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 Game of luck (Posted on 2004-01-27)
4 people play a game of chance. They each take turns until everyone has taken a turn, then they begin a new round. They stay in the same order every round. Every time a player takes a turn, they have a certain chance of winning. When someone wins, the game ends. They all have even odds of winning a game. The chance of someone winning in any given round is 3/5.

What is the probability for each person to win during their turns?

 See The Solution Submitted by Tristan Rating: 3.4000 (5 votes)

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 solution | Comment 3 of 22 |
Let 'a' be the probability the first person wins a round, 'b' for the second, 'c' for the third, and 'd' for the fourth.

Since all four players have the same probability of winning in any round,
a = (1-a)b = (1-a)(1-b)c = (1-a)(1-b)(1-c)d

This yields b = a/(1-a), c = a/(1-2a), d = a/(1-3a)

The proability of no one winning the round is (1-a)*(1-b)*(1-c)*(1-d) = (1 - 3/5)

(1 - a)*(1 - a/(1-a) )*(1 - a/(1-2a) )*(1 - a/(1-3a) ) = 2/5

(1 - a)*( (1-2a)/(1-a) )*( (1-3a)/(1-2a) )*( (1-4a)/(1-3a) ) = 2/5

1 - 4a = 2/5

a = 3/20
b = (3/20)/(1-3/20) = 3/17
c = (3/20)/(1-6/20) = 3/14
d = (3/20)/(1-9/20) = 3/11
 Posted by Brian Smith on 2004-01-27 13:56:34

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