Two men are playing Russian roulette using a pistol with six chambers.
A single bullet is used and the chamber is spun after every turn.
What is the probability that the first man will lose?
In general if p is the probability associated with the gun going off then
Shooter 1 has p probability of shooting himself in the round
Shooter 2 has (1-p)p probability of shooting himself in the round.
The first man is therefore always p/(1-p)p (cancelling gives) 1/(1-p)more likely to 'lose'.
Probability first man will lose is given by
And for the second man;
(1-p) / (1-p)+1
So in general the probability that the first man will lose is 1/ (2-p)
Posted by Lee
on 2003-11-10 11:45:12