For two and three, this is six. And finally, we label our leaves. The fair coin now splits into six equally likely leaves, three heads and three tails. For the biased coin, we now have two tail leaves and four head leaves, and that is it. When Bob shows the result, Heads! this new evidence allows us to trim all branches, leading to tails, since tails did not occur. So the probability that he chose the biased coin, given heads occur?