So I saw the following riddle:
"Alice a Bob play a game using two coins. They flip both coins. If the results of the flips are HT, Alice wins, and if the results of the flip are TH, then Bob wins, but if the results are TT or HH they flip again. They do this until one of them wins. What is the chance that Bob wins?"
Spoiler
The answer is $\frac{1}{2}$
Can someone provide some mathematical evidence as to why this is the case?
I must be looking at it wrong, surely looking at a single game (flip two coins) the results are he wins, she wins or neither win? So how can him winning be a $\frac{1}{2}$?
Let $B$ be the event that Bob wins. Assuming that the coins are fair, conditioning on the first flips yields $$ \mathbb{P}(B)=\frac{1}{4}\mathbb{P}(B\mid HT)+\frac{1}{4}\mathbb{P}(B\mid TH)+\frac{1}{2}\mathbb{P}(B\mid HH,TT)=\frac{1}{4}+\frac{1}{2}\mathbb{P}(B\mid HH,TT)$$
But $\mathbb{P}(B\mid HH,TT)=\mathbb{P}(B)$ since if the flips are $HH$ or $TT$ then the game starts over. Hence $$ \frac{1}{2}\mathbb{P}(B)=\frac{1}{4} $$ or $\mathbb{P}(B)=\frac{1}{2}$.