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?
Round 1
The first shooter has a 1/6 chance of losing (or 6/36)
The second has a 5/6 chance of getting the gun, then a 1/6 chance of losing (or 5/36)
So the first shooter in the round is 6/5 times more likely to lose.
This follows in any round - since a round without the gun going off can be viewed as one event, like a coin toss, and can be ignored for future probabilities.
Since the first shooter has 6 chances to lose for every 5 chances the second shooter has, it follows the probability the first man will lose is 6/11
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Posted by Lee
on 2003-11-10 10:29:36 |