Russian Roulette with multiple cartridges

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DataGenetics
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Russian Roulette with multiple cartridges

Postby DataGenetics » Sat Feb 06, 2016 8:35 pm UTC

You're probably familiar with the vanilla variant of Russian Roulette (where a single round is placed in a six cylinder revolver, it's spun and the trigger pulled).

If, instead of one round, two rounds were placed in the revolver, the cylinder spun and the player survives the first trigger pull. Given a choice, is it safer to pull the trigger again immediately, or spin the cylinder a second time before pulling the trigger a second time?

Would your strategy change if you knew the rounds were placed in adjacent chambers instead of entirely at random?

Cauchy
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Re: Russian Roulette with multiple cartridges

Postby Cauchy » Sun Feb 07, 2016 12:58 pm UTC

Let A and B be the consecutive chambers that are fired when the trigger is pulled the first and second time, respectively. The question is: Given that chamber A is empty, what is the probability that chamber B is empty, and specifically how does this compare to the chance of chamber A being empty (which is the chance of surviving on a respin)?

a) Assuming that all possible initial configurations of bullets are equally likely, then there are 6C2 = 15 possible patterns for the bullets, of which 5C2 = 10 miss chamber A, and 4C2 = 6 miss both chambers A and B. The probability of surviving after a respin is 10/15 = 2/3, and the probability of surviving by pulling a second time without respinning is 6/10 = 3/5. Since 2/3 > 3/5, you should spin a second time.

b) If the bullets are in consecutive chambers, then there are 6 possible patterns for the bullets, of which 4 miss chamber A and 3 miss both chambers A and B. The probability of surviving after a respin is 4/6 = 2/3, and the probability of surviving by pulling a second time without respinning is 3/4. Since 2/3 < 3/4, you should not spin a second time.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
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mward
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Re: Russian Roulette with multiple cartridges

Postby mward » Mon Feb 08, 2016 5:52 pm UTC

The solution with no numbers:
Spoiler:
With random bullet positions: If you spin again, you could get the same chamber again and survive.
If don't spin, there is no possibility of getting the same chamber again: all the bullets are still in the smaller number of remaining chambers. So it is better to spin again.

With consecutive bullets: If you spin again, you could get a bullet. If you don't spin you will only get a bullet if your first spin happened to land at the end of the sequence of spaces. If there is more than one space, it is better to not spin again.
The argument seems to work for any number of chambers and bullets.

CharlieP
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Re: Russian Roulette with multiple cartridges

Postby CharlieP » Thu Feb 18, 2016 2:21 pm UTC

I was just about to point out that your chances of surviving if you spin again are actually 5/6, since you've just emptied one chamber by discharging the round that was in it. Oh, wait...
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jasondrake
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Re: Russian Roulette with multiple cartridges

Postby jasondrake » Thu Feb 18, 2016 5:12 pm UTC

Cauchy wrote:b) If the bullets are in consecutive chambers, then there are 6 possible patterns for the bullets, of which 4 miss chamber A and 3 miss both chambers A and B. The probability of surviving after a respin is 4/6 = 2/3, and the probability of surviving by pulling a second time without respinning is 3/4. Since 2/3 < 3/4, you should not spin a second time.

Except for one edge case.

Spoiler:
DataGenetics wrote:... the player survives the first trigger pull.

The riddle does not specify that you survive because the chamber was empty. You may be a lousy shot, have an exceptionally thick skull, or (more probably) encounter a dud bullet. In any case where you survive a loaded chamber (with two bullets in consecutive chambers), you should of course spin again.

Unless pulling on a loaded chamber means you won and the game is over. I'm really not clear on the victory conditions.


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