Russian Roulette

A simple game. On each turn one chance in six to lose, players take turns until someone loses. How likely is it that the player who starts loses?

...one in six? probably wrong

Chances of first player losing on first shot = 1/6

Changes of first player losing on second shot = 1/6 * the chance of other players not losing before the gun returns to the first player. Let 'p' be the number of players. Chances of first player losing on his second shot = 1/6 * (5/6)^(p-1)

Chances of first player losing on third shot = 1/6 * (5/6)^(2p-1)

And so on.

The total probability of the first player losing the game is the sum of the series
E(n=1 to infinity) 1/6 * (5/6)^(p(n-1)-1)

I'm sure that expression can probably be simplified, but it's beyond the limits of my meagre mathematical ability.

You are right, of course.

It's a geometric sequence, and they would solve like this.. if S is the sum then

S = 1/6 + 1/6 * (5/6)^2 + 1/6 * (5/6)^4 + ...
S * (5/6)^2 = 1/6 * (5/6)^2 + 1/6 * (5/6)^4 + ...
-------------------
S - (5/6)^2 S = 1/6

as almost all the terms on the left side cancel out when the lower equation is subtracted from the upper.

11/36 S = 1/6
S = 36 / (6*11) = 6/11



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Another way to do it, without referring to sums of infinite geometric sequences:

The first player loses if the 1st, 3rd, 5th etc. shot is the unlucky one.
The second player loses if the last shot is 2nd, 4th, 6th etc.

Every shot being the unlucky one comes with a probably of 5/6 of the one before; that is, one more "click" before boom. 5/6 can be taken as a common term, so the probability of player 2 losing is 5/6 is the probability of player 1 losing. Thus,

P(1st loses) + P(2nd loses) = 1
P(1st loses) + 5/6 * P(1st loses) = 1

from which P(1st loses) = 6/11