I've had a few thoughts on this issue, and want to expore this using more english and less math-ese.
I agree that as the question is stated, if you are brought into the room your odds of survival are 1/36. It doesn't matter how many people are in the room with you, the odds don't change. So the question, as stated, is solved. But, as Alberto and Erik found, there's another similar question that has a different answer. I think a good way to think about it is to imagine 3 rooms. First a waiting room, containing N people. For convenience, N will always be 1, 11, 111, 1111, 11111, etc...just so we can keep the math simple. But as we will see, the number we choose for N is very important. Next is the shooting room. Groups people are brought into the room, two dice are rolled, and if 36 comes up they are shot. If no one is shot, then ten times as many people are brought in and the dice are rolled again. When a group is shot, the experiment is over. Then there is a debriefing room, where the people who have been taken into the shooting room and weren't shot wait. So...the question is, what are the odds for an individual to be in either the waiting room or the debriefing room when the experiment ends? And of course, this depends critically on the number of people in the experiment. You must have a very large number of people in the waiting room if you want to ensure that a good chance that the experiement will end before you run out of people. But if you have very large numbers of people in the waiting room, the odds that any one particular person in that group will be shot become smaller and smaller. So, another way to ask the question is: Given a certain number of people--N--in the waiting room, what is the average number of people that will be killed in the experiment? Without knowing N there is no way to figure this out. Because many of these trials are going to terminate due to lack of people before anyone gets shot. So, the average number of people killed for N=1 is 1/36, or .0277778, about 3% And I believe the number for N=11 is ((1/36) + ((10/36)*(35/36)))/2, or .14888889. Therefore, if you are in group N, your chances of being killed are .1488889/11, or 0.0135, a little more than 1%. The number for N=111 is ((1/36) + ((10/36)*(35/36)) + ((100/36)*(35/36)*(35/36)))/3, or about .973, so your odds of being killed are .973/111, or .00877, less than 1%. The number for N=1111 ((1/36) + ((10/36)*(35/36)) + ((100/36)*(35/36)*(35/36)) + ((1000/36)*(35/36)*(35/36)*35/36)))/4, or 7.11, so your odds of being killed is 7.11/1111, or .0064, or a .64%. So...it seems that the odds of being killed get smaller and smaller the larger N gets. Or am I doing the math wrong? Maybe someone can set up a spreadsheet and calculate the numbers for each N, I don't know how to do that and I'm tired of doing it by hand. But the trend is clear. The more people in the waiting room, the more will be killed on average, but more will also survive. And the reason is that 36 is larger than 10. If you would have to increase the sample taken into the room by a factor of 36 rather than 10 if you want to have the same percentage of people killed regardless of N. ===== Darryl Think Galactically -- Act Terrestrially __________________________________________________ Do You Yahoo!? Yahoo! Greetings - Send FREE e-cards for every occasion! http://greetings.yahoo.com
