Earlier quoted context omitted.
The murder question sounds quite interesting, but I'm not sure I understand it. You have n people (initially n=13) in a circle, firing shots that are fatal with probability p (here we apparently fix p=1/6, but I guess that for large n the final answer doesn't depend on p provided it's neither 0 nor 1?), and then 1 fires at 2, then 2 (if alive) or 3 (if not) fires at 3/4, and so on. And you're person 1 (this isn't sta…
Ok, I talked to the engineer. I got it mostly right but everyone shoots to the right as soon as the lights go out. The question is "what's the probability you die?". Edit: You can also challenge people to think about the problem where everyone fires at exactly the same time OR random order since people have different reaction times. Edit 2: "The probability p(n) that a permutation of n things is a derangement -- whic…
If everyone shoots simultaneously (so in particular everyone does get the chance to shoot) then I die iff the one person shooting at me hits me. Probability equals probability that a given shot hits (so in this case 1/6). No dependence at all on the number of people.
If everyone shoots sequentially, this seems just like what I described above. Probability of death is now p/(1+p) instead of p, at least if you're first to shoot and n is very large. (Unless something's very broken in the heuristic argument I gave. Let's try another. First approximation says a fraction p of people die. But that's not quite right because people who die don't get to shoot, so next approximation says we get p(1-p). Next approximation says we get p(1-p(1-p)). Etc. We can either solve the obvious equation, or else notice that we're getting more and more terms of the binomial expansion of p/(1+p).
I don't see anything here that doesn't look, in a crude approximation, like a fraction p of people dying (p, again, is probability that a given shot hits, which in this case is 1/6).
I must be misunderstanding something in the problem statement here. Perhaps it would be clearer if I'd seen the movie?
Oh, what about this version? You shoot first, things proceed cyclically, and we keep going until just one person is left. What's the chance that it's you? Naively it seems like this should be approximately 1/n no matter what p is; shooting first could confer some advantage but surely it can't be much for large n. So this can't yield anything like 1/e either. Drat.