals door 3, then switching (to 2) is 50% win.
I think it's a crucial unstated assumption that Monty does choose randomly among available goat doors.
It could be random, it could be the highest, it could be the lowest, it could depend on the position of the stars. The point is that you don’t know so the answer cannot depend on it.
There are 1,000 doors - you are passed to pick one.
Monty removes all the doors except for your door and one other. There is money behind one of the doors. Do you stick, or change?
If you had picked a goat door, he would have just opened the door you picked or opened the door that has the money to show you that you got the wrong answer.
The assumption that Monty Hall will always offer you a chance to switch is what is broken in the problem statement and the reason why so many people think that the correct answer is unintuitive.
All true... but if you toss a coin to choose whether to switch or not, the odds are 50:50...
Choosing between any two outcomes with a coin flip where one outcome is good X% of the time and the other is good 1-X% (the rest) of the time will always give you a 50/50 good outcome...
https://en.m.wikipedia.org/wiki/Doomsday_argument
Just the fact that you’re doing an experiment is already extra information!