In Mitzenmacher's Probability and Computing there is the example of:
You are presented three coins. One of them is biased and will show heads with probability 2/3, the other two are fair, and show heads probability 1/2. Say that all three are flipped and the out come is HHT.
First what is the probability that the first coins is biased, call this $E_1$? The second coin is biased, call this $E_2$? The third coin is biased, call this $E_3$? The answers are:
$Pr(HHT | E_1) = \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{6}$
$Pr(HHT | E_2) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{6}$
$Pr(HHT | E_3) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{12}$
Applying Bayes' Law we get:
$Pr(E_1 | HHT) = \frac{Pr(HHT | E_1)Pr(E_1)}{\sum_{i=1}^{3} Pr(HHT | E_i)Pr(E_i)} = \frac{2}{5}$
$Pr(E_2 | HHT) = \frac{2}{5}$
$Pr(E_3 | HHT) = \frac{1}{5}$
My question is, what if we had another coin flip permutation of HTH. So the question is something like, what is probability of $E_1$ given HTH AND given HHT? How would I solve this?