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Q: Value of an N-Player game with Mixed Strategies ( No Answer,   0 Comments )
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 Subject: Value of an N-Player game with Mixed Strategies Category: Computers > Algorithms Asked by: tjj-ga List Price: \$10.00 Posted: 23 May 2006 16:49 PDT Expires: 26 May 2006 17:55 PDT Question ID: 731821
 ```Consider the following game: N people each hold a different weighted M-sided die. They each truthfully declare the weighting of their die, then play the following game: 1. Each player contributes \$1.00 to a pot of winnings 2. Each player rolls their die 3. Whoever rolls the highest number wins. If more than one person rolls the same highest number, the pot is split between them. Given the number of players, and the weighting of each player's die, is there a polynomial-time algorithm to determine each player's expected value for the game (and if so, what is it)? Example: 3 players, 2-sided die Player 1 rolls a '1' with probability 1.0 Player 2 rolls a '2' with probability 1.0 Player 3 rolls a '2' with probability 1.0 In this case, players 2 and 3 always tie when they roll a '2', and player 1 always loses when he rolls a '1'. Total pot is \$3.00 (each player contributed \$1.00) Expected Value for player 1 is 0 Expected Value for player 2 is 1.5 Expected Value for player 3 is 1.5```