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Russian Roulette

What is the probabillity that the person taking the first shot in a game of Russian Roulette will lose? I am assuming the the gun is a six shooter and the barrel is spun after every trigger pull.

james
Monday, February 21, 2005

1/6 first person, 5/6 * 1/6 second person, 5/6 * 5/6 * 1/6 third person, ...

tobi
Tuesday, February 22, 2005

So what number does this series converge upon?

james
Tuesday, February 22, 2005

I would say it converges to 0.

But if it's your turn to shoot no matter how many people have shot before you your chance is always 1/6.

tobi
Wednesday, February 23, 2005

I should have clarified that the game was to be played with only two people.

james
Wednesday, February 23, 2005

Here is one way to determine the answer: Let p represent the probabillity that the guy taking the first shot eventually loses. P= 1/6 + 5/6*(1-p), solving this gives p= 6/11

John
Wednesday, February 23, 2005

james - you mean two people play russian roulette until one loses and you want to know the probability that the guy who is starting loses? hmm... i will have to think about it.

john - i don't get it. can you explain your solution?

tobi
Thursday, February 24, 2005

The person taking the first shot can lose in the 1st shot, 3rd shot, 5th shot, 7, 9, and so on

1st shot, probability of losing, p, is
    1/6

3rd, p is 5/6 * 5/6 * 1/6 =
    1/6 * 25/36

5th, p is 5/6 * 5/6 * 5/6 * 5/6 * 1/6 =
    1/6 * (25/36) ^ 2

7th, p =
    1/6 * (25/36) ^ 3
...

sum them all up
total p = 1/6 * ((25/36)^0 + (25/36)^1 + (25/36)^2 + ...)

use infinite geometric series summation
1 + r + r^2 + r^3 + ... = 1 / (1 - r), for -1 < r < 1

let r = 25/36

we get p
= 1/6 * ((25/36)^0 + (25/36)^1 + (25/36)^2 + ...)
= 1/6 * 1 / (1 - 25/36)
= 1/6 / (11/36)
= 1/6 * 36/11
= 6/11

This is how I would do it, which is no where near as elegant as John's solution.

His solution must use some rules of probability that I forgot.

If p is the probability that the first guy loses, then 1-p is the probability that the first guy wins.  It is also the probability that the second guy loses.

5/6*(1-p) turns out to be the probability that the first guy loses after he survives the first shot.

JHY
Friday, February 25, 2005

In fact, if the russian roulette is played by the rules, the probability of getting shot is significantly less than 1/6.

Reason for this is that when you spin the barrel, you must let it spin freely. If you do this, gravity will make sure that the bullet is most likely found in the lower position. This, of course, assumes that you hold the gun as you usually do.

Here in Finland we had one accident due to a couple of guys playing the Russian roulette, but holding the gun such that the spin of the barrel was inhibited. I don't know if they have been nominated for the Darwin awards yet.

L
Thursday, March 03, 2005

Wow, I like John's solution. Quick and recursive at the same time.

1/6 of the time, the first guy loses.
5/6 of the time, the game becomes identifical to the initial problem with the roles reversed.

WanFactory
Monday, March 07, 2005

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