On Sat, Feb 23, 2002 at 06:07:35PM -0000, Alberto Monteiro wrote: > 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.
No, not chosen, "taken". In my previous post, I referred to "Alberto's solution". I meant his original solution, not the one below. > 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: Also, if we change the problem to "chosen" as you have done, then the manner of choosing is important. Do they just choose a group of N people? If so, then the assumption of infinite people needs to be modified. If we want to look at people outside the room, I think the best way to restate the problem is as I mentioned in my previous post, as an infinite sequence of chosen people, with every person having a number 1,2,3,4... > 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? It seems to me that the group you are calculating the survival probability of is changing from line to line. 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? 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 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. > 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? -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.com/
