> Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result (Vazsonyi 1999).
What's even wilder is that the people working for game shows utilized this strategy long before it was a known, publicly verified thing. So someone who wasn't a world famous mathematician figured it out and was confident in their work - the same work that a renowned, industry-leading professional refused to believe.
Once you figure it out, it's pretty obvious though. And the game show people don't need a rigorous mathematical proof. You can simulate it with some playing cards and
understand it in a hurry.
What made it very clear to me is if you just increase the number of incorrect doors that you open by a lot. (Making the assumption that the host knows where the prize is.) In that case, it's obvious that you're improving the odds by showing where the prize isn't.
Well, it's also amusing to me that people want to jump to simulations that generate a large number of examples. In the original 3-door version of it, it's reasonable to exhaustively list all the possible sequences and just tally up the outcomes under each strategy. The simulation is just sampling from this not-that-long list.
Basically, Wikipedia does not mandate a citation style, it just has to be consistent throughout an article. And once set, another editor can't change it simply for personal preference. Presumably the original author, or whichever editor first added citations, was more familiar with this style, and so it has stuck.
I found the easiest way to convince non-mathematical types is by suggesting we do the million-door version: “Ok, pick a door between 1 and 1,000,000. We’ll call that door N. Now, I’ll open 999,998 bad doors. So, now only door N and door 223,205 are still closed, and one of them has the good prize. Want to keep door N, or swap?”
This is exactly how I conceived of it when I first heard of the problem. Pretty sure I remember demonstrating it to people using 10+ playing cards and flipping over all but one of the cards the person didn’t choose.
The flaw with this thought experiment as a way of teaching the original problem is that it's not at all the same to reveal 998,000 bad doors as it is to reveal just 1, and when revealing just 1 in this situation the effect is much smaller than just revealing 1 of 3 doors.
Please explain to me why it is different whether to switch doors after opening all but one door vs after opening all but one door?
(note: when you do this in real life, you stop talking now. let them explain it.)
When you have N>2 doors, the host will with 100% certainty open N-2 doors with goats behind them. It doesn't matter whether N is 3 or 1,000,000. After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2.
The key phrase is "host's hidden knowledge". Once the host starts opening the wrong doors, the fundamental nature of the problem is changed from being an independent probability problem to a dependent probability one.
> After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2.
No, your numbers are wrong.
In your initial choice, the probability of picking the winning door is 1/N.
When Monty opens doors so that there is a single door remaining, the probability of that door containing the prize is the complement of the probability that you picked the winning door, i.e. 1-(1/N).
So switching doors increases your probability from 1/3 to 2/3 if N=3, not to 1/2.
If N=1000, switching doors increases your probability from 1/1000 to 999/1000, and that is why increasing the number of doors makes the correct solution more intuitive for some people.
And the point I'm making is that it's not clear at all why opening 998 doors out of 1,000 is analogous to opening 1 door out of 3 -- why wouldn't you still be opening just the 1 door? If you pick one of 1,000 doors, and then the host reveals a goat behind a single door, there's a slight (but only very slight) advantage to switching. The advantage is most pronounced with the fewest possible number of doors, i.e. 3.
That's why this approach to teaching it has never made sense to me -- it's not clear at all why opening 998 doors should be analogous to opening 1 door, which it's much more obvious that opening 1 door is analogous to 1 door.
Again you didn't say if you opened 999998 bad doors due to chance or by conscious decision. If you did it by chance then the answer would be completely different.
Those runs get ignored since that isn't the scenario. Specifically we're looking at the case where Monty opens a goat by chance (ie. he could have opened the car, but did not). It also helps that that result doesn't matter to the overall question, is it better to switch or to stay, since both options are irrelevant when Monty opens the car door.
By deleting the universe 1/3 of the time, you change the nature of the problem. It starts to resemble a quantum immortality argument. And if I can delete universes where certain events happen then screw probability, I can guarantee a win despite picking randomly.
"Picks a random door out of all closed doors, but that door is never the winner" is a contradiction. It doesn't give you a different chance of winning because it's not a coherent scenario in the first place.
Conditional probability is a coherent concept with no need to talk about deleting universes. You can ask "If it's raining in the morning how likely is it to still be raining in the afternoon?" even though it's sometimes not raining in the morning.
Conditional probability is fine. But you have to make the condition part of the question. You can't subtly omit certain results because it screws up the narrative.
This is a game where you start off with a certain layout, and then proceed forward. If you want conditions, they have to be conditions that you can apply before the game starts. You can't retroactively remove a significant chunk of samples.
If you added an actual outcome to him picking the winner, you could make a valid filter where the answer actually is 50:50. Perhaps they restart the game. Perhaps they never air the episode. And if you calculate only for finished/aired games then it's very clear then that you're solving a different math problem. You can't apply that answer to the original problem.
It doesn't seem that weird. That just means your code is choosing to not count half of the scenarios where you would have won in the original simulation.
Yes, it does! The whole problem revolves around whether Monty Hall was giving you more information by his choice of which doors to open, or whether he was just opening doors randomly.
If he was opening doors randomly then you're left with a one-in-two chance. But if he was giving you information by opening doors he knew were bad, you're left with two possibilities: the door you started with just happened to be the right door (a tiny chance), or (much more likely) the door you started with was not the right door - in which case Monty was forced to open every other door so as to avoid showing you the car, and he has eliminated all the non-car options.
But that means that something knows which door was the winning one. Either Monty knew which one it was, or whatever mysterious Agent is sampling from all the possible outcomes knew (because the Agent only sampled the ones where Monty happened to open a non-winning door). Either way, you're given the choice between having the 1 door you initially chose, or both of the other doors.
> Either way, you're given the choice between having the 1 door you initially chose, or both of the other doors.
i initially thought this too, but in both cases you're gonna see 2/3 doors. the thing that makes the difference is when he has the car (2/3 of the time), he has to tell you where it is. so the choice becomes between your unknown door and his 2/3 selectively chosen car door.
this is exactly what tripped me up when I first heard the problem and it took me a long time to understand the explanation because I didn't understand that Monty always opened the door that did not have the car behind it.
But it’s from a game show. Like, an actual show. 1/3 of the time they can’t just stop filming and say "whoops let’s pretend that didn’t happen." That interpretation of the problem makes no sense whatsoever.
> If he was opening doors randomly then you're left with a one-in-two chance.
No. Or, well, it depends on what happens if Monty opens a door with the car. If I lose if he does that, then switching doors doesn't change the probability.
It's easy to prove. Let's say that I always pick door #1, and Monty always picks door #2. Then we get the following probabilities:
car-goat-goat:
I stay, I win.
I switch, I lose.
goat-car-goat:
I stay, I lose.
I switch, I lose.
goat-goat-car:
I stay, I lose.
I switch, I win.
Staying: 1/3 chance of winning.
Switching: 1/3 chance of winning.
Imagine you're taping episodes of this game show for broadcast. The contestant picks a door, and then Monty picks one of the two remaining doors at random. Monty has 1/3 chance of accidentally picking the car, and if he does, we're scrapping the entire episode.
So to enforce your constraints, we have to scrap 1/3 of all episodes, we're only ever broadcasting the 2/3 of all episodes taped where Monty, by chance, didn't pick the door with the car behind it.
Your numbers are correct, but that interpretation of the problem doesn't make a lot of sense if you think it through.
Move the scenario from Monty Hall to a street trickster, then this interpretation makes a lot more sense than the original one since a street trickster is unlikely to try to increase your chance of winning.
As for the game show, nothing says that he needed to give this choice every time, it was just some random thing he did to make it more interesting not an important part of the show.
We don't have to scrap the remaining episodes. Suppose that the situation is the Monty always picks randomly and in the situations where he reveals a car the episode just ends then. Then if when you go on the show Monty happens to reveal a goat should you switch? The answer is that it doesn't matter whether you do or not.
Yes, but that is not what you claimed. You claimed:
> The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2.
If the episode just ends when Monty picks the car, then I lose if he picks the car. I don't just vanish into thin air such that the universe erases my run. I was on the show, and I lost. And that means my odds of winning are still just 1/3, not 1/2 as you claimed. Yes, the game has changed such that it doesn't matter if I switch doors or not with this addition, but the underlying odds don't change.
You are wrong here, the problem statement states "Monty Hall opened a door and it had a Goat", so the situation you describe where he selected a car didn't happen. Since he didn't select a car you now have 50% chance behind both doors. But you are right that you have 1/3 chance from the beginning, but the situation described in the problem is not at the beginning.
No, that is not how the original Monty Hall problem was formulated.
The original said: "the host, who knows what's behind the doors, opens another door, which has a goat."
Your formulation of the problem changes it from the original Monty Hall problem to something more similar to the Tuesday Boy problem, where the trick is that you have pre-selected a bunch of outcomes and discarded others, changing the probabilities.
And going back to the comments above this, it's not the case that Monty choosing a door by random is the thing that matters, it's the thing that in this version he chose a door by random and you discarded all the outcomes where Monty chose the car.
It's not the case that people are confused by the problem because they interpret the problem like you did. People understand that a goat door is always eliminated as part of the game, and they still don't agree that switching is better.
I feel like you're fighting the meaning of the word given here. You might as well say that your probability of winning is tiny because you're unlikely be picked as a contestant, or even less likely than that because you might live somewhere than the US.
