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

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

#31

Post by tqi »

posix86 wrote: post #22
tqi wrote: post #12

I think this is less about Bayesian thinking and more about misinterpreting the question, as another poster mentions: https://twitter.com/farrwill/status/1751788706355392639 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 di…

Why would n be "uniformly randomly chosen" but then suddenly be known? I would mark this mis interpretation simply as wrong if I was a teacher.

That's definitely fair, and I assuming in a classroom setting you'd be very consistent and precise about what that means. But this is a twitter poll, and I doubt people are that consistent. Ultimately I'm just pushing back on the conclusion that this is because "Bayesian thinking is really foreign to people."

I think if the question had instead been "would you bet on the urn having started out with more red balls, green balls, or equal number of balls" the distribution would have been very different.

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

#32

Post by addandsubtract »

russdill wrote: post #8
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.

No, the chance of it being red is much higher. If there's one green ball in the jar, than all previous draws would've needed to avoid the green one. The chances of that happening are way less than 50%.

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

#33

Post by eigenvalue »

anigbrowl wrote: post #25

I think the wording is also biasing the outcome. The next ball is more likely to be green than the previous ball , and a lot of people will regard that posterior probability statement as implicit. If the question had been phrased differently eg, 'what is the most likely color for the next ball selected' I suspect more people would have voted for equal likelihood.

Yes, I think you're right about that. Just look at all the disagreement about it right here in this discussion!

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

#34

Post by danielmarkbruce »

thfuran wrote: post #29
danielmarkbruce wrote: post #27

Earlier quoted context omitted.

You are basically asking "how could it possibly be the wrong answer?". Yes, under that definition of "more likely", it cannot be more likely. It can be equally likely or less likely. And then weighted it would be less likely. And that would be the answer. Which is one of the choices.

No, I'm saying "Your interpretation of the sentence is obviously wrong". With that interpretation, parts of the question are nonsense. The more straightforward interpretation does not have that problem.

1 - I said it's poorly worded, I didn't interpret it that way. I could see how people would misinterpret it (as pointed out by the parent comment) and people did misinterpret it that way. Look at the other posts here. If someone posts a question in english and a bunch of english speakers misinterpret the question, it's poorly worded by definition.

2 - Exactly what parts of the question are nonsense under that interpretation?

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

#35

Post by danielmarkbruce »

eigenvalue wrote: post #33
anigbrowl wrote: post #25

I think the wording is also biasing the outcome. The next ball is more likely to be green than the previous ball , and a lot of people will regard that posterior probability statement as implicit. If the question had been phrased differently eg, 'what is the most likely color for the next ball selected' I suspect more people would have voted for equal likelihood.

Yes, I think you're right about that. Just look at all the disagreement about it right here in this discussion!

And if you change the distribution, that answer is close to correct.

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

#36

Post by eximius »

mitthrowaway2 wrote: post #15

Uniform n is what drives a lot of the interesting properties here. Other distributions can behave differently. For example if balls were inserted into the urn with their colors determined independently by coin flips, then instead of n being uniform it would be binomial. In that case, observing the color of a previously drawn ball would tell you nothing about the next one.

Is that true? You'd still have evidence that the distribution of balls tilts one way, wouldn't you?

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

#37

Post by jncfhnb »

It’s not Monty Hall. Monty Hall does not open doors randomly. To think about this problem right you need to acknowledge that it could have gone the other way with some probability but it didn’t, and so that tells you something more about the state of things.

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

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

#38

Post by tines »

I think this is easiest to see if you imagine the urn is filled with three balls. You draw one, and it's red. The possibilities are that the urn originally contained:

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?

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

#39

Post by danielmarkbruce »

eximius wrote: post #36
mitthrowaway2 wrote: post #15

Uniform n is what drives a lot of the interesting properties here. Other distributions can behave differently. For example if balls were inserted into the urn with their colors determined independently by coin flips, then instead of n being uniform it would be binomial. In that case, observing the color of a previously drawn ball would tell you nothing about the next one.

Is that true? You'd still have evidence that the distribution of balls tilts one way, wouldn't you?

I think with a binomial the tilt is perfectly offset by the tilt you get from taking the ball.My math isn't fresh enough to write a nice dense statement showing why.

If you change their simulation formula to pull from a binomial instead of uniform... it looks 50/50.

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

#40

Post by [deleted] »

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.

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