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Paradoxes of Probability and Other Statistical Strangeness

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Re: Paradoxes of Probability and Other Statistical Strangeness

#91
post #89

Earlier quoted context omitted.

ok ok... let me try to get this straight. Just as kind of a mental process for trying to understand whether or not something passes the smell test, I typically try to take the basic premise and turn it up to 11 and see if that still makes sense. In this problem, as you've described it, we're enumerating "ways to assign gender and birth-day-of-week." We can do this because there are a countable number of "days of the…

You can't assume "the boy is k1". The original 2 * 7 * 2 * 7 cases were indeed all equiprobable different cases. And we're not given that K1 is a boy. We're given that at least one child is a boy. If K1 is a girl and K2 is a boy born on Tuesday, this still counts as the family (Mr. Jones, if you like) having a boy born on Tuesday. There are 27 cases that count as the family having a boy born on Tuesday, all equiproba…

Frankly, though, I'd prefer no one ever used the "conventional language of probability", because it leads to precisely these miscommunications.

If the question had been phrased "Out of two-children families that have at least one boy born on Tuesday, what proportion have a girl? [on natural assumptions about lack of biases or correlations concerning the distribution of children's genders and days]", would you agree that the answer was 14/27?

That was the question the author intended to ask. The dispute may simply be as to whether the question which the author did ask is equivalent to the above; if that is indeed our only disagreement, we can still investigate that dispute further, if you like. But let's first see if the dispute is linguistic or mathematical.

Re: Paradoxes of Probability and Other Statistical Strangeness

#92
post #90

Earlier quoted context omitted.

ok ok... let me try to get this straight. Just as kind of a mental process for trying to understand whether or not something passes the smell test, I typically try to take the basic premise and turn it up to 11 and see if that still makes sense. In this problem, as you've described it, we're enumerating "ways to assign gender and birth-day-of-week." We can do this because there are a countable number of "days of the…

Let's see if I can help you understand this a bit better. First, let's clarify the problem being asked. There are 2 different problems with different solutions and it helps to explicitly separate them. problem 1) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy born on Tuesday. They say yes. What is the probability that they have a girl? problem 2) You go up to a pers…

:) Thanks for taking the time. I've realized a few things, and found it helped to get a bit more formal.

Jones has 2 kids. Let A be "he has a girl" and B be "he has a boy born on tuesday." First thing I realized is A and B are NOT independent - this is key. P(A) includes the option of Jones having two girls. But if B is true, then the two girls option isn't on the table anymore, which affects P(A). Realizing this helped me start to better understand what kind of problem we're dealing with.

Second was realizing that P(A&B) is not at all the same thing as P(A|B) - the probability of A given B - when A and B aren't independent. The problem is asking for P(A|B), and by the rule of conditional probability: P(A|B) = P(A&B)/P(B)

P(B) can be solved for without too much fuss: solve 1-P(!B). For each kid you have 2 genders and 7 days of the week, or 2 * 7 = 14 options. 13 of those are not "Boy & Tuesday." So you have P(!B) is (13/14) * (13/14) = 169/196. P(B) = 1 - 169/196 = 27/196.

This leaves us trying to figure out P(A&B). I can't think of any other way to do it other than enumerating all options. We can take a shortcut and just look at all 27 possible scenarios where B is true. This seems to be the method of choice ;) As others have shown, we see that 14 of those satisfy A. So P(A&B) = 14/196.

Now, we can solve: P(A|B) = P(A&B)/P(B) = (14/196)/(27/196) = 14/27

So I'm now part of the "math checks out" club. Thanks for all the help people!

Re: Paradoxes of Probability and Other Statistical Strangeness

#93
post #91
post #89

Earlier quoted context omitted.

You can't assume "the boy is k1". The original 2 * 7 * 2 * 7 cases were indeed all equiprobable different cases. And we're not given that K1 is a boy. We're given that at least one child is a boy. If K1 is a girl and K2 is a boy born on Tuesday, this still counts as the family (Mr. Jones, if you like) having a boy born on Tuesday. There are 27 cases that count as the family having a boy born on Tuesday, all equiproba…

Frankly, though, I'd prefer no one ever used the "conventional language of probability", because it leads to precisely these miscommunications. If the question had been phrased "Out of two-children families that have at least one boy born on Tuesday, what proportion have a girl? [on natural assumptions about lack of biases or correlations concerning the distribution of children's genders and days]", would you agree t…

> Out of two-children families that have at least one boy born on Tuesday, what proportion have a girl?

OH YES.

I finally figured it out (see the other comment). Thanks for all the explaining, but this statement right here was the best.

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