The whole premise of the problem is that you are already on the show looking at an open door with a goat behind it.
Upthread I see that you are correctly writing about the different ways you can model Monty's behaviour, and how each different way changes the nature of the problem, the outcomes, and the odds of switching vs. staying.
So I don't understand what you are arguing against right now?
I'm saying that "Monty reveals a goat" is what happens in the correct interpretation of the problem, i.e. Monty knows where the car is, and always chooses a door with a goat in every single instance of game play. In this case switching gives you 2/3 odds of winning the car.
And I'm saying that "Monty revealed a goat" is what happens in the incorrect interpretation of the problem, i.e. Monty does not know where the car is, and in this single instance of game play that you happen to be participating in, he happened to pick a door with a goat. And if you have gotten to this point, switching is 50/50. Just like you say.
Do you understand my way of describing these two different models of Monty?
I was arguing against reveals vs revealed because I thought you were making a different point. I understand what you meant about that difference now.
I'm still not sure what the difference between these models has to do with my point about goat-car-goat. That can't happen in either model when we say that I'm picking door 1 and Monty is picking door 2.
Ok, right, the point I was arguing against the parent poster to my initial post in this thread is that he said that "Monty picks a door at random" implicitly means that we have to discard all the outcomes where Monty picked the car.
But I'm saying that since that is a wrong interpretation of the original problem, any other wrong interpretation of the problem is equally valid. For example, if Monty picks the car, it's game over and the contestant never gets to choose. Or, if Monty picks the car, the show goes "oops", closes the door, shuffles the prizes, and lets Monty pick again until he picks a goat.
There's not enough information in "Monty picks a door at random" to infer a single interpretation of the problem.
"Monty picks a door at random AND he picked a goat". That's enough information to get down to your universe where goat-car-goat could never happen.
Nitpicky semantics, sure.
But to roll back to the original, original question: I don't think people get the Monty Hall problem wrong because they're interpreting it as "Monty picks a door at random and he picked a goat". They interpret the problem correctly, i.e. "Monty knows where the car is and always picks a goat", and they still get it wrong.
Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat.
I'm not saying that goat-car-goat could never happen. I'm saying that when you're evaluating what to do after Monty Hall has revealed a goat, the one piece of information you know is that goat-car-goat didn't happen.
I guess you're analyzing from the point of view of someone who's trying to figure out their odds of winning with different strategies before playing.
I agree with you that most people don't get the problem wrong because of interpretation. I'd go even further and say that many people who think they understand it don't get it right because of the interpretation.
Most arguments I see for the correct answer don't use the assumption that Monty knows which door has the goat, and therefore work equally well under a model of Monty picking the door at random. They're just picking different ways to model the problem and assuming a uniform distribution. You get the answer right if you picked a model where the distribution is uniform under the assumptions in the problem.
The real issue seems to be much simpler. Probability is not intuitive. Most people don't have the tools to figure out problems like this reliably, and those that do tend not to apply their tools when they think they see a simple obvious answer.
> Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat.
No, if Monty picks a door by random, but the show shuffles the doors until he gets it "right" and picks a goat, that is exactly the same scenario as if Monty knows where the goats are and picks one straight away, which means my odds of winning when switching is 2/3.
So that is a scenario where "Monty picks a door by random" doesn't lead to the odds changing to 50/50.
If by shuffle you mean that Monty just gets to keep trying until he picks a goat, you're right.
If instead you mean that they randomize all three doors and start over, the probability is 1/2 because half the time that the player has a goat you start over and get to pick again, which perfectly cancels out the double probability of having a goat.
This can be analyzed pretty easily using the three possibilities you outlined earlier. Suppose the player gets door 1 and Monty always picks door 2. Each time they shuffle, you get one of the following with equal probability:
A. car-goat-goat
B. goat-car-goat
C. goat-goat-car
A means the player wins, B means we re-shuffle, C means the player loses. It's pretty easy to see that A and B are equally likely outcomes, so they become 50/50 odds after all the re-shuffling is done.
Yes this makes sense to me. The way I thought about it is if you win if he picks the door with the car, then you’re in essence getting two choices. So the total probability of winning is 1/3 (probability of Monty picking car) + 2/3 * 1/2 (prob of monty picking goat and you picking car). The reason I commented is I can compute the total probability of winning in your scenario by saying Monty has 2/3 chance of picking a goat and I have 1/2 chance of picking a car after that. The result then is 2/3 * 1/2 = 1/3
The key insight is that which door is picked actually communicates information to you about the other doors because you know those other doors were not picked. You now know more about the probabilities of various outcomes which means you need to change your choice. Also note the question is about "do you change your choice?" not about the chances of success with any specific door.
If it's unspecified, the odds that Monty Hall opened all 999,998 bad doors out of 999,999 randomly is pretty damn low. You still switch if it's unspecified.
If it is unspecified we should probably assign some positive probability that the host did it on purpose. At that point, it is better strategy to switch doors.
p=1. This is the scenario. The host teases you by opening doors that do not have a prize. If he opened one with a prize then you couldn't win it which would break the game.
He knows exactly where the prize is. He just had a good poker face and head game. Which is why he hosted the show for 15+ years.
It’s in the original statement of Monty’s Paradox: it is strictly the case where Monty Hall knows which is the good door, and always opens bad doors, then offers you to swap the one remaining door for yours. Anything else isn’t Let’s Make A Deal’s Monty Hall problem.
Yes, it does, because the problem statement specifies that he always opens a bad door. He can't do that unless he is using the information about what's behind the doors to choose which door to open.
To be fair the original problem statement is a bit ambiguous:
> Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
The "opens another door, say No. 3, which has a goat" could be misinterpreted as opening the goat door being an example rather than a requirement.
Even in the 70s, you couldn't tell a contestant on a show that they could win a prize if they could not actually win the prize. There is something worse than being needlessly pedantic - it's being needlessly pedantic and wrong.
You are correct. Anything else is not the problem for which Marilyn Vos Savant's solution is correct. But her original presentation of the problem in Parade magazine did not state whether or not Monty always opens a bad door. If he does not always open a bad door or does not use some method uncorrelated with whether or not the contestant has already selected the door with the car to determine whether or not he will open a bad door, Marilyn's analysis is wrong.
While this is true, this also doesn't make sense in the context of the Monty Hall game. If Monty opened up the winning door, what then? "Oh look, there's the new car. Guess we're done here."
Exactly. If Monty doesn't know what's behind the doors, then sometimes he'll reveal the winning door, and of course you'll choose to switch to that door.
The bizarre scenario some people try to propose is that Monty doesn't know what's behind the doors, but if he happens to reveal the winning door, then we...ignore that entire Universe, or...rewind the simulation and try again, until he "by chance" reveals a non-winning door.
No need to talk about ignoring Universes or rewinding simulations, that just sounds confusing to most people.
Enough to say that if Monty opens the winning door then he apologizes, closes it, and his helpers place the car to a new random place and they try again from the start.
Or say, ignorant Monty may open the winning door accidentally, but here I have a stack of DVDs of all lucky episodes where Monty got a goat. When binge watching these videos, would you expect to see switchers win at a higher rate?
Answer to both: it becomes 50-50 if Monty was ignorant, even if we post hoc filter out those cases where he got it right.
Bonus philosophical question: you don't know whether Monty knew the car's location. He may say he does, or he may say he doesn't but you don't trust him. Or maybe he says "phew, how lucky to be a goat, I actually had a short amnesia an bluffed it". How can such a murky thing whether a human knows something or not impact the very, financially real world, even when all outward appearances are equal: I saw with my eyes that Monty opened a goat door with his hands. Yes it might have been an accident/luck but what is the precise definition of luck anyway? How can it have an influence whether his brain contained this information, given that he ended up opening this goat door anyway. How can counterfactual Universes influence ours? I have my own answers but it's an interesting thing to ponder.
> No need to talk about ignoring Universes or rewinding simulations, that just sounds confusing to most people.
It should sound confusing. People who justify their claim that switching doors doesn’t change your odds in this way are proposing a very bizarre and confusing revision of the problem.
The original statement of the problem is a very straightforward description of a simple and realistic game show, that happens to have a solution that is counterintuitive to many people.
> Answer to both: it becomes 50-50 if Monty was ignorant, even if we post hoc filter out those cases where he got it right.
It's not 50/50 because he is ignorant, it's 50/50 because you've discarded 1/3 of the outcomes, and all the outcomes you've discarded were ones where the contestant initially picked a goat.
That's true, but I discarded it based on what I think Monty knew and counterfactual reasoning. The question is why "knowledge" or "luck" or "just accidentally" have an effect on what I should discard.
Another "more serious" related "paradox" is in scientific hypothesis testing. You can have different p values even if you observed the exact same experimental outcome but your stepping criterion was different. The only difference is the speculation of what you would have done if the experimental outcome had been different. [0]
These things are quite unexpected unless you've already worked through such examples.
If Monty knows what's behind the doors, we arrive at the scenario immediately where the door that's left has 2/3 chance of hiding the car, because the door you picked has 1/3 chance of hiding the car.
If Monty doesn't know what's behind the doors, and goes into a cycle of opening doors and shuffling the prizes if it was the car, or rewinding the universe, or whatever, at the end of that sequence, you will still hold on to the same door you initially picked that has 1/3 chance of hiding the car. And the door that you get the offer to switch to therefore has a 2/3 chance of hiding the car.
