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?
Russian Roulette with multiple cartridges
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Re: Russian Roulette with multiple cartridges
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.
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}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Russian Roulette with multiple cartridges
The solution with no numbers:
The argument seems to work for any number of chambers and bullets.
Spoiler:
Re: Russian Roulette with multiple cartridges
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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Re: Russian Roulette with multiple cartridges
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:
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