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 Grand Anticipation (Posted on 2008-12-24)
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.

 No Solution Yet Submitted by K Sengupta Rating: 4.5000 (2 votes)

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 A better guess... maybe Comment 4 of 4 |

Okay, as near as I can figure it, if Riley bets \$997 on the first game then \$1 on the 2nd, 3rd, and 4th games and then, on each subsequent game, bets ABS(1000-ABS(CurrentBalance)) he should end up winning or losing \$1000 approximately 80% of the time.....  I think (i'm still working on proving it) ;-)

Hey, what do you expect?  It's Christmas Eve!

MERRY CHRISTMAS EVERYBODY!

 Posted by Sing4TheDay on 2008-12-24 17:45:03

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