According to [1], Monty didn't always give the players the opportunity to switch. If they picked the wrong door, sometimes he would just end the game there.
"For example, he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door."
What made me convince friends is the following (using 3 cards, one red and two black):
- Let them pick one card of three
- Offer to switch their card for the other two (closed!) cards. Most people take this deal, in my experience everyone takes this deal.
- Then, show the "goat"/useless card of the two remaining cards
- Offer them (again) to switch their original choice of card for the remaining card.
If they accept to switch, let them explain why (this is when people get it). If they don't switch, tell them that one of the two cards they could have switched for must have been a goat, so I could always show them one goat between these two.
After this, most people (all my test subjects, anyway) understand that in the long run, they're best off switching to the "two card" deal. They now understand that one must be a goat but that the probability of the "car" being among the two-card offer is twice as likely.
I've had similar experiences with my friends. The concept is hard to grasp in the abstract, but if you actually play it out and keep track of how often they win and lose, people are generally willing to accept the correctness of the result, even if they don't grasp the proof behind it.
I think the fault is one of false temporal equivalency.
When a casual observer assesses the situation they see an equal set of unknowns behind each outcome, since that matches the set of conditions.
While the probability would have remained the same, it's the elimination of all (but one) remaining false choices that tips the scale. The weight that the selected door was the correct one remains at 1/3rd, but the probability that the only remaining door was the correct one is the inverse, 2/3rds.
Meanwhile in the 1,000,000 doors set, if instead one door was picked, one other door was opened, and the choice was between keeping the current door and picking one of the other 999,998 doors, the odds increase so infinitesimally that both ego and a need to know would bias me to staying the course.
My method of explaining it is that you can think of it as getting to pick 1 of 3 doors or 2 of the 3 doors... your original pick has a 1/3 chance... 2/3 of the time it will be one of the other two... if it was not the door you picked, it HAS to be the one that you didn't pick and wasn't opened... meaning the non-picked, non-opened door has a 2/3 chance.
Yes this is the exact explanation I give. Mentally pick 2 doors, but say out loud the other 1, when offered the switch you take it as it means you are back on your secretly pre-selected 2. Hence 2/3 chance by always switching.
I was debugging my monty-hall simulator to try and work out why my intuition was so wrong when I came across my comment "Iterate through doors, open first door without prize" when this point finally struck me.
With more than 3 doors, it is essentially a different problem. Imagine an evil host who only opens a door and offers to switch when the guest chose the door with the car, and not offer to switch during other episodes. Or, more evil, infrequently open a bad door and offer to switch, just to confuse the candidate. With 1,000,000 doors such an evil host will never open and offer the switch.
Alternatively, the way I think of it is as follows: An initial random guess has a 1 out of 3 chance of being correct. Being shown a goat doesn't change this probability. Staying with the first choice doesn't change this probability. Only making another choice, and choosing from the smaller pool of two doors increases the odds.
Or perhaps better stated in the Wikipedia article:
> "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
But being shown a goat can change the probability. If Monty picks randomly (which is not the original problem but is interesting to consider) then there is no benefit to switching.
Very true! It depends on the particular variation[1] of the game being played. Which is one difficulty of MH - the ambiguity. Probabilities change when the initial conditions are slightly different or when the problem is interpreted differently. I've never assumed Monty picks at random. But it's entirely possible someone else would make that assumption, in turn changing the optimal strategy.
As others have pointed out, that doesn't quite work because you have not specified that you know where the prize is and make sure to not open that door when opening your 999 998 doors.
You can sidestep this by playing a slightly different but equivalent game.
The player chooses a door. The host then gives the player the choice of sticking with that door, or instead taking all of the other doors.
Of course, that’s the definition of Monty’s Paradox, and exactly how ”Let’s Make A Deal” worked: Monty knows where the one good prize is, and always shows you only bad doors. Anything else is indeed a different story, and a different problem.
I'm a mathematician, but I've always wondered when this silver bullet is suggested: am I the only person for whom this exaggerated strategy offers no additional explanatory value? I know, and can prove mathematically, the right answer, but it's no more intuitive to me with 1000000 doors than with 3.
The missing part of the explanation (imo), at least the part that helps me explain it in my own brain, is said but not explicit. Imagine the actual mechanics of Monty opening the doors, one by one, and when he gets to that 223,205 door as stated, he skips it, and keeps skipping it. Why _that_ door? What makes it special? I don't think expanding the number of doors without that bit of intuition gives you much.
That's interesting - for me it was the first time seeing it explained that way, and it immediately made perfect sense.
I think the intuition behind is that, when you initially choose one door out of 1000000, you are almost sure not to have picked the right one, so you ignore the fact that you even chose one. Then, the host opens 999998 doors, but leaves one closed. That door is now really special!
Or a different way to look at it: The host gives you partial information about what's between the 999999 doors you didn't choose. Doesn't it seem really unlikely that your intial choice matters given that there are so many doors?
Also a mathematician; also think this is not a real silver bullet. However, I think for those who don't prove the result, the visual intuition of opening 999 998 doors is sensible.
The problem to me is more that when you go back down from 1M to 3 doors one can just as well say again "but now we only have three doors". Meaning that a proof of the original problem is the only satisfactory answer; the 1M comparison is maybe for people that don't like proofs and want "evidence".
You want to have an ugly argument with other developers, bring up the Monty Hall problem. People who are wrong will not let it go.
We had it beaten into us not to use truth tables to solve problems in college, because it doesn't scale. But the Monty Hall problem has a finite set of states you can easily write on a piece of paper. And then you'll see that it jumps from 2/6 successes to 3/6.
I tried to replicate that table under duress once (see initial comment) and I couldn't get it sorted. I should try it again and write it down this time.
Edit: Why do work when you can find someone else who already has:
Don’t have the debate; just break out a deck of cards and simulate 30ish rounds. Should be done in 10 minutes, or however small p needs to become for the winds to shift. Then, answer questions.
You damn kids and your Monte Carlo’s. When I was a kid we did everything deterministically and we were grateful! And we didn’t have no fancy tablets either. You wanted a mobile computer, you had to steal a cart from Larry. And Larry didn’t like people stealin’ his carts, no sir.
I was skeptical after hearing the correct answer, and just couldn't wrap my head around it. I'm no developer, but I wrote a simple Python program to simulate multiple plays of the game. It seems to be human nature to argue a problem like this even when one has the tools to solve it.
You have to be careful with the truth table method. It's easy to get it wrong, as your events might not have equal probability.
For example, we'll label the three doors P, L, R, for the prize door, the leftmost goat door and the rightmost goat door. This gives possibilities
1. You choose P, Monty chooses L - switching loses
2. You choose P, Monty chooses R - switching loses
3. You choose L, Monty chooses R - switching wins
4. You choose R, Monty chooses L - switching wins
This seems to indicate that switching gives you a 50/50 likelihood of winning. In fact, this is the correct truth table to use if Monty doesn't know which door has the prize. In that case, these are all equally likely.
Instead, if he does know where the prize is and won't open that door, this truth table is misleading because the cases where switching loses are less likely. If you choose L, he'll definitely choose R, but if you choose P, he has two options to split the probability.
Similarly, if you model Monty as not knowing what's behind the doors, then the truth table you're thinking of doesn't work. Either way, the truth table on its own isn't convincing.
The thing I find more clear is using Bayes' rule, as it shows exactly how the model of Monty's behavior affects the probability that switching is beneficial.
Let H be the event that your door has a prize and G be the event that Monty shows you a goat door. Then P(H|G) = P(G|H) * P(H) / P(G). Obviously, P(H) = 1/3 and (assuming Monty must pick a door other than yours) P(G|H) = 1, so it just comes down to what P(G) is.
If he always shows a goat, then P(G) 1 and you should switch. If he doesn't know what's behind the doors, it's 2/3 and it doesn't matter whether you switch. If he knows but prefers to spoil the prize, it's 1/3 and you definitely have the prize door.
I've always preferred to draw it out as a tree of what your your possible choices are at each stage, I find easier to get right and I find that people understand it more intuitively.
┌─────────────┼─────────────┐
G G C
┌┴┐ ┌┴┐ ┌┴┐
Stay│ │Switch Stay│ │Switch Stay│ │Switch
G C G C C G
That has similar drawbacks of hiding the subtlety behind how Monty is being modeled. The tree you've drawn will tend to lead people to the wrong answer if you ask whether switching improves your odds when Monty doesn't know which door has the car.
The problem with this logic is that it applies regardless of how we model Monty, and not every model of Monty has you doing better than expected. This reasoning is just as wrong as expecting that your chances are 50/50 in the standard form of the problem.
I think my statistics prof would have said that they are dependent variables.
That was the part that always tripped me up, and I see it trip up plenty of others so I'm in good company, but it's definitely part of my mathematical heartbreak. I loved math in high school and a series of unfortunate events and terrible curricula ruined calculus for me. All I have to show for it is that I have more sympathy for people who 'hate' math (even though I think it's fear sublimated into frustration), and a deep and abiding hatred for Stephen fucking Wolfram.
Those italics on Mathematica became my proxy nemesis and I figured out how to break the autocorrect by editing the word just so.
That's how I prove it with playing cards. Except I use a handful of cards, and not a million of them.
