Oops. Ignore my last message and replace by: Richard Baker wrote: > >Alberto said lots of things I agree with and then: > Ah, ok. I forgot the _if_ part. So, the problem is, really, what is the chance of survival of one of those unhappy people that were chosen to enter the room.
Of course this problem depends on the number of runs that it took before the machine gun was fired - which is another variable, with distribution: P(F = 1) = p (p = 1/36, but I like to state the problem in general terms) P(F = 2) = (1-p) p P(F = 3) = (1-p)^2 p <===== here I wrote it wrongly on past message etc P(F) = (1-p)^(F-1) p For each F, the probability of survival is different, namely: PS(F = 1) = 0 PS(F = 2) = 1 / (1 + k) (k = 10) PS(F = 3) = (1 + k) / (1 + k + k^2) etc PS(F) = (1 + k + ... + k^(F-2)) / (1 + k + ... + k ^(F-1)) = = (1 - k^(F-1)) / (1 - k^F) So, the correct answer is the combination of the two, namely the sum of P(F) * PS(F) = (1 - p)^(F-1) p (1 - k^(F-1)) / (1 - k^F) I don't think this sum can be expressed in familiar functions, but I guess it converges to something between 1/(k-1) and 1/k Alberto Monteiro
