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In the game of Russian roulette (not recommended by the author), one inserts a single cartridge into the drum of a revolver, leaving the other five chambers of the drum empty. One then spins the drum, aims at one's head, and pulls the trigger.

(a) What is the probability of being still alive after playing the game N times?

(b) What is the probability of surviving (N-1) turns in this game and then being shot the Nth time one pulls the trigger?

(c) What is the mean number of times a player gets the opportunity of pulling the trigger in this macabre game?

Community
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mehrdad
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  • Is the drum spinned everytime, or just once? – AnalysisStudent0414 Nov 06 '15 at 11:41
  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please consider rewriting your post. – Jesse P Francis Nov 06 '15 at 11:46
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    When you spin the drum, does it tend do come to rest with the chambered round in one of the lower positions (that is, not aligned with the firing mechanism) due to gravity? That might upset the naive probabilities a lot. – hmakholm left over Monica Nov 06 '15 at 11:47
  • Too many "whats". What have you tried??? – barak manos Nov 06 '15 at 12:10

3 Answers3

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The player has a $\dfrac56$ chance of survival per game.

a) To survive $N$ games is $\left(\dfrac56\right)^N$

b) $\left(\dfrac56\right)^{N-1}\dfrac16$

c) The answer to (b) is a geometric distribution, with $p$ being the chance of success, (in this case failure!), so the mean is $\dfrac{1}{\frac16}=6$.

JMP
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(1) $({5\over6})^N$ you need to not get shoot for $N$ consecutive turns.

(2) $({5\over6})^{N-1}{1\over6}$

(3) The expected number is $6$ as the probability is $1\over6$.

As for a rigorous proof using the definition, the expected number is

$1({1\over6})+2({5\over6})({1\over6})+3({5\over6})^2({1\over6})+...$

$=(({1\over6})+({5\over6})({1\over6})+({5\over6})^2({1\over6})+...)+(({5\over6})({1\over6})+({5\over6})^2({1\over6})+...)+(({5\over6})^2({1\over6})+...)+...$

$={1\over6}({1\over1-{5\over6}})+{1\over6}({5\over6})({1\over1-{5\over6}})+{1\over6}({5\over6})^2({1\over1-{5\over6}})+...$

$={1\over6}({1\over1-{5\over6}})({1\over1-{5\over6}})$

$=6$

cr001
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  • would you mind explaining the third part?what is your reasoning? – mehrdad Nov 07 '15 at 15:53
  • The expected number of times needed for an event of probability $x$ to happen is basically $1\over x$. This is a very intuitive result (i.e. the expected number of tosses needed for a fair coin to become a head is $2$) but a rigorous proof using the definition of expected number is possible. I will update the answer to include that. – cr001 Nov 07 '15 at 16:19
  • Edited the answer to include a rigorous proof with the definition of expected number. – cr001 Nov 07 '15 at 16:25
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Another way to understand what needs to be done at letter (c):

On (b) we found out the chance of surviving $N$ turns is:

$p(N) = (\frac{5}{6})^{N-1}(\frac{1}{6})$

By definition, the expected value of a variable is:

$<N> = \sum_{N=1}^\infty p(N) * N$

Here we go from one (died on first shot) to infinity, because the game can go on forever.

So,

$<N> = \sum_{N=1}^\infty (\frac{5}{6})^{N-1}(\frac{1}{6}) N = \frac{1}{6} \cdot \sum_{N=1}^\infty (\frac{5}{6})^{N-1} N$

That's the series you needd to calculate; then just use any method (like this or any other suggested) and find that:

$<N> = \frac{1}{6} \cdot 36 = 6$

(Wolfram confirms it)

Luan Nico
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