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 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)

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 A footnote to this problem | Comment 8 of 23 |
Had DJ stated that "the chamber is NOT spun after every turn" (as in that infamous scene in the movie "Malcolm X"), then the odds change. (And you'd have to have a hole in your head to play it this way!) The decreasing likelihood of getting to succeeding turns is offset by the fewer and fewer possible empty chambers.....

1st turn: 1/6 for 1st guy to lose, same as before
2nd turn: (5/6)*(1/5) = 1/6 for 2nd guy to lose
3rd turn: (4/6)*(1/4) = 1/6 for 1st guy to lose
4rth turn: (3/6)*(1/3) = 1/6 for 2nd guy to lose
5th turn: (2/6)*(1/2) = 1/6 for 1st guy to lose
6th turn: 2nd guy loses if it gets this far

First guy's chance of losing in this game is 1/2.
Edited on November 10, 2003, 11:53 pm
 Posted by Dan on 2003-11-10 15:59:16

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