I have a basic question in probability theory. The question is about a Russian roulette where we spin the chambers before each turn.

The question asks "what is the probability that the player who started will be killed, in the long run?", and similarly for the second player.

The thing that confuses me a little bit is the meaning of "in the long run". The probability of staying alive after a long period of time is quite low (distributed geometrically). But how do I get the probability from that?

Thank you.

  • $\begingroup$ Just to clarify, you mean Russian Roulette between two players, where the gun is passed back and forth? And there's a random, geometric chance to be shot? $\endgroup$ – theREALyumdub Oct 31 '18 at 0:42
  • $\begingroup$ Yes, the game is between two players, where the gun is passed back and forth. I just meant that, in my understanding, the probability of staying alive after the k-th trial is distributed geometrically $\endgroup$ – Mr. Tea Oct 31 '18 at 0:47

The meaning must be that the players take turn pulling the trigger until one of them hits a live round. The survivor then stops playing. We're after the probability of the first player being the overall loser. Call this $q$.

Now the conditional probability $$ P(\text{second player dies}\mid\text{first player survives the first round})$$ must be $q$ too, because the situation is symmetric after the trigger has been pulled once without anyone losing. But this means that $$ P(\text{second player dies}) = \frac 56 q $$ We also know that exactly one player will eventually die, so $$ q + \frac56 q = 1 $$ This linear equation is easy to solve, giving $$ q = \frac 6{11} $$

  • $\begingroup$ Thank you for the answer. Can you please explain why Pr(the second player dies)=5/6q? $\endgroup$ – Mr. Tea Oct 31 '18 at 7:35
  • $\begingroup$ How does this solve the problem in the infinite "long run" case? If both the first and second player survive their respective first rounds, does this probability account for the first player dying after being safe once? $\endgroup$ – theREALyumdub Nov 1 '18 at 12:51
  • $\begingroup$ @theREALyumdub: Yes, as you can see because $\frac{6}{11} \ne \frac{1}{6} $. $\endgroup$ – Henning Makholm Nov 1 '18 at 12:57
  • $\begingroup$ @Henning Makholm I perceive this calculation as representing the results after both players have been under the gun once. What makes the argument distinct, the claim of symmetry? $\endgroup$ – theREALyumdub Nov 1 '18 at 13:59
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    $\begingroup$ @theREALyumdub: The calculation derives the probability before starting the game that the first player will be the eventual loser. $\endgroup$ – Henning Makholm Nov 1 '18 at 14:01

I think you can solve it like this - first note that it isn't an even game. You want to be the second person behind the barrel, I'll call them the first player (do they shoot the other?).

To be specific: the game ends when someone dies, with say a random probability 0 < p < 1 on each fire attempt. Therefore, on the nth "turn" after the first, the chance that is the killing shot is p^n (keep in mind p is less than one, so this is smaller than p^1 in the long run)

If you split the turns between the players, you can unravel the infinite geometric series

$$ \sum_{n=1}^\infty p^n $$ $$ = \sum_{k=1}^\infty p^{2k} + \sum_{k=0}^\infty p^{2k+1} $$

Where the first player (who fires on odd terms, and thus has a chance to, uh, win on those terms) gets the right sum, and the other gets the left sum.

So the problem is to compare the size of each sum. That would give you the odds for each player. The thing to realize is that the right sum, times p, is actually the same as the left (add one to each odd number exponent. What do you get?) Since p is less than 1, the right sum is the bigger value, and thus the higher chance of survival.

If you assume a 6 shot revolver, with one bullet, I guess that means the relevant probability for the first under the gun is a sixth of the other guy's. So he loses one out of seven times.

  • $\begingroup$ The first player loses immediately one out of six times. The probability for him to lose at all cannot possibly be "one out of seven" which is smaller than that. $\endgroup$ – Henning Makholm Nov 1 '18 at 14:03

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