>> 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: > >The survival probability of the people that have entered the room is >always the same: 35/36. It cannot be otherwise, by definition. > Err... No. This would be true if the experiment didn't end when the machine fires. But when the machine fires, there's a bigger number of dead bobies than survivors. That's the problem with _conditional_ probability. A simpler example: both your grandmothers have dark eyes, both your grandfathers have blue eyes. Both your parents have dark eyes. You have dark eyes. Your wife has dark eyes. Her mother has blue eyes. Your first kid has dark eyes. What is the probability that your _second_ kid has blue eyes? If I had asked about your _first_ kid, the answer would be a trivial 1/4. But it's not 1/4 for the second :-) > >But I think the answer to my question is that you aren't calculating >the probability of a specific group in the sequence, but rather some >strangely varying group. It looks like an ill-defined system to me. > No, it's not. Given that the probability that the machine never fires is lower than any number [IOW, it converges to zero], we have one well-defined set for each firing-index number. > >I don't think so. Maybe you should try to say it more explicitly. > Parrdom mee, buth I am nought a naytive English speekar. I kennt make it moor zplissit Alberto Monteiro
