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is odd, chances
of losing are (n+12nn. n"12ns) - it's the disadvantage to go first? In Russiaian
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Russian Roulette, the game you play with a revolver that you have a 5/6 chance of survival, 1/6 chance of being swept up with a broom. No, in fact it's illegal and is usually a form of torture.
The order of blank chambers and bullet are determined in the very beginning of the game. For 2 players and n chambers player 1 loses if the round is in the odd-numbered chamber. Hence if n is even, chances are 50/50 and if n is odd, chances of losing are (n+12n,n"12n) - it's a disadvantage to go first.
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