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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

#151

Post by danielmarkbruce »

tines wrote: post #145
danielmarkbruce wrote: post #140

Earlier quoted context omitted.

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 binom…

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 i…

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.

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

#152

Post by layer8 »

Consider a 100 x 101 grid (101 rows and 100 columns). Each row is a possible urn (from 0 green balls to 100 green balls). The columns correspond to the 100 balls in each urn. The rows are sorted by ascending number of green balls, and the columns are sorted by red balls first. Reduced to 10 x 11, this would be:

    RRRRRRRRRR
    RRRRRRRRRG
    RRRRRRRRGG
    RRRRRRRGGG
    RRRRRRGGGG
    RRRRRGGGGG
    RRRRGGGGGG
    RRRGGGGGGG
    RRGGGGGGGG
    RGGGGGGGGG
    GGGGGGGGGG
The rows are parallel universes, each of them equally likely. When you draw the first ball, the universes each split into 100 also equally likely universes, corresponding to which of the 100 balls you drew. So each of the 10100 grid cells represents a possible universe you can find yourself in, each equally likely.

When you draw red, you know that you are somewhere in the upper left triangle of Rs, which each R being equally likely.

    RRRRRRRRRR
    RRRRRRRRR
    RRRRRRRR
    RRRRRRR
    RRRRRR
    RRRRR
    RRRR
    RRR
    RR
    R
It’s clear that you are more likely to be in the upper half of that triangle, because it contains more Rs. In fact, because there are 5050 Rs, and the upper 30 rows have 2565 Rs (I’m applying the triangle-number formula n * (n + 1) / 2), you’re more likely to be within the upper 30 rows (urns) than the lower 70 rows (urns), and hence it is clearly more likely that the second ball you draw is again red rather than green. (Though visualizing this further split of universes would require a 3rd dimension.)

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

#153

Post by danielmarkbruce »

j7ake wrote: post #137
danielmarkbruce wrote: post #18

Change it to drawing from a normal distribution... the answer will change dramatically and get very close to what one would likely consider "common sense", ie that it remains close to 50/50. I think binomial gets it to 50/50. Humans aren't wired for uniform distribution. 100 reds seems crazy unlikely compared to 50/50, yet this trick question makes those equally likely.

How would normal distribution change the result? I am imaging a random number generated from truncated normal distribution with mean 50, clipped 0 and 100: the answer will still lean right. How would binomial work? You have 101 urns from which to choose.

Normal: it is very close to 50/50. It does still lean, but only just. It's close enough that in practice any guess is fine.

Binomial: it is 50/50.

The reason these change the situation dramatically is that under normal or binomial, you are waaaay more likely to be in a situation that is close to 50/50 to begin with because they are way more likely to occur. The quirk of this question is that a 0 green 100 red urn is just as likely as a 50-50 red/green urn using a uniform distribution - hence if you get red on the first draw you are much more likely to be in a heavily tilted red urn.

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

#154

Post by tines »

danielmarkbruce wrote: post #151
tines wrote: post #145

Earlier quoted context omitted.

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 i…

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.

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

#155

Post by danielmarkbruce »

tines wrote: post #154
danielmarkbruce wrote: post #151

Earlier quoted context omitted.

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 de…

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

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

#156

Post by tines »

danielmarkbruce wrote: post #155
tines wrote: post #154

Earlier quoted context omitted.

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 de…

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.

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

#157

Post by AtNightWeCode »

bitshiftfaced wrote: post #129
tines wrote: post #38

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% proba…

Why are we excluding red, green, red?

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

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

#158

Post by danielmarkbruce »

tines wrote: post #156
danielmarkbruce wrote: post #155

Earlier quoted context omitted.

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"

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

#160

Post by tines »

danielmarkbruce wrote: post #158
tines wrote: post #156

Earlier quoted context omitted.

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.

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