I have the following problem:
A coin is tossed until it gives either 10 heads or 10 tails. Player A bets on 10 heads and player B bets on 10 tails. The game is unexpectedly interrupted after 15 tossings with 8 heads and 7 tails observed. What would be the fair ratio to split the prize pool between player A and B?
I know the answer to that question and I know how to calculate the probability of 8 heads after 15 tosses. However I do not understand how to calculate the answer. Any help is highly appreciated
Update: as Lulu mentioned below those 15 toses are irrelevant. We have to try 4 toses more. If we enumerate them we will get 11 possibilities to get 2 H and 5 possibilities to get 3 T. Thus we should divide the prize fund in 11/5 proportion.
The idea is that you should divide the prize according to the probability that a player would eventually have won. If, say, the probability that $A$ would eventually have won was $p$, then you give $p$ times the prize money to $A$.
Thus: The question here is: "What is the probability that you get $2$ Heads before you get $3$ Tails?"
Note that this must be decided in $4$ tosses. It follows that it is equivalent to the question "What is the probability that you get at least $2$ Heads in $4$ tosses?" That is easily seen to be $\frac {11}{16}$, so that is the portion of the prize you should give to $A$.