All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Quit While You're Ahead (Posted on 2003-11-10)
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?

 See The Solution Submitted by DJ Rating: 3.5556 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 completeness | Comment 6 of 23 |
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
1/ (1-p)+1
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

 Search: Search body:
Forums (0)