Richard Baker wrote: > >This is true (and thanks to Alberto for pointing this out off-list too). >However, what's wrong with the argument when restricted to a particular >run of the game that terminates after a finite number of batches? > >Rich, who genuinely doesn't understand this. > Ah, I thought you knew the answer, that's why I answered in private.
The truth is that when you create those kind of random variables, sometimes you get one that has _no_ mean. For example, let's imagine this (simple) experiment: You start with number 1. The loop is: you throw a dice. If it gets 6, the experiment ends. If you get any other number, the number doubles. This generates a random variable, X, defined as the number you get when the experiment ends [and it can be proved that for every probability P, there is a time T such that the probability that the experiment ends before T is greater than P -- this is the best we can affirm mathematically that comes closest to saying the "the experiment must terminate one day"] Ok, back to X. What is the distribution of X? P(X = 1) = 1/6 P(X = 2) = 5/6 * 1/6 P(X = 3) = 0 P(X = 4) = 5/6 * 5/6 * 1/6 etc [[ btw, the machine gun problem is that same, for different probabilities and a different rate ]] Now, for _this_ random varible, we can speculate on its mean, or on its standard deviation, etc. So, rewriting the problem in general terms [p = chance that the experiment ends, k = multiplication factor] we have: P(X = 1) = p P(X = k) = (1 - p) p P(X = k^2) = (1 - p)^2 p and, in general P(X = k^n) = (1 - p)^n p This variable is well defined for all purposes. However, calculating its mean is equivalent to computing sum from n = 0 to infinite of (k^n (1-p)^n p) It's immediate that this sum will diverge if k (1-p) >= 1. Now, back to Argument 1: Argument 1. You are killed if two sixes are thrown. This happens 1 in 36 times. Therefore your chance of getting out alive is 35/36 = 97%. 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 Argument 2. Most people who are taken into the room are killed, therefore you are very likely to die. For example, suppose the third batch are killed. Then 100 people who go into the room die and 11 survive. The chance of getting out alive is then 11/111 = 9.9%. (Working out the true probability is left as an exercise.) This argument is wrong, too, because it does not consider the chance that you will not be chosen to enter the room. The correct answer is that, since there is an infinite number of people, and you have _no_ idea how the peope are sorted to enter the room, you have no idea about the chance of being shot. In the worst case [if Murphy made it so that you are the first to enter the room] your chance of dying is 2.8%.But it can be as good as you wish; if you are the googol-th people in the queue, you have a much smaller chance of being killed [easily computed from the formula above, it's P(X = 10^100) = (1-1/36)^100 * 1/36 = 0.17%] Alberto Monteiro
