Richard Baker wrote: > 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.) If you're taken into the room, what is the probability that you > get out alive? > > 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%. > > 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.)
By this logic, supposing the first batch is killed, the probability of surviving is zero! In other words, you are calculating the probability of getting out alive _given_ that the third batch dies. The prob that anyone in a batch dies is still 1/36, regardless of the size of the batch. Ray (probably) Ludenia.