The trick is in the wording. With 3 you are literally saying "let me remove all the cards except for a random one, the one you chose, and the one I chose, of which, one is the correct card to choose. Do you want to change your choice?" So it plays out as expected with 3, and the obviousness of the solution when done with 52 cards.
The brilliance of the problem is that you're led to think that the odds have changed.
I think there's a psychological component too. So many probability questions are about random events so you (very naturally) take a position that the revealed door is randomly chosen. It is NOT - you always see one of the goats.
Yeah exactly. The Monty Hall problem is about the probability of being correct with your first pick. It doesn’t really have anything to do with "changing probabilities".
I struggled with this because I didn't see the first choice as relevant to the final outcome. It is easy to see the game as restarted somehow when the 50/50 choice is presented.
Yep, same for me. Even though I understand (and believe!) the explanation of why switching is advantageous, a little voice still says "there's a 50% chance it's in one of the two remaining doors, because there are two doors and one prize!".
The swap doesn’t change your chances, it’s the host opening one of two other doors that changes your chances. Because the host can open one of two other doors, there are two configurations in which swapping will result in a win but one where not swapping will win.
The host opening the door before the swap is just for show. Ignore that part. The swap is really just picking the two other doors instead of the one you picked first.
The swap is the same as picking all the doors you didn't pick at first. The probability of winning when swapping is the complement of the probability of you winning on your initial pick. There is no third outcome.
This is the famous Feynman method for gaining intuition of many physical problems: just set one of the variables to an absurdly huge number and see what happens.
He illustrated it with the problem of what happens to the level of water on a pool when you throw overboard a stone from a floating boat. Does the water level rise or drop? Well, imagine that your small stone is made of neutron star material...
Forgive me if I'm missing something obvious, I have a newborn and am underslept.
I assume neutron star material is incredibly dense and so the rock would be unbelievably heavy. To the point that the mass would actually likely sink the boat, and then we can't throw the rock out.
OK but let's assume that the boat is somehow massive enough to support the weight of the rock. Now we've changed the scale of the other variables of the experiment in proportion to the rock/neutron star, and we're back to ground zero, are we not?
A 1-ton neutron-star stone will be the size of a dust speck. When it is aboard, only its mass matters, and it will push the boat downwards and displace 1 ton of water. When it is at the bottom of the pool, only its volume matters, and it will displace a negligible amount of water.
That didn't work for me. Because if you had been choosing randomly (i.e., did not know which door had the prize), my understand is that the last two doors are still 50/50?
It's only when the person opening the door knows where the prize is, that I should always switch?
"Your brain will compute monty hall problem in fast casual causal mode AND/OR compute slow mode formally for the exact answer. Probability predictions that accurately map reduce to efficient intuitive representations in our brain dictate direction in our evolutionary biology."
"[since] we don't solve the monty hall problem effectively with our intuition, an evolving life form would be poor at resource collection from such problem sets without knowledge of axiom representation and formal probability."
"Sire. Shall we hunt in the valley, in the forest, or in the mountains?"
We shall hunt in the forest.
"Sire! Our scouts have just returned from the mountain. They report that the herds are not there at the present time! Shall we try the valley, or continue on to the forest?"
A) Stick to the plan, and hunt in the forest
B) Hunt in the valley instead
The Monty Hall paradox hinges on the fact that the host knows in advance where the car is. If your scouts don't know where the herds are, but just picked an area at random to recon, then it is a different problem, and switching makes no difference.
And I think you stumbled upon the crux of the problem. I think we have a hard time grasping the fact that Monty not just picks one of the other two doors, but also specifically guarantees that it won't be the winning door.
Without that guarantee, it makes no difference.
Additionally, there are not many situations in real life that would replicate the conditions of that guarantee. So our "intuition" is simply geared towards situations where an actor can't read our thoughts (door pre-selection) before we act them out, and in such a scenario, pre-selecting/switching selection of doors doesn't matter.
But it doesn’t depend on Monty knowing ahead of time. Monty can learn after you pick, but before he reveals (to be sure he doesn’t reveal a car).
The scouts can depart after you’ve chosen the valley, find nothing in the mountain, give that information to you, and you should still switch, even if the scouts didn’t confirm where they are.
Monty(or the scouts), knowing if your original pick was right has no impact. The critical piece is that they are forced to reveal to you a bad choice. 2/3 times it’s the only bad choice to reveal because you’ve picked the other.
If the herds might have been in the valley and don't happen to be, you learn something. Your odds rise from 1/3 to 1/2. If the scouts know where the herds are and specifically checked the valley because the herds weren't there, you learn more - your odds rise to 2/3. But not as much as if they just told you where the herds are, so we're back to it being an unlikely scenario.
Knowing that Monty could not have revealed the car (vs happened not to) has an impact.
If you are objecting to how I described that, in terms of who knows what when, then you may be right and I probably could have been more precise.
But if you are disagreeing that scouts sent to investigate an area and return an answer (which this time happens to be negative) differs materially from the original question then you are wrong.
> Knowing that Monty could not have revealed the car (vs happened not to) has an impact.
Nope, it does not. The fact that he revealed a dud is what matters. The notion of him maybe revealing a car is senseless anyway because you would just then pick the car or the game would be over.
> Nope, it does not. The fact that he revealed a dud is what matters.
That's simply wrong. This has been discussed countless times on this forum. The most effective way to proceed is for you to spend 5 minutes writing the simulation that you think will prove you right. When I was on the other side of this, that experience is what convinced me.
> The notion of him maybe revealing a car is senseless anyway because you would just then pick the car or the game would be over.
That's not a very strong objection - game shows are weird sometimes and could often apparently be easily improved.
But more importantly, it's an objection to the wrong thing. My point was that the presented "find the herds" problem differed from the game show. "That would have been a bullshit game show" is just confused.
>> But it doesn’t depend on Monty knowing ahead of time. Monty can learn after you pick, but before he reveals (to be sure he doesn’t reveal a car).
Right, of course. I just meant he knows ahead of opening his door, it doesn't matter if he knows when you pick your door.
The scouts did not know that the herds are not in the mountain when they chose to scout the mountains.
Monty, on the other hand, does know that the car is not behind door C, when he chooses to open door C.
The difference is that Monty will never pick the door where the car is, whereas your scouts will sometime recon a region and find the herds there.
(Unless you assume that your scouts know where the herds are, and are purposefully not telling you and scouting empty areas, which would make them terrible scouts. We can expect this sort of crafty behaviour from a game show host, sure, but not from your own scouts.)
This results in different odds. With the scouts, switching makes no difference. You can run simulations or draw a full probability tree to convince yourself of it.
Something that really adds to the confusion, is that "in a not mathematically perfect" world, the observer doesn't know whether the host will always open a door. When the candidate chooses the winning door, the host could always open another door, and when the candidate chooses a loosing door, only open another door with a reduced probability. Life teaches people to always be suspicious when free money is promised.
The "Monty from Hell" scenario (Monty only offers the choice if you picked the car, and otherwise does nothing), which the original statement of the problem allows, is IMO the most likely in real life because it would make for the most entertaining television.
You have real world knowledge of what makes entertaining TV, and perhaps how this particular show actually operated. And you are correct. In the real world, Monty Hall always revealed the worst of the two prizes that the contestant did not select. In the real world, you can use common sense and assume that Monty Hall will use that algorithm.
But in a mathematical probability problem, you can't do that. For example, if there's a slightly smaller object and a slightly larger object concealed in two different sized boxes, common sense says that a person would put the larger object in the larger box. But if it's a abstract mathematical probability problem, and you have no information on the algorithm used, then you have to assume the concealer flipped a coin when they decided which box to use.
If I'm the one setting up the big-small game, and I use the random algorithm, the smaller object is going to be in the bigger box about half the time.
If my mom is the one setting up the game, the smaller object is going to be in the smaller box every time.
In abstract mathematical probability problems, if you don't know what algorithm was used, you aren't allowed to bring human behavior in, unless it's stated in the problem.
The problem depends on human behavior, so as originally stated it is unsolvable. You need to clarify that Monty always opens a door, and always reveals a goat.
> But in a mathematical probability problem, you can't do that. For example, if there's a slightly smaller object and a slightly larger object concealed in two different sized boxes, common sense says that a person would put the larger object in the larger box. But if it's a abstract mathematical probability problem, and you have no information on the algorithm used, then you have to assume the concealer flipped a coin when they decided which box to use.
I think that this misrepresents mathematical reasoning.
For solving a problem in a textbook (which a mathematician does intensively at the beginning of his or her career, and seldom thereafter), you must assume neither—that the larger object goes in the larger box, nor that the concealer flipped a coin; you must use only the information contained in the problem. (From this point of view, many problems in most probability textbooks are ill posed, because they force you to make some assumption while solving them; and it then becomes a mind-reading exercise of whether you can successfully make the same assumptions as the poser, rather than a mathematical exercise.)
For a mathematician who actually wishes to use theoretical probability to do something useful, then you must again make assumptions that allow you to translate the messy real world into an idealised mathematical object; but to say that you must, or must not, assume any particular thing is otiose—different assumptions will lead to different results, and the only guide to whether you made the 'right' assumptions is whether or not the idealised mathematical results you get match the real world (to within whatever bounds of error are tolerable).
This is only a paradox due to very poor definition.
"You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." Where does it state that the host always opens a door with a goat? From the statement it looks like the goat was accidental. No wonder Paul Erdős got confused by this too. The translation of the paradox to his language could have been even more confusing.
