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The problem is the one talked about here. As far as I understood it, the conclusion is that if I take a large number of families of two children with one boy born on Tuesday, the empirical probability will approach $13/27$ instead of $1/3$.

But the probability of having two boys given one boy is born on an Xday is 13/27 for Xday= any day of the week. So suppose I am told that one of the child is a boy. Why can't I just assume that that boy was born on some Xday and come to the probability that both are boys to be 13/27 even though I don't know what day Xday is?

RobPratt
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Brain Stroke Patient
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  • Naïve question maybe, but why isn't Tuesday irrelevant? – Ritam_Dasgupta Jul 05 '21 at 16:45
  • In fact you can assume any day, instead of Tuesday, since any other day has exactly same probability as Tuesday, that is 1/7 – Nikos M. Jul 05 '21 at 16:48
  • The given link provides a better explanation that I can ever give but simply put it changes sample size from {boy & boy, boy & girl, .... } to {boy-tuesday & girl-sunday, ...} – Brain Stroke Patient Jul 05 '21 at 16:48
  • But a consistent choice of day is required to arrive at the final result. So he same problem where the only change is Tuesday -> Friday, has exactly same result as well as using any other day, but the same day should be used throughout – Nikos M. Jul 05 '21 at 16:49
  • You need to know what will be said in different circumstances, such as there being a boy and girl both born on a Tuesday, or there being two girls of which one was born on a Friday and the other on a Sunday etc. – Henry Jul 05 '21 at 16:51

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The problem with your proposed answer is the sentence "Why can't I just assume that that boy was born on some Xday ..." There is no "that boy," because it is possible that there are two boys.

Here is a more detailed explanation. Let $A$ be the event that both children are boys. Let $B_1$ be the event that at least one child is a boy born on a Sunday, $B_2$ the event that at least one child is a boy born on a Monday, and so on for $B_3$ through $B_7$. We have $\bigcup_{i=1}^7 B_i = B = $ the event that at least one child is a boy.

As you have observed, the reasoning for the Tuesday problem would apply to any day of the week, so $P(A \mid B_1) = P(A \mid B_2) = \cdots = P(A \mid B_7) = 13/27$. It is tempting now to reason as follows: The sets $B_1$, ..., $B_7$ are disjoint, since a child can't be born on two different days of the week. And using this disjointness, it is not hard to show that $P(A \mid B)$ is the average of $P(A \mid B_1)$ through $P(A \mid B_7)$ (just use the definition of conditional probability), so it is also 13/27. But we know that $P(A \mid B) = 1/3$, so something is wrong.

The mistake is that the sets $B_1$ through $B_7$ are not disjoint, because there could be two boys born on different days. So the argument that $P(A \mid B)$ is the average of the $P(A \mid B_i)$ doesn't work.

For more on this, see my book with Stan Wagon, Bicycle or Unicycle, problem 49.

Dan Velleman
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  • I read that particular section of the book. There you bring up an extreme case where one of the sons is born on January 1. You say in there "The chance of her having such a special son is much greater if she has two sons as opposed to one". But isn't that that P(1boy Jan1 | 2 boys) instead of P(2 boys | 1 boy Jan1) ? – Brain Stroke Patient Jul 25 '21 at 18:22
  • Take a look at the computation at the top of p. 153, and the discussion afterwards. It shows how P(2 boys | 1 boy Jan 1) depends on the ratio P(1 boy Jan 1 | 1 boy, 1 girl)/P(1 boy Jan 1 | 2 boys). The sentence you quote shows that this ratio will be small. – Dan Velleman Jul 25 '21 at 21:31