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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 $\frac{1}{2}$?

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    $\begingroup$ Full stops (.) and commas (,) are free, why don't use them? $\endgroup$ Commented Nov 21, 2016 at 18:57
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    $\begingroup$ The result set is = {she wins, he wins} since if neither wins, they nullify the round and move on. $\endgroup$
    – Paichu
    Commented Nov 21, 2016 at 18:57
  • $\begingroup$ if you believe that HT and TH are equally likely then it can be seen by symmetry that neither of the HT or TH target is preferable to the other, so by symmetry it has to be a 1/2 chance of winning $\endgroup$
    – Cato
    Commented Nov 22, 2016 at 9:59

1 Answer 1

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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 $\mathbb{P}(B)=\frac{1}{2}$.

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