Yes. If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching.
If the problem does not state what the host's algorithm is, and you have no knowledge of the host's algorithm (example: you have never seen the TV show before), then the logical action is to assume the host chooses a door to reveal randomly.
I don't think you are correct. The host does not give you any useful information by opening a goat door. Opening a goat door only serves to confuse you into not realizing that the switch-choice is actually between one of the doors and a grouping of the remaining two doors. Opening a goat door in the grouping of the other two only changes the groups appearance, making it look like the single door that you picked. It does not change the functional state of that group in terms of its probability of containing a car.
It doesn't change the state of the group of the two remaining doors, correct (ie. the probability remains at 2/3). But you can't choose two doors, you can only choose one. The host is effectively telling you which door to switch to--the one he didn't pick.
The only scenario in which the host doesn't give you any useful information by opening a door is when the host chooses the door at random. Assuming the host didn't choose the door with the prize, your odds are 50:50.
Of course all of this hinges on you being aware of the host's decision making paradigm, and that you want a car more than you want a goat.
I'm not sure what your point is. Assuming the host always opens a goat door, you can simply restate the problem as being allowed to pick two doors at the same time. You are literally choosing two doors, because if the car was behind the first he would have opened the second, and if the car was behind the second he would have opened the first. It would have looked like you are switching to one other door but you can simply look at it as switching to the other two doors, where the elimination of the dud door can be done before or after.
You also have to trust the host to behave fairly. For instance, a malicious host might decide to open a door with a goat only when your first choice was correct. If the rules of the game are such that the host must always open a door with a goat regardless of whether your first guess was right or wrong, then it makes sense to switch doors. If not, it might be a trap; probably the best strategy is just to flip a coin.
I love discussing the Monty Hall problem with people who haven't been exposed to it before. I always make it clear that the host knows what's behind each door and will open a door with a goat. People still, most of the time, get the answer wrong.
The first time I read about this problem, it was formulated in a way which made it clear that the host opens a door with a goat behind it. If you learn about it like this, you literally cannot believe how anyone (even Paul Erdős) could believe that switching is not better. After all, you choose a door with a goat behind it with probability 2/3 on your first try, and if you do this, than switching will always win you the car. It is so trivial that the only "paradox" here is that it is classified as a paradox.
E: the explanation on Wikipedia is unnecessarily elaborate and hard to follow:
"The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 that was not available at the beginning of the game, when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. "
There are plenty of people who keep getting tripped up by this problem even when it's correctly formulated. Statistics and probabilities are sometimes incredibly unintuitive and hard for a lot of people.
I only took a single class of statistics at university, but my biggest take-away from that class is that you absolutely cannot trust your gut, because the human bias is crazy strong.
Humans naturally over-connect. We see causation where only correlation is present. Gamblers believe that something random that has happened frequently (like roulette landing on a specific number) is more likely to happen again. Educated people learn about dependent and independent probability, which corrects this impulse. But sometimes it over-corrects.
The Monty Hall problem is hard because people see can't explain the dependent probability link clearly and conclude that the probabilities are independent and their brain is acting up - first guess 1 in 3, second guess 1 in 2. After all, the game host is always capable of opening a door with a goat, and the player doesn't know anything. Put differently, what did the player learn between the first and the second guess that should make the player now believe one door is more likely than the other of the remaining two?
It also takes advantage of our natural thinking. The problem is artificial, because, outside of a game show, the host could also just not offer to switch. Emotional then, brains consider the information that the host made the second offer even if it's not logically part of the problem.
I wouldn't say that's the core of the paradox. The thing is that people don't have a good understanding what probability is. We think that when we choose a gate at the beginning, with a chance of winning being 1/3 it means that the universe "rolls" a dice and decides whether we win or not. It is then very confusing to realize that someone else, just by revealing the content of one of the other gates - after our decision has been made - can somehow make the universe roll a different dice with 2/3 odds.
The paradox disappears when you think about probability as a tool for reasoning from incomplete information, not as anything to do with "physical" property of the system under investigation. It then makes perfect sense that after receiving new information from the host we should reassign our probabilities.
As a math professor recalling reactions back in 1990, there were various forms of "Marilyn vos Savant" bias in play. A good cover for gender bias was the absurdity of the world's highest IQ writing newspaper columns. There are plenty of other examples of narrow spikes in particular forms of intelligence having a crippling affect, while the most creative people I know work to change the world with less spectacular but more broad-spectrum cognitive advantages. Still, it's a bit of a leap to conclude she was actually wrong about this problem.
The key was... did Marilyn vos Savant state in her version of the problem that the host always uses an algorithm to open a door with a goat? (The alternative was that the host randomly picked a door, and this one time it happened to be a goat.) I read some of her articles on this topic, and my opinion was that in some of her articles, she did not state what the host's algorithm was.
So, yes, even a person with the world's highest IQ can make a mistake.
Her column was written for an audience that was familiar with the show. Monty Hall had, at that point, been opening doors and revealing goats-but-not-cars to contestants five days a week for more than twenty years.
But as Monty Hall himself notes, the game show rules did not actually require him to open a door.
> Hall clarified that as a game show host he did not have to follow the rules of the puzzle in the vos Savant column and did not always have to allow a person the opportunity to switch (e.g., he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door).
If the host uses an adversarial strategy, such as only offers a switch when the remaining door is a goat, switching is not necessarily wise.
I think most people get it that prize is twice as likely to be behind one of the two other doors rather than behind the door you pick.
So one time out of three you'll get it right while two times out of three the presenter indirectly indicates the winning door.
Edit. Be nice if my downvoter would explain what's wrong with the above.
Here is the scam version to make money out of math people who know the Monty-Hall problem.
You look at the 3 cards then put them on the table and propose a fair triple or nothing, if they can find the red queen.
You make them point toward the card they chose.
If they point a wrong card you go toward the card and reveal -> you won.
If they have pointed the right card, you return one of the card they didn't pick and ask them if they are sure or want to change. Because they think they are facing a Monty Hall problem they will switch.
-> you won again. :)
This is not cheating, you don't have to hide the fact that you know where the good card is. It is entertainment teaching a valuable lesson.
It is, as I said before, a fair game so it has zero expectation value for both players, better odds than a casino for the player and there is no reason for a gambler to refuse this bet.
In expectation you won't lose money, but people who think they know better will happily give you their money by making a bad move while being convinced they are getting an edge.
Most math-type people don't know the hypothesis of the Monty-Hall Problem and can't state it properly, but most have seen it played on TV-shows and know they must switch.
In the heat of the moment, especially if you have played a few follow the red queen games before, the brain is still in fast mode, and will jump to using heuristics. Mind trick.
It's simple: choosing again switches from a losing outcome to a winning outcome and vice versa, thus the probability of winning by choosing again is 2/3.
No, it becomes less obvious, since the host's help has less influence the more doors there are.
EDIT: My reply is based on a misunderstanding. I was assuming that the host continues to open one door, whereas parent intended that the host would open all-but-one doors.
Clearly you're assuming that with more than 3 doors, the host will open all but one, and they're assuming that the host will continue to open just one door.
Instead of asking yourself "should I switch?", ask yourself "At the time of making my first guess, would I have bet that it got the car (1/3) or a goat (2/3)?"
Obviously the answer is that your first guess is more likely to have been a goat. So if you had to bet, you'd have bet on the car being in one of the ones you didn't choose.
After the host reveals a goat, it's no longer "one of the ones you didn't choose", but rather "the one you didn't choose" (since the host has taken the other out of consideration). So the choice is easy now (as is the probability calculation).
Since my brain could not accept this long ago, I wrote a tiny script to see what the result would be if this was to be emulated. Works out perfectly! (Might need to update script for python 3). It's weird. Still feels so unintuitive. Which tells a lot about intuition I guess :)
276 comments
[ 5.7 ms ] story [ 238 ms ] threadthis tidbit is pretty wild
What made it very clear to me is if you just increase the number of incorrect doors that you open by a lot. (Making the assumption that the host knows where the prize is.) In that case, it's obvious that you're improving the odds by showing where the prize isn't.
Basically, Wikipedia does not mandate a citation style, it just has to be consistent throughout an article. And once set, another editor can't change it simply for personal preference. Presumably the original author, or whichever editor first added citations, was more familiar with this style, and so it has stuck.
(note: when you do this in real life, you stop talking now. let them explain it.)
When you have N>2 doors, the host will with 100% certainty open N-2 doors with goats behind them. It doesn't matter whether N is 3 or 1,000,000. After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2.
The key phrase is "host's hidden knowledge". Once the host starts opening the wrong doors, the fundamental nature of the problem is changed from being an independent probability problem to a dependent probability one.
No, your numbers are wrong.
In your initial choice, the probability of picking the winning door is 1/N.
When Monty opens doors so that there is a single door remaining, the probability of that door containing the prize is the complement of the probability that you picked the winning door, i.e. 1-(1/N).
So switching doors increases your probability from 1/3 to 2/3 if N=3, not to 1/2.
If N=1000, switching doors increases your probability from 1/1000 to 999/1000, and that is why increasing the number of doors makes the correct solution more intuitive for some people.
That's why this approach to teaching it has never made sense to me -- it's not clear at all why opening 998 doors should be analogous to opening 1 door, which it's much more obvious that opening 1 door is analogous to 1 door.
