Things were getting confused, so I am going to recap from my perspective. ***
Here's the original statement of the puzzle: There is a room with a machinegun and a guy with two dice. A person is taken into the room and the dice thrown. If they come up with two sixes then the person is shot. Otherwise he or she is let out and a group of ten people brought in. Again the dice are thrown and if they both come up sixes then the people are shot. Otherwise a group ten times bigger is brought into the room. This "game" goes on until a group is shot. (There is an infinite supply of people. Nobody goes into the room twice.) *** Okay, just for some clarification, I am going to propose an equivalent system for the operation of the room. Inside the room is a computer connected to two room-sized paint sprayers, one containing red paint, one containing blue paint. Every hour on the hour, the computer generates a uniformly distributed random integer, 1 to 36 inclusive. If the number comes up 36, then the computer fires the red paint sprayer. Otherwise the computer fires the blue paint sprayer. The computer has no RAM, no memory, only its program to accomplish the above. (Assume it really is a random number, not pseudo-random; perhaps the computer is connected to a Geiger counter). There is no human intervention with the paint sprayer. *** Here's the original question: If you're taken into the room, what is the probability that you get out alive? The key phrase here is, "if you're taken into the room...". This is equivalent to asking, "if you are in the room on the hour, what is the probability that you get painted blue". It does NOT state, "if you are about to be taken into the room..." *** Here's the first argument: 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%. Since the question is equivalent to, "if you are in the room on the hour, what is your probability of being painted blue", the answer depends only on the computer. The computer will shoot the red paint with a probability of 1/36, and the blue paint with a probability of 35/36. So argument 1 is true, by definition of the problem and definition of probability. *** Here's the second argument: 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 assumes that the "chance" of getting painted blue is equivalent to the total number of people who got painted blue divided by the total number of people who got painted. This is an incorrect assumption. This is sort of like the gambler's fallacy, which assumes that because the die has not come up 6-6 for a while, that a 6-6 is more likely because it is "due". Of course, for fair dice, it is not more likely. The probability is still 1/36 of getting 6-6, no matter what happened in the past. Therefore, this argument is not answering the question asked, but rather another question, "what is the chance of NOT being painted red, for people who haven't yet entered the room and haven't been painted? [question B]" But the system being calculated is ill-defined. The problem is in the "infinite supply of people" and how they are chosen to enter the room. This MUST be defined before a rational calculation of question B can be made. The definition, since it wasn't supplied in the question, is somewhat arbitrary, and someone else may very well like to use a different definition. But we must be clear, this is NOT answering the original question asked, but rather another question (question B). The simplest way to define the people outside the room is that they are an infinite sequence, a queue, waiting their turn. Number the person who is next to enter the room 1, and the next after that, 2, 3, 4... Then the probability that person N is NOT painted red is equal to 1 - ( ( 35 / 36 ) ^ ( int_ceiling[ log10[ 9*N + 1 ] ] - 1 ) ) / 36 Here are a few numbers from that equation N P[not red] ------------------ 10^0 0.972 10^1 0.973 10^2 0.974 10^10 0.979 10^100 0.998 -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.com/
