Distribution of reward
Two players play against each other up to 5 wins. Whoever wins 5 games gets the whole reward. That means: there can be played maximum 9 games. Every game is equally likely to be won by each player. Due to some problems (weather conditions, lack of time, etc.) the competition cannot be finished. The game stopped at the score 4:2 (4 wins for player A, 2 wins for player B). How should they fairly distribute the reward?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Maximum number of games that can be played at this momment is 3 (before one of the players wins). There are 8 options how this 3 games may end: BBB (player B wins 3 games in a row), BBA, BAB, BAA, ABA, AAB, ABB, AAA. Only 1 one these options leads to win of the player B (i. e. BBB). In every other case the player A would get the fifth win. So, the probability that player A wins is 7:1. Therefore the reward should be distributed in ratio 7:1 in favor of player A.