I didn't believe it until I simulated it myself.
Monte has to show you what's behind a different door than the one you initially picked. The game doesn't make sense otherwise.
Corrected simulation gives 50/50 when Monty picks at random among 3 doors.
switchWins=33044 switchLoses=33581
#include <stdio.h> #include <stdlib.h> #include <time.h> #include <string.h>
#define NUM_RUNS 100000 #define NUM_DOORS 3 int main(int argc, const char* argv[]) { srand(time(NULL)); char doors[NUM_DOORS]; int switchLoses = 0; int switchWins = 0; for(int i = 0; i < NUM_RUNS; i++) { memset(&doors, 0, sizeof(doors)); doors[rand() % NUM_DOORS] = 1; int myPick = rand() % NUM_DOORS; int montysPick = rand() % (NUM_DOORS - 1); if(montysPick >= myPick) montysPick++; if(doors[montysPick]) continue; if(doors[myPick]) switchLoses++; else switchWins++; } printf("switchWins=%d switchLoses=%d\n", switchWins, switchLoses); return 0; }
EDITED TO ADD:
Just for further info:
3 doors: switchWins=33111 switchLoses=33470
4 doors: switchWins=50182 switchLoses=24698
8 doors: switchWins=74876 switchLoses=12687
100000 doors: switchWins=99998 switchLoses=2
"Picks a random door out of all closed doors, but that door is never the winner" is a contradiction. It doesn't give you a different chance of winning because it's not a coherent scenario in the first place.
This is a game where you start off with a certain layout, and then proceed forward. If you want conditions, they have to be conditions that you can apply before the game starts. You can't retroactively remove a significant chunk of samples.
If you added an actual outcome to him picking the winner, you could make a valid filter where the answer actually is 50:50. Perhaps they restart the game. Perhaps they never air the episode. And if you calculate only for finished/aired games then it's very clear then that you're solving a different math problem. You can't apply that answer to the original problem.
If he was opening doors randomly then you're left with a one-in-two chance. But if he was giving you information by opening doors he knew were bad, you're left with two possibilities: the door you started with just happened to be the right door (a tiny chance), or (much more likely) the door you started with was not the right door - in which case Monty was forced to open every other door so as to avoid showing you the car, and he has eliminated all the non-car options.
i initially thought this too, but in both cases you're gonna see 2/3 doors. the thing that makes the difference is when he has the car (2/3 of the time), he has to tell you where it is. so the choice becomes between your unknown door and his 2/3 selectively chosen car door.
No. Or, well, it depends on what happens if Monty opens a door with the car. If I lose if he does that, then switching doors doesn't change the probability.
It's easy to prove. Let's say that I always pick door #1, and Monty always picks door #2. Then we get the following probabilities:
car-goat-goat: I stay, I win. I switch, I lose.
goat-car-goat: I stay, I lose. I switch, I lose.
goat-goat-car: I stay, I lose. I switch, I win.
Staying: 1/3 chance of winning. Switching: 1/3 chance of winning.
Imagine you're taping episodes of this game show for broadcast. The contestant picks a door, and then Monty picks one of the two remaining doors at random. Monty has 1/3 chance of accidentally picking the car, and if he does, we're scrapping the entire episode.
So to enforce your constraints, we have to scrap 1/3 of all episodes, we're only ever broadcasting the 2/3 of all episodes taped where Monty, by chance, didn't pick the door with the car behind it.
Your numbers are correct, but that interpretation of the problem doesn't make a lot of sense if you think it through.
As for the game show, nothing says that he needed to give this choice every time, it was just some random thing he did to make it more interesting not an important part of the show.
> The point is that given that the door Monty opened had a goat behind it your probability of winning would then be 1/2.
If the episode just ends when Monty picks the car, then I lose if he picks the car. I don't just vanish into thin air such that the universe erases my run. I was on the show, and I lost. And that means my odds of winning are still just 1/3, not 1/2 as you claimed. Yes, the game has changed such that it doesn't matter if I switch doors or not with this addition, but the underlying odds don't change.
The original said: "the host, who knows what's behind the doors, opens another door, which has a goat."
Your formulation of the problem changes it from the original Monty Hall problem to something more similar to the Tuesday Boy problem, where the trick is that you have pre-selected a bunch of outcomes and discarded others, changing the probabilities.
And going back to the comments above this, it's not the case that Monty choosing a door by random is the thing that matters, it's the thing that in this version he chose a door by random and you discarded all the outcomes where Monty chose the car.
It's not the case that people are confused by the problem because they interpret the problem like you did. People understand that a goat door is always eliminated as part of the game, and they still don't agree that switching is better.
The whole premise of the problem is that you are already on the show looking at an open door with a goat behind it.
One little letter and the whole thing is a completely different problem.
Upthread I see that you are correctly writing about the different ways you can model Monty's behaviour, and how each different way changes the nature of the problem, the outcomes, and the odds of switching vs. staying.
So I don't understand what you are arguing against right now?
I'm saying that "Monty reveals a goat" is what happens in the correct interpretation of the problem, i.e. Monty knows where the car is, and always chooses a door with a goat in every single instance of game play. In this case switching gives you 2/3 odds of winning the car.
And I'm saying that "Monty revealed a goat" is what happens in the incorrect interpretation of the problem, i.e. Monty does not know where the car is, and in this single instance of game play that you happen to be participating in, he happened to pick a door with a goat. And if you have gotten to this point, switching is 50/50. Just like you say.
Do you understand my way of describing these two different models of Monty?
I'm still not sure what the difference between these models has to do with my point about goat-car-goat. That can't happen in either model when we say that I'm picking door 1 and Monty is picking door 2.
But I'm saying that since that is a wrong interpretation of the original problem, any other wrong interpretation of the problem is equally valid. For example, if Monty picks the car, it's game over and the contestant never gets to choose. Or, if Monty picks the car, the show goes "oops", closes the door, shuffles the prizes, and lets Monty pick again until he picks a goat.
There's not enough information in "Monty picks a door at random" to infer a single interpretation of the problem.
"Monty picks a door at random AND he picked a goat". That's enough information to get down to your universe where goat-car-goat could never happen.
Nitpicky semantics, sure.
But to roll back to the original, original question: I don't think people get the Monty Hall problem wrong because they're interpreting it as "Monty picks a door at random and he picked a goat". They interpret the problem correctly, i.e. "Monty knows where the car is and always picks a goat", and they still get it wrong.
I'm not saying that goat-car-goat could never happen. I'm saying that when you're evaluating what to do after Monty Hall has revealed a goat, the one piece of information you know is that goat-car-goat didn't happen.
I guess you're analyzing from the point of view of someone who's trying to figure out their odds of winning with different strategies before playing.
I agree with you that most people don't get the problem wrong because of interpretation. I'd go even further and say that many people who think they understand it don't get it right because of the interpretation.
Most arguments I see for the correct answer don't use the assumption that Monty knows which door has the goat, and therefore work equally well under a model of Monty picking the door at random. They're just picking different ways to model the problem and assuming a uniform distribution. You get the answer right if you picked a model where the distribution is uniform under the assumptions in the problem.
The real issue seems to be much simpler. Probability is not intuitive. Most people don't have the tools to figure out problems like this reliably, and those that do tend not to apply their tools when they think they see a simple obvious answer.
No, if Monty picks a door by random, but the show shuffles the doors until he gets it "right" and picks a goat, that is exactly the same scenario as if Monty knows where the goats are and picks one straight away, which means my odds of winning when switching is 2/3.
So that is a scenario where "Monty picks a door by random" doesn't lead to the odds changing to 50/50.
If instead you mean that they randomize all three doors and start over, the probability is 1/2 because half the time that the player has a goat you start over and get to pick again, which perfectly cancels out the double probability of having a goat.
This can be analyzed pretty easily using the three possibilities you outlined earlier. Suppose the player gets door 1 and Monty always picks door 2. Each time they shuffle, you get one of the following with equal probability:
A. car-goat-goat
B. goat-car-goat
C. goat-goat-car
A means the player wins, B means we re-shuffle, C means the player loses. It's pretty easy to see that A and B are equally likely outcomes, so they become 50/50 odds after all the re-shuffling is done.
He knows exactly where the prize is. He just had a good poker face and head game. Which is why he hosted the show for 15+ years.
> Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
The "opens another door, say No. 3, which has a goat" could be misinterpreted as opening the goat door being an example rather than a requirement.
The bizarre scenario some people try to propose is that Monty doesn't know what's behind the doors, but if he happens to reveal the winning door, then we...ignore that entire Universe, or...rewind the simulation and try again, until he "by chance" reveals a non-winning door.
Enough to say that if Monty opens the winning door then he apologizes, closes it, and his helpers place the car to a new random place and they try again from the start.
Or say, ignorant Monty may open the winning door accidentally, but here I have a stack of DVDs of all lucky episodes where Monty got a goat. When binge watching these videos, would you expect to see switchers win at a higher rate?
Answer to both: it becomes 50-50 if Monty was ignorant, even if we post hoc filter out those cases where he got it right.
