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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.
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…
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I'd argue that it's not so much a difference of opinion than it is just a reasoning error given the question as stated. That's sort of the whole point of this post-- this is a case where "doing the math" in the expected way gives the wrong answer, because the state of the system is cast in stone (in terms of the ratio of red/green balls in the urn) when you first start out, so it's all about leveraging the informatio…
It's definitely a poorly worded question. "More likely" is ill-defined. More likely than what? Than the last draw? Than drawing the other color just on this pick?
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It's definitely a poorly worded question. "More likely" is ill-defined. More likely than what? Than the last draw? Than drawing the other color just on this pick?
More likely than the other option, given the information. Getting an accurate grasp on the information is the whole point of the question.
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More likely than the other option, given the information. Getting an accurate grasp on the information is the whole point of the question.
Why is that the question? Why not "more likely than the last draw" ?
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Why is that the question? Why not "more likely than the last draw" ?
How could the next draw be more likely to be red than the draw already known to be red?
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.
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.
But if there is a green ball, it is unlikely to be the last picked. So the fact that you saw 99 red balls makes it unlikely there is a green ball.
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How could the next draw be more likely to be red than the draw already known to be red?
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.