Comment by tines | Hacker News Reader

Comment by tines | Hacker News Reader

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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 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?

And the underlying cause of people's intuition failing them is caused by those configurations being equally likely, which is a quirk of the question.

Why are we excluding red, green, red?

You have just highlighted the actual problem with the question. Uniform distribution.

Not 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.

I 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.

I 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.

It 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.

I think the thing that clarifies it for me is that if the balls are chosen in such a way that they have a binomial distribution, then, intuitively, the question would be implying retrocausality by saying that N was also "chosen":

> n is chosen uniformly at random

If the balls follow a binomial distribution, then your intuition would say that N wasn't the thing that was chosen; the balls were chosen, and then N was determined afterwards by counting the balls that you picked. This is because you cannot choose N and then draw balls binomially and get that many green or red balls. You'd have to choose 100 balls, see whether the number of Red matches your chosen N, and if it didn't, draw again, which is a tortuous interpretation given the phrasing, as I said above.

By saying that N was "chosen", it implies to me that N comes first, and the choice of balls comes after that, i.e. the balls are not binomially distributed.

Oh, I'm not suggesting you are wrong. I'm just saying it's easy to overlook the detail.

Sure, I guess I'm just arguing that even if you overlook that detail, then your intuition should tell you that the balls having a binomial distribution is suspect, i.e. you have to overlook a small thing and a big thing to misunderstand, and given that, I don't think the question is ambiguous. All good though.

I don't think that's right though, I don't think "chosen" implies what you are saying. An existing binomial distribution could exist and you just choose n from it. You wouldn't have to "generate" the distribution. To me "chosen" and "sampled" would mean the same, and it wouldn't be weird to say "n was sampled from a binomial distribution"

I think I see what you mean now, and I understand that that would be possible.

Wouldn't you need some extra parameters if it were a binomial distribution though? For a uniform distribution every item is equiprobable (i.e. there's only one uniform distribution for a given N) but there are many binomial distributions possible for a given N. (Disclaimer, I know very little about statistics.) So the fact that you aren't given those parameters in the question means you can't make much of a prediction if it were binomial.

Edit: I read that binomial distributions are parameterized on both the number of trials and the probability of the two outcomes, so that kind of demonstrates my point doesn't it? You're not given the probability of Red and Green for the binomial distribution, so you can't really answer the question that way. You could assume it's 50/50, but that's an assumption that isn't justified by the question.

The balls in the container are not ordered.

Because it is a trick question. That state is not in play.

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