Share a prize in a chess match Two friends, Alice and Bob, are playing chess. Bob leads with 2 games to 1. The winner is the first of them which wins 3 games and he/she will get 100 $. Suddenly the fourth game is intrerrupted and Alice and Bob decide to share the prize. How many dollars will get each of them? 
The answer is: Bob will get 75, Alice will get 25. I would like to know some hints or an elementary solution because I have just started the study of probabilities.
 A: Badly worded question but assuming that Bob was winning the 4th game and that by prize you mean the $100$ dollars and by sharing the prize you mean Alice will get a chunk for her winning of $1$ game, then it somewhat makes sense that Bob would get $75$ dollars and Alice would get $25$ dollars.
However there are many interpretations/variations of these "rules".  First you never stated what $3$ games they had to win.  If Bob and Alice played a game of checkers immediately before the chess games and Bob won, then after winning the 2nd chess game Bob would have won $3$ games thus be declared a winner.
Another problem with your wording is you said Alice and Bob decide to share the prize but what is the prize?  You didn't say how they would share it either ($50/50, 75/25$, x/y...).
Another interpretation of this (badly) worded question is it can be considered that they were "tied midway" in the 4th game so the "score" was $2.5$ games (Bob) to $1.5$ games (Alice), therefore, Bob would have only needed to win $0.5$ games to win the $100$ bucks but Alice would have needed to win $1.5$ games therefore it was $3$ times as likely for Bob to win the $100$ bucks as it was for Alice but this is an arbitrary interpretation.
Yet another answer that would make other answers here incorrect is the 4th chess game could be considered "dead even" and halfway thru so the effective score would have been $2.25$ games for Bob and $1.25$ games for Alice, thus NOT making it $3$ times as likely for Bob to have won the tournament, thus NOT making it a $75/25$ "split" pot but something like $70/30$ instead.
A: Suppose Alice wins any game against Bob with probability $p$, independently of all other games. The only way that Alice could have won the tournament is by winning the next two games, which occurs with probability $p^2$. Thus, it makes sense to say she should get $p^2\cdot100$ dollars, while Bob gets the rest, since this is the expected amount of money she would have gotten had they kept playing.
If we in addition assume that $p=\frac12$, so that Alice and Bob are equally skilled, then this means Alice gets \$25, and Bob gets the rest, \$75. 