Bonus philosophical question: you don't know whether Monty knew the car's location. He may say he does, or he may say he doesn't but you don't trust him. Or maybe he says "phew, how lucky to be a goat, I actually had a short amnesia an bluffed it". How can such a murky thing whether a human knows something or not impact the very, financially real world, even when all outward appearances are equal: I saw with my eyes that Monty opened a goat door with his hands. Yes it might have been an accident/luck but what is the precise definition of luck anyway? How can it have an influence whether his brain contained this information, given that he ended up opening this goat door anyway. How can counterfactual Universes influence ours? I have my own answers but it's an interesting thing to ponder.
It should sound confusing. People who justify their claim that switching doors doesn’t change your odds in this way are proposing a very bizarre and confusing revision of the problem.
The original statement of the problem is a very straightforward description of a simple and realistic game show, that happens to have a solution that is counterintuitive to many people.
It's not 50/50 because he is ignorant, it's 50/50 because you've discarded 1/3 of the outcomes, and all the outcomes you've discarded were ones where the contestant initially picked a goat.
Another "more serious" related "paradox" is in scientific hypothesis testing. You can have different p values even if you observed the exact same experimental outcome but your stepping criterion was different. The only difference is the speculation of what you would have done if the experimental outcome had been different. [0]
These things are quite unexpected unless you've already worked through such examples.
[0] https://en.m.wikipedia.org/wiki/Likelihood_principle
If Monty knows what's behind the doors, we arrive at the scenario immediately where the door that's left has 2/3 chance of hiding the car, because the door you picked has 1/3 chance of hiding the car.
If Monty doesn't know what's behind the doors, and goes into a cycle of opening doors and shuffling the prizes if it was the car, or rewinding the universe, or whatever, at the end of that sequence, you will still hold on to the same door you initially picked that has 1/3 chance of hiding the car. And the door that you get the offer to switch to therefore has a 2/3 chance of hiding the car.
"For example, he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door."
[1] https://en.wikipedia.org/wiki/Monty_Hall#Monty_Hall_problem
- Let them pick one card of three
- Offer to switch their card for the other two (closed!) cards. Most people take this deal, in my experience everyone takes this deal.
- Then, show the "goat"/useless card of the two remaining cards
- Offer them (again) to switch their original choice of card for the remaining card.
If they accept to switch, let them explain why (this is when people get it). If they don't switch, tell them that one of the two cards they could have switched for must have been a goat, so I could always show them one goat between these two.
After this, most people (all my test subjects, anyway) understand that in the long run, they're best off switching to the "two card" deal. They now understand that one must be a goat but that the probability of the "car" being among the two-card offer is twice as likely.
N<10 so YMMV
When a casual observer assesses the situation they see an equal set of unknowns behind each outcome, since that matches the set of conditions.
While the probability would have remained the same, it's the elimination of all (but one) remaining false choices that tips the scale. The weight that the selected door was the correct one remains at 1/3rd, but the probability that the only remaining door was the correct one is the inverse, 2/3rds.
Meanwhile in the 1,000,000 doors set, if instead one door was picked, one other door was opened, and the choice was between keeping the current door and picking one of the other 999,998 doors, the odds increase so infinitesimally that both ego and a need to know would bias me to staying the course.
I was debugging my monty-hall simulator to try and work out why my intuition was so wrong when I came across my comment "Iterate through doors, open first door without prize" when this point finally struck me.
https://en.m.wikipedia.org/wiki/The_Drunkard's_Walk
Or perhaps better stated in the Wikipedia article:
> "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
a) Staying wins (Monty opens to a goat) b) Swap wins (Monty opens to a goat) c) First pick lost, no chance to swap (Monty opens to the car)
each of equal probability.
[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Variants
You can sidestep this by playing a slightly different but equivalent game.
The player chooses a door. The host then gives the player the choice of sticking with that door, or instead taking all of the other doors.
I think the intuition behind is that, when you initially choose one door out of 1000000, you are almost sure not to have picked the right one, so you ignore the fact that you even chose one. Then, the host opens 999998 doors, but leaves one closed. That door is now really special!
Or a different way to look at it: The host gives you partial information about what's between the 999999 doors you didn't choose. Doesn't it seem really unlikely that your intial choice matters given that there are so many doors?
This explanation offered above makes sense to my "gut" however: https://news.ycombinator.com/item?id=21349295
The problem to me is more that when you go back down from 1M to 3 doors one can just as well say again "but now we only have three doors". Meaning that a proof of the original problem is the only satisfactory answer; the 1M comparison is maybe for people that don't like proofs and want "evidence".
We had it beaten into us not to use truth tables to solve problems in college, because it doesn't scale. But the Monty Hall problem has a finite set of states you can easily write on a piece of paper. And then you'll see that it jumps from 2/6 successes to 3/6.
I tried to replicate that table under duress once (see initial comment) and I couldn't get it sorted. I should try it again and write it down this time.
Edit: Why do work when you can find someone else who already has:
https://www.statisticshowto.datasciencecentral.com/probabili...
and particular to my statement, this table:
https://www.statisticshowto.datasciencecentral.com/wp-conten...
When switching the number of lose scenarios is lower than a naive expectation of outcomes.
Wait, really? When did you go to college?
In mid 2000s, we used truth tables every day studying boolean algebra, logic gates, circuits, etc.
Statistics? (and this is a statistics problem) no.
It jumps to 4/6! Even better.
Please stop spreading this misinformation. Leave the math up to the experts.
People argue about whether the odds go up at all, not about how much they go up.
I couldn't get my mind to accept the explanation. So I coded it and ran the simulation 100,000 times.
I got about 50,000 instances where switching won the prize. Biases confirmed.
... until I found the bug in my code where Monty was allowed to open the door containing the car as the one showed to the player.
Fixed the bug, and switching won the prize in about 66,666 instances.
For example, we'll label the three doors P, L, R, for the prize door, the leftmost goat door and the rightmost goat door. This gives possibilities
1. You choose P, Monty chooses L - switching loses
2. You choose P, Monty chooses R - switching loses
3. You choose L, Monty chooses R - switching wins
4. You choose R, Monty chooses L - switching wins
This seems to indicate that switching gives you a 50/50 likelihood of winning. In fact, this is the correct truth table to use if Monty doesn't know which door has the prize. In that case, these are all equally likely.
Instead, if he does know where the prize is and won't open that door, this truth table is misleading because the cases where switching loses are less likely. If you choose L, he'll definitely choose R, but if you choose P, he has two options to split the probability.
Similarly, if you model Monty as not knowing what's behind the doors, then the truth table you're thinking of doesn't work. Either way, the truth table on its own isn't convincing.
The thing I find more clear is using Bayes' rule, as it shows exactly how the model of Monty's behavior affects the probability that switching is beneficial.
Let H be the event that your door has a prize and G be the event that Monty shows you a goat door. Then P(H|G) = P(G|H) * P(H) / P(G). Obviously, P(H) = 1/3 and (assuming Monty must pick a door other than yours) P(G|H) = 1, so it just comes down to what P(G) is.
If he always shows a goat, then P(G) 1 and you should switch. If he doesn't know what's behind the doors, it's 2/3 and it doesn't matter whether you switch. If he knows but prefers to spoil the prize, it's 1/3 and you definitely have the prize door.
That was the part that always tripped me up, and I see it trip up plenty of others so I'm in good company, but it's definitely part of my mathematical heartbreak. I loved math in high school and a series of unfortunate events and terrible curricula ruined calculus for me. All I have to show for it is that I have more sympathy for people who 'hate' math (even though I think it's fear sublimated into frustration), and a deep and abiding hatred for Stephen fucking Wolfram.
Those italics on Mathematica became my proxy nemesis and I figured out how to break the autocorrect by editing the word just so.
Which is funny because that's a group that's very capable of simulating it for themselves.
The trick is in the wording. With 3 you are literally saying "let me remove all the cards except for a random one, the one you chose, and the one I chose, of which, one is the correct card to choose. Do you want to change your choice?" So it plays out as expected with 3, and the obviousness of the solution when done with 52 cards.
Your first choice is either right or it's wrong. If you switch, right becomes wrong and wrong becomes right.
Your first choice is 1/3 right 2/3 wrong.
I think there's a psychological component too. So many probability questions are about random events so you (very naturally) take a position that the revealed door is randomly chosen. It is NOT - you always see one of the goats.
You may be on to something.
Perhaps part of it is not wanting to switch from a winning position to a losing one. Thats harder to take so inertia sets in.
This is reminiscent of the endowment effect: https://en.wikipedia.org/wiki/Endowment_effect
I struggled with this because I didn't see the first choice as relevant to the final outcome. It is easy to see the game as restarted somehow when the 50/50 choice is presented.
He illustrated it with the problem of what happens to the level of water on a pool when you throw overboard a stone from a floating boat. Does the water level rise or drop? Well, imagine that your small stone is made of neutron star material...
Forgive me if I'm missing something obvious, I have a newborn and am underslept.
I assume neutron star material is incredibly dense and so the rock would be unbelievably heavy. To the point that the mass would actually likely sink the boat, and then we can't throw the rock out.
OK but let's assume that the boat is somehow massive enough to support the weight of the rock. Now we've changed the scale of the other variables of the experiment in proportion to the rock/neutron star, and we're back to ground zero, are we not?
It's only when the person opening the door knows where the prize is, that I should always switch?
A random wiki with a few points gets on the front page
Meanwhile all Jul I an As as ange wiki.... Leaks news is buried. Say hello to your masters.
https://twitter.com/judegomila/status/1183425802399969282
"Your brain will compute monty hall problem in fast casual causal mode AND/OR compute slow mode formally for the exact answer. Probability predictions that accurately map reduce to efficient intuitive representations in our brain dictate direction in our evolutionary biology."
