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 reply to
re: A footnote to this problem by Popstar Dave)
LOL !! Nice thought, Popstar Dave !! The humanitarian spirit is alive and well on perplexus.
And how about this alternate game, in the interests of life and peace ?
"Two men are playing Irish roulette using a pistol with six chambers. A single, harmless blank cartridge is used and the chamber is spun after every turn. What is the probability that the first man will lose his bet and have to pay the other guy a dollar?"
Then the answer is:
"As the chamber is spun after every turn the probability of losing a dollar by hearing the harmless sound of a blank going off on any turn is (1/6); the probability of a silent miss is (5/6). Player 1 could lose a dollar on the first turn. This has a probability of (1/6). If not, he could lose his dollar missing on the first turn, then the second player also misses, then the first player loses his dollar when the blank goes off harmlessly on their second turn. This has a probability of (5/6)²(1/6). If this still doesn't happen then they could lose their small dollar bill by both players missing on their first two turns, then player one hearing the little blank go off on their third turn. This has a probability of (5/6)^4(1/6). As this pattern continues indefinetely, the probability of player one hearing an innocuous bang and losing a dollar is the total of the infinite series:
(1/6) + (5/6)^2(1/6) + (5/6)^4(1/6) + ... + (5/6)^n(1/6).
This equates to 6/11 or a 54.545% chance of the first player being embarrassed by the exploding blank and feeling the chagrin of losing a dollar."
There we go, no dead people! :)
Edited on November 10, 2003, 8:31 pm

Posted by Dan
on 20031110 20:30:57 