Two teams play against each other in a tournament. The rules of the tournament do not allow any drawn games and the first team to win four games wins the tournament. Therefore, a minimum of four, and a maximum of seven games must be played in the tournament to determine the winner.
Riley supports one of the teams, and wants to end up having lost $1000 if his team loses the tournament, and having won $1000 if his team wins the tournament. Riley wants to wager an amount on each game, and in terms of betting rules he will win that amount of money if the team he bets on wins, and lose it if the team loses. The amount wagered by Riley may not necessarily be the same for any two games.
How much money should Riley bet on the first game?
Note: For the purposes of the problem, assume that Riley will bet on every game until the tournament ends, so that he cannot bet $0 on any game.
When the teams each have 3 wins, we want Riley to be at $0 and to bet $1000. Thus, to determine what he should have and bet at 3,2 we take the desired outcome at 4,2 and 3,3 and average it. This is $1000+$0/2 = $500. Thus, he should have $500 and he should bet $500, either getting to the desired $0 or the desired $1000. Going back this way we get the following, where the first number is the number of wins of the team he supports, then the number of wins of the opposing team, then the money he should have, then the money he should bet.
3 3 $0 $1000
3 2 $500 $500
3 1 $750 $250
3 0 $875 $125
2 3 $500 $500
2 2 $0 $500
2 1 $375 $375
2 0 $625 $250
1 3 $750 $250
1 2 $375 $375
1 1 $0 $375
1 0 $312.50 $312.50
0 3 $875 $125
0 2 $625 $250
0 1 $312.50 $312.50
0 0 $0 $312.50
Thus he should bet 312.50
Edited on December 24, 2008, 5:35 pm

Posted by Joe
on 20081224 17:34:59 