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How many chess games are required to be played if 9 players win 2 games against each other player?

Would the answer = 1152?

I got this because each player plays 8 games in 1 iteration (he cannot play against himself) and wins 1. To win 1 game against every player, he plays 64 games. To win 2 games against every player, he plays 128 games. However, there are 8 other players who do the same so there are (128*9) games played? Is this correct?

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To win a game against every player, (s)he plays $8$ games. –  André Nicolas Aug 16 '13 at 6:05
@andrénicolas so you're saying, 144 games are played? (8 games played for 1 player to win 1 game against everyone else, *2 to win 2 games which gives 16) There are 9 players so 16*9 = 144 –  user115422 Aug 16 '13 at 6:07

1 Answer 1

up vote 2 down vote accepted

Presumably we are asked what is the minimum possible number of games. Draws are a standard part of chess. And if Player A beats B, she is likely to beat B again the next time, or draw. So the actual number of games played until every player has beaten every other player is likely to be very large. After this nod to reality, we calculate.

Each player has $16$ wins. There are $9$ players. Since there are no draws, the number of wins is the number of games.

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So in effect 144 games at minimum are needed to be played. However, in a probabilistic scenario which includes draws, how many games would be needed to be played to ensure a 50% chance that everyone wins 2 games against everyone else? –  user115422 Aug 16 '13 at 6:10
Different problem. Hard to set up realistic assumptions. There are players who could give me a bishop handicap and beat me while drunk. Interesting question could be made, perhaps about a more probabilistic game. –  André Nicolas Aug 16 '13 at 6:14

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