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The sisters paradox is madenning example of counter-intuitive probability. The resolution is straightforward, but it's really easy to get tied up in knots.

by Vermin2000 1y ago

bell-cot 1y ago
> maddening example of counter-intuitive probability. Not how I'd describe it. The setup is mundane enough for people to just assume that their intuition will work fine. The difference between the naive and correct answers is too small to spot in a small-n dataset. And ~0% of the population is actually familiar with analyzing such situations, for their "intuition" to be applicable. It's a bit like Gell-Mann amnesia - people are too quick to apply an easy cognitive strategy, when (in theory) they know enough to rule that strategy out.
- spadros 1y ago
  Yes, I found this one easy. Was surprised my data management intuition came back after all these years since school. There’s really only three options: - boy - boy - boy - girl - girl - girl So it must be 1/3 chance. If you’re looking at permutations in order, that’s a different question.
  - AIPedant 1y ago
    This intuition is wrong even if turned out to get the right answer. The three unordered options do not have equal probabilities, boy+girl is twice as likely to occur as boy+boy and girl+girl. To get the right answer you must be careful about conditional probabilities (or draw out the sample space explicitly). The crux of the issue is that you are told extra information, which changes your estimate of the probability. (This question as written is very easy to misinterpret. The Monty Hall problem, which illustrates the same thing, is better since the sample selection is much more carefully explained.)
    - taeric 1y ago
      Oddly, this is a part I'm sticking with on this problem. Specifically, if you know that one is a girl, then the unordered options seem like they are back on equal footing? That is, it isn't twice as likely if you know that one ordering can't happen? (Or, stated differently, you don't know which version of two girls you are looking at.) So, for this one, you know that either the youngest is a girl (so, girl-boy is not possible) or that the oldest is a girl (so boy-girl is out). That puts you back to the rest of the possibilities. Boy-boy is out, sure, as you have a girl. But every other path remains? So, you have one of (boy-girl(known), girl-girl(known), girl(known)-boy, girl(known)-girl). Which drops you back to 50/50?
      - AIPedant 1y ago
        Like others in the thread have said, the question could have been phrased more precisely. Technically you are misreading it but in an annoying and trivial way. What the problem is really saying is this: 1) You have a large collection of families with two kids of varying genders. 2) You draw one of them at random. At this point, your only estimate of P(2 girls) is 0.25. 3) Someone tells you that the family you drew has at least one girl. 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. The way you are interpreting it is this: 1) You have a large collection of families with two kids, at least one of whom is a girl. 2) Then the probability that the other child is a girl is clearly 50%. As a reminder this is how the original post phrased the question: Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls? This is just too vague and admits both interpretations, they needed to be more specific about where the family "came from." That's why Monty Hall is a better illustration: it starts with you explicitly choosing a door at random. Here the family has been chosen at random from the pool of families with two children, but that's totally unclear.
        kgwgk 1y ago
        > 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. That’s not correct in general. It’s only correct if you assume that “3) Someone tells you that the family you drew has at least one girl.” was equally likely to happen whether or not there were two girls. That’s a quite strong assumption. One can make different assumptions and get answers different from 1/3. For example, 1/2.
EMM_386 1y ago
I don't understand at all how this is maddening or counter-intuitive. When you have a child, the odds are ~50% ... so the chance the next child is a boy or girl is almost equal. Is it because of the way it's framed that makes people think harder than they need to be? This is like when I (very rarely) play something like "pick six". I play 1, 2, 3, 4, 5, and 6. People think I'm crazy. They don't realize I have the same odds as any ticket they purchase.
- n4r9 1y ago
  If you have to share the prize money with other people that guessed the same numbers, then it seems sensible to avoid obvious patterns.
  - EMM_386 1y ago
    Good point, others have said the same. In the rare scenario where I buy one of these tickets, I'd be more than happy to share the winnings! One billion dollars goes a long way ... even after taxes!

The sisters paradox is madenning example of counter-intuitive probability. The resolution is straightforward, but it's really easy to get tied up in knots.

bell-cot 1y ago

> maddening example of counter-intuitive probability. Not how I'd describe it. The setup is mundane enough for people to just assume that their intuition will work fine. The difference between the naive and correct answers is too small to spot in a small-n dataset. And ~0% of the population is actually familiar with analyzing such situations, for their "intuition" to be applicable. It's a bit like Gell-Mann amnesia - people are too quick to apply an easy cognitive strategy, when (in theory) they know enough to rule that strategy out.

spadros 1y ago
Yes, I found this one easy. Was surprised my data management intuition came back after all these years since school. There’s really only three options: - boy - boy - boy - girl - girl - girl So it must be 1/3 chance. If you’re looking at permutations in order, that’s a different question.
- AIPedant 1y ago
  This intuition is wrong even if turned out to get the right answer. The three unordered options do not have equal probabilities, boy+girl is twice as likely to occur as boy+boy and girl+girl. To get the right answer you must be careful about conditional probabilities (or draw out the sample space explicitly). The crux of the issue is that you are told extra information, which changes your estimate of the probability. (This question as written is very easy to misinterpret. The Monty Hall problem, which illustrates the same thing, is better since the sample selection is much more carefully explained.)
  - taeric 1y ago
    Oddly, this is a part I'm sticking with on this problem. Specifically, if you know that one is a girl, then the unordered options seem like they are back on equal footing? That is, it isn't twice as likely if you know that one ordering can't happen? (Or, stated differently, you don't know which version of two girls you are looking at.) So, for this one, you know that either the youngest is a girl (so, girl-boy is not possible) or that the oldest is a girl (so boy-girl is out). That puts you back to the rest of the possibilities. Boy-boy is out, sure, as you have a girl. But every other path remains? So, you have one of (boy-girl(known), girl-girl(known), girl(known)-boy, girl(known)-girl). Which drops you back to 50/50?
    - AIPedant 1y ago
      Like others in the thread have said, the question could have been phrased more precisely. Technically you are misreading it but in an annoying and trivial way. What the problem is really saying is this: 1) You have a large collection of families with two kids of varying genders. 2) You draw one of them at random. At this point, your only estimate of P(2 girls) is 0.25. 3) Someone tells you that the family you drew has at least one girl. 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. The way you are interpreting it is this: 1) You have a large collection of families with two kids, at least one of whom is a girl. 2) Then the probability that the other child is a girl is clearly 50%. As a reminder this is how the original post phrased the question: Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls? This is just too vague and admits both interpretations, they needed to be more specific about where the family "came from." That's why Monty Hall is a better illustration: it starts with you explicitly choosing a door at random. Here the family has been chosen at random from the pool of families with two children, but that's totally unclear.
      - kgwgk 1y ago
        > 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. That’s not correct in general. It’s only correct if you assume that “3) Someone tells you that the family you drew has at least one girl.” was equally likely to happen whether or not there were two girls. That’s a quite strong assumption. One can make different assumptions and get answers different from 1/3. For example, 1/2.

EMM_386 1y ago

I don't understand at all how this is maddening or counter-intuitive. When you have a child, the odds are ~50% ... so the chance the next child is a boy or girl is almost equal. Is it because of the way it's framed that makes people think harder than they need to be? This is like when I (very rarely) play something like "pick six". I play 1, 2, 3, 4, 5, and 6. People think I'm crazy. They don't realize I have the same odds as any ticket they purchase.

n4r9 1y ago
If you have to share the prize money with other people that guessed the same numbers, then it seems sensible to avoid obvious patterns.
- EMM_386 1y ago
  Good point, others have said the same. In the rare scenario where I buy one of these tickets, I'd be more than happy to share the winnings! One billion dollars goes a long way ... even after taxes!