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A confusing probability question: Red and green balls in an urn

colab.research.google.com

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A confusing probability question: Red and green balls in an urn

#1

Post by eigenvalue »

A Twitter post made the rounds yesterday asking the following question in the form of a poll (https://twitter.com/littmath/status/1751648838501224790):

Daniel Litt @littmath

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 take a random ball out of the urn—it’s red—and discard it. The next ball you pick (out of the 99 remaining) is:

More likely to be red 22.6%

More likely to be green 37.1%

Equally likely 20.9%

Don’t know/see results 19.5%

I thought it was interesting (but not surprising) that so many people got this wrong. Basically, Bayesian thinking is really foreign to people. It's essentially similar to the famous Monty Hall problem, where the first bit of information tells you something about the world. In any case, I thought it might be illuminating to give a complete demonstration, using both exact calculations AND simulation, of the "wrong" and "right" approach, which you can see at the Colab link above.

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

Re: A confusing probability question: Red and green balls in an urn

#3

Post by Terr_ »

To try to put it into words:

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."

Re: A confusing probability question: Red and green balls in an urn

#4

Post by eigenvalue »

Terr_ wrote: post #3

To try to put it into words: 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…

Yes, that's exactly right. Paraphrasing what another poster wrote about this on Twitter: "You would rather go fishing in a lake where you just saw someone else catch a fish."

Also, the logic continues as you draw more balls from the urn. If the second ball is ALSO red, then you have even more evidence suggesting that N was selected in such a way as to make red the overwhelmingly more likely choice. Thus the chance of the third ball being red is even higher than the 66% probability in the second draw.

Re: A confusing probability question: Red and green balls in an urn

#5

Post by Terr_ »

eigenvalue wrote: post #4
Terr_ wrote: post #3

To try to put it into words: 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…

Yes, that's exactly right. Paraphrasing what another poster wrote about this on Twitter: "You would rather go fishing in a lake where you just saw someone else catch a fish." Also, the logic continues as you draw more balls from the urn. If the second ball is ALSO red, then you have even more evidence suggesting that N was selected in such a way as to make red the overwhelmingly more likely choice. Thus the chance of…

To extend the fishing metaphor, the premise of ~100 balls is also important: If each lake could only sustain one (good) fish, you would actively avoid anywhere someone had already caught one from.

Re: A confusing probability question: Red and green balls in an urn

#6

Post by orlp »

Follow up question:

> 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?

Re: A confusing probability question: Red and green balls in an urn

#7

Post by user_7832 »

Terr_ wrote: post #5
eigenvalue wrote: post #4

Earlier quoted context omitted.

Yes, that's exactly right. Paraphrasing what another poster wrote about this on Twitter: "You would rather go fishing in a lake where you just saw someone else catch a fish." Also, the logic continues as you draw more balls from the urn. If the second ball is ALSO red, then you have even more evidence suggesting that N was selected in such a way as to make red the overwhelmingly more likely choice. Thus the chance of…

To extend the fishing metaphor, the premise of ~100 balls is also important: If each lake could only sustain one (good) fish, you would actively avoid anywhere someone had already caught one from.

I guess the counterpoint to this is as the number of balls goes from 100 to infinite, it becomes increasingly harder to predict, correct?

Re: A confusing probability question: Red and green balls in an urn

#8

Post by russdill »

orlp wrote: post #6

Follow up question: > 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?

50%. With current knowledge N is either 100 or 99 with an equal probability.

Re: A confusing probability question: Red and green balls in an urn

#10

Post by cornel_io »

user_7832 wrote: post #7
Terr_ wrote: post #5

Earlier quoted context omitted.

To extend the fishing metaphor, the premise of ~100 balls is also important: If each lake could only sustain one (good) fish, you would actively avoid anywhere someone had already caught one from.

I guess the counterpoint to this is as the number of balls goes from 100 to infinite, it becomes increasingly harder to predict, correct?

No, the higher the count goes the closer it becomes to the single removed ball not mattering at all, so the only thing that matters is the information you got about which urn you're selecting from.
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