"[since] we don't solve the monty hall problem effectively with our intuition, an evolving life form would be poor at resource collection from such problem sets without knowledge of axiom representation and formal probability."
We shall hunt in the forest.
"Sire! Our scouts have just returned from the mountain. They report that the herds are not there at the present time! Shall we try the valley, or continue on to the forest?"
A) Stick to the plan, and hunt in the forest B) Hunt in the valley instead
Without that guarantee, it makes no difference.
Additionally, there are not many situations in real life that would replicate the conditions of that guarantee. So our "intuition" is simply geared towards situations where an actor can't read our thoughts (door pre-selection) before we act them out, and in such a scenario, pre-selecting/switching selection of doors doesn't matter.
The scouts can depart after you’ve chosen the valley, find nothing in the mountain, give that information to you, and you should still switch, even if the scouts didn’t confirm where they are.
Monty(or the scouts), knowing if your original pick was right has no impact. The critical piece is that they are forced to reveal to you a bad choice. 2/3 times it’s the only bad choice to reveal because you’ve picked the other.
The entire gain in odds is predicated on Monty being forced to reveal a bad choice after you have chosen.
Monty knowing ahead of time has absolutely no impact on the outcome.
If you are objecting to how I described that, in terms of who knows what when, then you may be right and I probably could have been more precise.
But if you are disagreeing that scouts sent to investigate an area and return an answer (which this time happens to be negative) differs materially from the original question then you are wrong.
Nope, it does not. The fact that he revealed a dud is what matters. The notion of him maybe revealing a car is senseless anyway because you would just then pick the car or the game would be over.
That's simply wrong. This has been discussed countless times on this forum. The most effective way to proceed is for you to spend 5 minutes writing the simulation that you think will prove you right. When I was on the other side of this, that experience is what convinced me.
> The notion of him maybe revealing a car is senseless anyway because you would just then pick the car or the game would be over.
That's not a very strong objection - game shows are weird sometimes and could often apparently be easily improved.
But more importantly, it's an objection to the wrong thing. My point was that the presented "find the herds" problem differed from the game show. "That would have been a bullshit game show" is just confused.
Right, of course. I just meant he knows ahead of opening his door, it doesn't matter if he knows when you pick your door.
The scouts did not know that the herds are not in the mountain when they chose to scout the mountains.
Monty, on the other hand, does know that the car is not behind door C, when he chooses to open door C.
The difference is that Monty will never pick the door where the car is, whereas your scouts will sometime recon a region and find the herds there.
(Unless you assume that your scouts know where the herds are, and are purposefully not telling you and scouting empty areas, which would make them terrible scouts. We can expect this sort of crafty behaviour from a game show host, sure, but not from your own scouts.)
This results in different odds. With the scouts, switching makes no difference. You can run simulations or draw a full probability tree to convince yourself of it.
A random wiki with a few points gets on the front page
Meanwhile all Jul I an As as ange wiki.... Leaks news is buried. Say hello to your masters.
A random wiki with a few points gets on the front page
Meanwhile all Jul I an As as ange wiki.... Leaks news is buried. Say hello to your masters.
But in a mathematical probability problem, you can't do that. For example, if there's a slightly smaller object and a slightly larger object concealed in two different sized boxes, common sense says that a person would put the larger object in the larger box. But if it's a abstract mathematical probability problem, and you have no information on the algorithm used, then you have to assume the concealer flipped a coin when they decided which box to use.
If I'm the one setting up the big-small game, and I use the random algorithm, the smaller object is going to be in the bigger box about half the time. If my mom is the one setting up the game, the smaller object is going to be in the smaller box every time.
In abstract mathematical probability problems, if you don't know what algorithm was used, you aren't allowed to bring human behavior in, unless it's stated in the problem.
I think that this misrepresents mathematical reasoning.
For solving a problem in a textbook (which a mathematician does intensively at the beginning of his or her career, and seldom thereafter), you must assume neither—that the larger object goes in the larger box, nor that the concealer flipped a coin; you must use only the information contained in the problem. (From this point of view, many problems in most probability textbooks are ill posed, because they force you to make some assumption while solving them; and it then becomes a mind-reading exercise of whether you can successfully make the same assumptions as the poser, rather than a mathematical exercise.)
For a mathematician who actually wishes to use theoretical probability to do something useful, then you must again make assumptions that allow you to translate the messy real world into an idealised mathematical object; but to say that you must, or must not, assume any particular thing is otiose—different assumptions will lead to different results, and the only guide to whether you made the 'right' assumptions is whether or not the idealised mathematical results you get match the real world (to within whatever bounds of error are tolerable).
https://pubs.aeaweb.org/doi/pdfplus/10.1257/jep.33.3.144
"You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." Where does it state that the host always opens a door with a goat? From the statement it looks like the goat was accidental. No wonder Paul Erdős got confused by this too. The translation of the paradox to his language could have been even more confusing.
If the problem does not state what the host's algorithm is, and you have no knowledge of the host's algorithm (example: you have never seen the TV show before), then the logical action is to assume the host chooses a door to reveal randomly.
Isn't it the other way round?
The only scenario in which the host doesn't give you any useful information by opening a door is when the host chooses the door at random. Assuming the host didn't choose the door with the prize, your odds are 50:50.
Of course all of this hinges on you being aware of the host's decision making paradigm, and that you want a car more than you want a goat.
The first time I read about this problem, it was formulated in a way which made it clear that the host opens a door with a goat behind it. If you learn about it like this, you literally cannot believe how anyone (even Paul Erdős) could believe that switching is not better. After all, you choose a door with a goat behind it with probability 2/3 on your first try, and if you do this, than switching will always win you the car. It is so trivial that the only "paradox" here is that it is classified as a paradox.
E: the explanation on Wikipedia is unnecessarily elaborate and hard to follow:
"The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 that was not available at the beginning of the game, when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities. "
What?
I only took a single class of statistics at university, but my biggest take-away from that class is that you absolutely cannot trust your gut, because the human bias is crazy strong.
Humans naturally over-connect. We see causation where only correlation is present. Gamblers believe that something random that has happened frequently (like roulette landing on a specific number) is more likely to happen again. Educated people learn about dependent and independent probability, which corrects this impulse. But sometimes it over-corrects.
The Monty Hall problem is hard because people see can't explain the dependent probability link clearly and conclude that the probabilities are independent and their brain is acting up - first guess 1 in 3, second guess 1 in 2. After all, the game host is always capable of opening a door with a goat, and the player doesn't know anything. Put differently, what did the player learn between the first and the second guess that should make the player now believe one door is more likely than the other of the remaining two?
It also takes advantage of our natural thinking. The problem is artificial, because, outside of a game show, the host could also just not offer to switch. Emotional then, brains consider the information that the host made the second offer even if it's not logically part of the problem.
The paradox disappears when you think about probability as a tool for reasoning from incomplete information, not as anything to do with "physical" property of the system under investigation. It then makes perfect sense that after receiving new information from the host we should reassign our probabilities.
So, yes, even a person with the world's highest IQ can make a mistake.
> Hall clarified that as a game show host he did not have to follow the rules of the puzzle in the vos Savant column and did not always have to allow a person the opportunity to switch (e.g., he might open their door immediately if it was a losing door, might offer them money to not switch from a losing door to a winning door, or might only allow them the opportunity to switch if they had a winning door).
If the host uses an adversarial strategy, such as only offers a switch when the remaining door is a goat, switching is not necessarily wise.
Edit. Be nice if my downvoter would explain what's wrong with the above.
You look at the 3 cards then put them on the table and propose a fair triple or nothing, if they can find the red queen.
You make them point toward the card they chose.
If they point a wrong card you go toward the card and reveal -> you won.
If they have pointed the right card, you return one of the card they didn't pick and ask them if they are sure or want to change. Because they think they are facing a Monty Hall problem they will switch. -> you won again. :)
Even if you didn't cheat them, this game has zero expectation value... I guess if you are working with people who wouldn't know this you are fine.
It is, as I said before, a fair game so it has zero expectation value for both players, better odds than a casino for the player and there is no reason for a gambler to refuse this bet.
In expectation you won't lose money, but people who think they know better will happily give you their money by making a bad move while being convinced they are getting an edge.
Most math-type people don't know the hypothesis of the Monty-Hall Problem and can't state it properly, but most have seen it played on TV-shows and know they must switch.
In the heat of the moment, especially if you have played a few follow the red queen games before, the brain is still in fast mode, and will jump to using heuristics. Mind trick.
Maybe needs one introductory sentence: "Note that your first choice has probability 2/3 of being a losing choice".
EDIT: My reply is based on a misunderstanding. I was assuming that the host continues to open one door, whereas parent intended that the host would open all-but-one doors.
Obviously the answer is that your first guess is more likely to have been a goat. So if you had to bet, you'd have bet on the car being in one of the ones you didn't choose.
After the host reveals a goat, it's no longer "one of the ones you didn't choose", but rather "the one you didn't choose" (since the host has taken the other out of consideration). So the choice is easy now (as is the probability calculation).
Had it in a browser reading list (that I'd forgotten was there), and came across it by chance yesterday.
https://gist.github.com/kiriappeee/4316871