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 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 >[Argument 1] > >> This argument is wrong, because there is a chance that you will >> _not_ be chosen to enter the room any time, because the experiment >> ends when the machine shoots > >If the argument is wrong, it's not wrong for that reason because the >question I posed was: "If you're taken into the room, what is the >probability that you get out alive?" > >[Argument 2] > >> This argument is wrong, too, because it does not consider the >> chance that you will not be chosen to enter the room. > >Again, notice the "if" in the question I asked. > >Now, if we restrict attention to a particular run then (with probability >1) only a finite number of batches will be taken into the room. If we >just look at such a run, then what is wrong with argument (2)? > See above. Argument(2) is wrong because we must compute a series. Alberto Monteiro
