A confusing probability question: Red and green balls in an urn (colab.research.google.com)

We don't actually know the mix-rate of what's in the urn. It's possible we just pulled the only red ball and all the others are green... However the fact that our first n=1 sample happened to be red (and not something else) gives a small (and varying) amount of confidence towards red-heavier mixes rather than the red-scarce ones.

We can use that to figure out which mixes are slightly more-likely to be in there (based on our tiny amount of data after assuming they begin equally-likely) and then for each of those determine "if it's mix X, then what are the odds of our next draw", and combine the two layers of probability.

In this case, the bias towards presuming the mix is fundamentally reddish outweighs the presumption of "removing one red makes the greens the majority."

> You are given an urn containing 100 balls; n of them are red, and 100-n are green, where n is chosen uniformly at random in [0, 100]. You select 99 balls at random and find that they are all red. What is the probability the last ball is red?

4901:4900 in favor of red. (100*98/2 = 100*49. Plus one in favor of red)

If the number of distribution were odd, the odds would be even.

Am I right?

ie i think many of the More Likely to be Green people are doing the math of if you pull from an urn with n/100 odds of getting red, your second pull will have odds (n-1)/99, which is less than n/100 for all n except n=100. Which is obviously a different question than should you bet on Green or Red.

Which is obviously a different question than "would you bet on the next ball being red or green"

Humans aren't wired for uniform distribution. 100 reds seems crazy unlikely compared to 50/50, yet this trick question makes those equally likely.

Would the question be valid if the first ball was green? If not, then it's equivalent to the standard answer to the Monty Hall problem. Analyze it as if the first ball was not picked randomly (i.e. Monty intentionally picked the wrong door).

If monty opens doors randomly, and shows you a goat, you do not benefit from switching. Edit: assuming he doesn’t open your door of course

1. Red, Red, Red

2. Red, Red, Green

3. Red, Green, Green

(The Green, Green, Green case is impossible because you drew one Red.)

Drawing a Red from urn configuration (1) was a 100% probability; from (2) was a 66% probability, and from (3) was a 33% probability. If these configurations were equally likely, then the probability of Red and Green on the second draw would be the same.

However, and this is the crux: we are more likely to be in a configuration which shows us what we have observed with a higher likelihood [1]; so we're more likely in configuration (1) or (2) than (3), and as (3) is the only one that favors Green for the next draw (and only by as much as (1) favors Red), the next ball being Red is more probable.

[1] Imagine, for example, that you have 100 coins, and 99 of those coins are biased so that they only show heads once every trillion tosses, while the remaining one coin is fair. If you pick a coin randomly and flip heads, which is more likely: that you got a biased coin to show a one-in-a-trillion event, or that you picked the fair coin?

What is 3 x 2?

. . .

. . .

1, 2, 3, 4, 5, 6 . . . six?

Edit: To add to that, check out the Erdos book from the 90s on innumeracy (here's Wikipedia instead of an AMZN affiliate link):

https://en.wikipedia.org/wiki/Innumeracy_(book)

1. More likely to be red because the urn has a greater chance of having more red balls

2. Equally likely by considering all remaining urn possibilities to be equally likely

3. More likely to be green because obviously there is one less red ball than before

I wrote a Python program to simulate the problem. It tells me that the correct answer is (2) - it's equally likely that the next ball is red or green.

I wonder if I've made a mistake, or if that's really the real answer!

    #!/usr/bin/python
    import random
    from collections import Counter
    
    def experiment():
        urn = random.choices(["R", "G"], k=100)
        if urn.pop() != "R":
            return
    
        return urn.pop()
    
    results = (experiment() for i in range(1_000_000))
    results = (result for result in results if result is not None)
    print(Counter(results))

    #!/usr/bin/python
    import random
    from collections import Counter
    
    def experiment():
        num_red = random.randint(1, 100)
        num_green = 100 - num_red
        urn = ["R"] * num_red + ["G"] * num_green
        random.shuffle(urn)
        if urn.pop() != "R":
            return
    
        return urn.pop()
    
    results = (experiment() for i in range(1_000_000))
    results = (result for result in results if result is not None)
    print(Counter(results))

In this specific case, P(redOnKthSample) = (numberOfRedSamples + 1) / (totalNumberOfSamples + 2) = (1+1)/(1+2) = 66% red.

On the second draw, if the ball is green, then you get P(redOnThirdSample) = (1 + 1) / ( 2 + 2) = 50%

"More likely to be red"

More likely than what?

"The next ball you draw from this urn is more likely to be red than green?"

One is 0% red, one 1% red, one 2% red, etc.

If I choose one urn at random, and then the first ball I pick is red, odds are good I picked a red-heavy urn.

I’d be interested to see if the general public answers differently when it is framed more concretely.

An obvious visual proof: Line up 101 marbles. Choose one to be the divider; to the left is red, to the right is green. Now draw two more marbles. Or just pick three, then decide which one is the divider.

`You take a random ball out of the urn—it’s red—and discard it.`

How normal people read it: Given this specific instance where you just discarded a red ball from this urn, what's the probability of the next ball?

How it expects you to read it: Given infinitely many random samples from the urn. For cases where you get red, remove it, then take a second sample. What's the probability of the next ball, given all the samplings?

Two ways to think about this (same conclusion):

- Without loss of generality, take the case of just one red ball, rest are green. If you pick the one red ball on your first draw the chances are now 100% that the next ball is green (since there aren't any red ones left) whereas before you had picked the red one you had just a 99% chance of picking a green one.

Hence after picking a red ball you are slightly more likely to pick a green ball next.

- Another way to think of it, again without loss of generality consider if there was just one green ball and 99 red, then if you pick every one of them in some order there are a hundred different positions the green one could end up as. If you consider the one where it's last, after you have picked all but one (and they were all red), now you have a 100% chance to pick the green ball, even though you started out with 1% chance on the first draw: without loss of generality in the case of drawing until you reach the green ball when it is 1 in 100, each red ball you have drawn so far increases the chance of drawing the green ball next, up to a 100% chance when it is the only one left. Visualizing the chance going from 1% to 100% in this case as you draw more red balls should drive home the point that it becomes more and more likely you'll draw a green one next, in the cases that you've made it that far drawing red ones.)

We have N+1 people, also numbered 0 through N. Person k is holding jar k.

We ask each person to draw a ball at random from their jar. If they draw a green ball they take their jar and leave. If they draw a red ball they stay.

We expect on average (N+1)/2 people will remain.

We now ask them to draw another ball.

The expected number of red balls on this second drawing is (N+1)/3.

The expected number of green balls on this second drawing is (N+1)/6, half the expected number of red balls.

That's because on the first drawing, the more green balls in your jar the more likely you will draw green on the first round and thus take your jar and go home. The second round is biased toward people who have jars with more red balls.

Here's the math behind those results. The probability that person k makes it to round 2 is k/N. The expect number that will make it is sum {k=0, N) = (N+1)/2.

Once in round 2 the probability they draw red is (k-1)/(N-1). The probability they draw green is (N-k)/(N-1).

The probability then that they make it to the second round and then draw red is k/N (k-1)/(N-1).

The probability that they make it to the second round and then draw green is k/N (N-k)/(N-1).

The expected number of reds we see in the second round is sum {k=1, N} k (k-1) / N / (N-1).

The expected number of greens is sum {k=1, N} k (N-k) / N / (N-1).

Use the well known formulas 1 + 2 + 3 + ... + M = M (M+1) / 2 and 1^2 + 2^2 + ... + M^2 = M (M+1) (2M+1) / 6 and simplify and you'll get the (N+1)/3 expected red and (N+1)/6 expected green I stated above.