Ok, let me re-state the pseudo-paradox. One person enters a room. Now a loop begins: 2d6 are rolled, and if it gets 12, everybody in the room is killed, and the experiment ends. Otherwise, those people in the room become free, and 10x their number enter the room.
For any person that entered the room, what is the probability that he left it alive? Erik Reuter wrote: > >> 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 > >Looks good so far. > >> 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) > >Whose survival probability are you calculating here? > For each number F - the index number of the firing of the machine gun - there is a different survival probability for the people that entered the room, namely, the number of people that entered the room in the earlier rounds divided by the total number of people that entered the room. >It seems to me that >the group you are calculating the survival probability of is changing >from line to line. > It is. For each F >= 2, the "universe" grows by k^(F-1), while the set of people that survived grows by k^(F-2) >In order to clarify whose survival we are talking about, could you state >it in terms of my notation of an infinite sequence of people? > I can, but only implicitly. >Are you >talking about person N in the sequence? (I don't think so). Are you >talking about the first (1 + k + ... + k ^(F-1)) people? That's what it >looks like to me, which is why I said your chosen group is changing with >F. > The N-th person in the queue can only be killed if (1 + F + F^2 + ... + F^(N-2)) < N <= (same sum) + F^(N-1) Explicitly, for each N the killing number F is: N = 1 -> F = 1 N = 2 -> F = 2 up to N = 11 -> F = 2 N = 12 to N = 111 -> F = 3 etc >The answer to my restatement of the question, the probability of >survival of person N in the sequence, is > > 1 - ( ( 35 / 36 ) ^ ( int_ceiling[ log10[ 9*N + 1 ] ] - 1 ) ) / 36 > >which is the approach taken in Alberto's first email. > Whatever [I didn't check the numbers] :-) But this is a different problem, namely: what is the chance that the N-th person of the queue will survive? >> 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 > >That is the number that argument 2 seemed to be headed for. But to me it >is a strange number. I can't come up with a well-defined question that >has that as its answer. Can you? > Yes - as I did in the introduction. Alberto Monteiro
