Why is my logic incorrect for this probability question? What is the probability that two people were born on the same day of the week if it is known that one of two was born on a Saturday
So I split it up into two cases: one where person A is born on Saturday and one where person B is born on Saturday. So for the cases where both are born on Saturday, it would be $\frac{1}{7}$ for the first case ie $\frac{1}{7}$ chance that person B is also born on Saturday (fixing that person A is born on Saturday), and also $\frac{1}{7}$ for the second case. So case one or case two $\frac{1}{7}+\frac{1}{7}$ is the probability. Now I know the answer is $\frac{1}{13}$ and I get it correctly when I draw the dot diagram, but I don't see why I get it wrong with the above logic.
 A: You can apply Bayes' theorem given the condition that one of them is born on Saturday.
For explanation, given the information that one of them is born on Saturday, your sample space is now restricted to all the cases where one of them is born on Saturday or both of them are born on Saturday which is $13$.
$6 + 6 + 1 = 13$ possibilities out of $49$.
Now there is only $1$ of the $13$ possibilities where both are born of Saturday. So you get the probability of $\frac{1}{13}$.
Applying Bayes' theorem,
If $P(A)$ is probability of both of them being born on Saturday and $P(B)$ is probability of at least one of them being born on a Saturday.
$P(B) = 2 \times \frac{1}{7} \times \frac{6}{7} + \frac{1}{7} \times \frac{1}{7} = \frac{13}{49} \, \,$.   The first term is one of them having birthday on Saturday and second term is where both of them have birthday on Saturday. Can you see why $\frac{2}{7}$ is wrong even for $P(B)$?
$P(A\cap B) = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49}$
So your answer is $P(A|B) = \displaystyle \frac {\frac{1}{49}}{\frac{13}{49}} = \frac{1}{13}$
