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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 cont
by tines 3y ago
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?
- danielmarkbruce 3y agoAnd the underlying cause of people's intuition failing them is caused by those configurations being equally likely, which is a quirk of the question.
- bitshiftfaced 3y agoWhy are we excluding red, green, red?
- danielmarkbruce 3y agoYou have just highlighted the actual problem with the question. Uniform distribution.
- tines 3y agoNot really; he's hypothesizing a potential draw order, but what I was listing were the possible contents of the urn before any are drawn. Since the contents of the urn aren't ordered, they form a set, not a list, and shuffling elements doesn't add to the list of possibilities.
- danielmarkbruce 3y agoI don't think he/she is talking about that. I think the question is about the probabilities of each of the three options you lay out. You lay it out as though it's obvious the three are equally likely possibilities in the urn. That's only true using uniform distribution to generate the number of red(/green) in the urn. The contents of the urn would still be a set, but if the contents were initially drawn from a binomial distribution, calling out the possible combinations is required for purposes of counting, ie figuring the probability of each combination. People intuitively know that there are more ways to have two green and one red than three reds. That maps to most things in the real world too. This question sort of tricks people.
- tines 3y agoI thought that it was pretty clear from the part where it describes how the urn is filled: > n of them are red, and 100-n are green, where n is chosen uniformly at random in [0, 100]. N is chosen first, and that many red balls are put in. You don't flip a coin 100 times and then determine N from how many heads you flipped, that would be a tortuous interpretation from my pov, and would violate the requirement that N is chosen uniformly, so it seems like it's right there in the question.
- danielmarkbruce 3y agoIt is there. But it's just a detail many will breeze over and then their intuition fails them. The way you can know this is: ask people the same question with "binomial" instead of "uniform" and see if you get a different answer. My guess is people will say the same regardless.
- tines 3y agoThe balls in the container are not ordered.
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- AtNightWeCode 3y agoBecause it is a trick question. That state is not in play.