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Russian Roulette - probability of winning

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Jonathan-Joe | 19:11 Wed 25th Apr 2012 | Science
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I thought it was easy to calculate the probability for each player of winning a game of Russian roulette but my answer is not the as that given by a couple of books I have.
I would be be interested to see the reasoning of others on this; will they agree with me or my books; maybe both are wrong!
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Hi J-J- my reference to JJ in my post at 21:05 was for the other JJ - Joggerjayne - not you.
Yes, spinning makes a difference
There are two of us now ???
There's only one joggerjayne... but there is a Jonathan-Joe
According to my calculations, if the chamber is not spun after each round, each player has an equal chance of winning...

going 1st: 5/6 x (1/5 + 4/5 x 3/4 x (1/3 + 2/3 x 1/2)) = 1/2
going 2nd: 1/6 + 5/6 x 4/5 x (1/4 + 3/4 x 2/3 x 1/2) = 1/2

to explain, being the 1st player that is probability that:
you lucky AND (opponent unlucky OR opponent lucky AND you lucky AND... etc, etc...

I guess it must be equal overall if the chamber is respun too, but it would be better to not go first so that at no point in time have you taken more turns than the other guy.
Like any form of gambling, everyone except the bookies loses in the long run.

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