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Going through MITx: 6.041x Introduction to Probability on edX and came across the King's sibling problem. I could comprehend the official answer of $\frac{2}{3}$. However, I tried a different approach and ended up with a different answer:

$$P(\text{sibling is G}) = P(K_1) \cdot P(G_2 | K_1) + P(K_2) \cdot P(G_1 | K_2) = 0.5 \cdot 0.5 + 0.5 \cdot 1 = \frac{3}{4}$$

where:

$K_1$: King is the $1^{st}$ child
$K_2$: King is the $2^{nd}$ child
$G_1$: $1^{st}$ child is a girl
$G_2$: $2^{nd}$ child is a girl

$\bullet$ Since we have no other information on the King's sibling, I suppose its fair to assume
    P($K_1$) = P($K_2$) = 0.5
$\bullet$ Also, since the gender of the second child should be independent of the first, P($G_2$ | $K_1$) = 0.5
$\bullet$ And assuming the first boy is crowned as the king, if the king is the $2^{nd}$ child, P($G_1$ | $K_2$) = 1

I suspect there's definitely something fishy here but I can't put my finger on it. Is it something to do with the assumption $K_1 = K_2 = 0.5$?

jvdhooft
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Aligator
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  • We have to assume that both children of the previous monarch are still alive, and that they are using the old sexist rule that a male child has precedence over a female (which, btw, no longer applies in the British royal family for those born after 28 October 2011). – Robert Israel Sep 29 '16 at 05:40

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It's wrong to assume $P(K_1) = P(K_2) = 1/2$. If the children are a boy and a girl, the king is the boy and equally likely to be first or second born. But if the children are both boys, the king must be the first. So $P(K_1) > P(K_2)$.

It's also wrong to assume that the second child's gender is independent of the first. Since we are given the fact that there is a king, if the first child is female the second must be male.

Robert Israel
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  • I get what you're onto Robert. But wouldn't P($K_1$) > P($K_2$) actually be P($K_1$ | given info on sibling) > P($K_2$ | given info on sibling). For absolute probabilities P($K_1$), P($K_2$) assuming no other events, shouldn't P($K_1$) = P($K_2$) = 0.5? – Aligator Sep 29 '16 at 06:05
  • No, it wouldn't. There are three equally likely cases where there is a king: $(G_1 B_2), (B_1 G_2), (B_1 B_2)$, and $K_1$ occurs in two of them. – Robert Israel Sep 29 '16 at 07:03