Derived from example 2.5.10 from Blitzstein and Hwang (page 65).
Assume $P( heads | fair )=1/2$ and $P( heads | biased )=1/4$.
If we let F be the event we draw the fair coin, $A_1$ be the event the first flip is heads, and $A_2$ be the event the second flip is heads, how can I find $P(A_2|A_1)$?
I find myself running into a lot of recursion when trying to parse these conditional probability problems, i.e., $$P(A_2|A_1)={P(A_1|A_2)P(A_2)\over P(A_1)}$$
Which doesn't really get me anywhere.
By LOTP, $P(A_i)=P(A_i|F)P(F) + P(A_i|F^c)P(F^c)$, of course, but the logical jump that I need to do to incorporate the information about F from $A_1$ is escaping me.
Any help?
Let $B$ be the event that the coin is biased and $F$ that the coin is fair. We also assume that the probability of choosing a biased coin or a fair coin are equal.
Then $$P(A_2|A_1)=\frac{P(A_2,A_1)}{P(A_1)}=\frac{P(A_2,A_1|F)P(F)+P(A_2,A_1|B)P(B)}{P(A_1)}=\frac{\frac{1}{4}\frac{1}{2}+\frac{1}{16}\frac{1}{2}}{P(A_1)}=\frac{5}{32P(A_1)}$$
Plugging in the values for $$P(A_1)=P(A_1|F)P(F)+P(A_1|B)P(B)$$ gives $P(A_1)=\frac{3}{8}$
Combining everything gives $P(A_2|A_1)=\frac{5}{12}$