On Thu, Feb 28, 2002 at 10:40:21AM -0800, Darryl Shannon wrote: > So, the average number of people killed for N=1 is 1/36, or .0277778, > about 3%
Yup. > 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%. Nope. That final 2 is incorrect. To find the average number of people killed, sum up the products of the probabilities with the value you are looking at. In this case, there are three possibilities: 1/36 : 1 person is shot 35/36 * 1/36 : 10 people are shot 1 - ( 1/36 + 35/36 * 1/36 ) [call this x] : 0 people are shot avg. # killed = 1/36 * 1 + 35/36 * 1/36 * 10 + x * 0 When you calculate the probability of being killed, you are assuming that the people in the waiting room are randomly selected to enter the shooting room every time new people are needed to enter. I will assume the same, but I just wanted to note that this assumption was implicit (and SOME assumption is necessary to calculate the probability). > 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%. Again, no divide by 3. Should be 3 times your calculations above. > 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%. Lose the 4. > So...it seems that the odds of being killed get smaller and smaller > the larger N gets. Despite the mistake, the conclusion is true. In fact, as N goes to infinity, the probability of being killed does approach zero. If p=1/35 and q=1-p=35/36, M=10, and j = log_M[ (M-1)*N +1 ], and assuming N is such that j is an integer, then the average number of people killed is ( p / ( M*q - 1 ) ) * ( (M*q)^j - 1 ) and the probability of being killed is ( (M-1)*p / ( M*q - 1 ) ) * ( (M*q)^j - 1 ) / ( M^j - 1 ) Note that as N goes to infinity, j goes to infinity, and the part of that expression which depends on j is ( (M*q)^j - 1 ) / ( M^j - 1 ) Note that M*q is greater than 1 for M=10 and q=35/36, so as j goes to infinity, this is equal to (M*q)^j / M^j = q^j which since q < 0, goes to 0 as j goes to infinity. > The more people in the waiting room, the more will be killed on > average, but more will also survive. Yes. > 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. No. The probability of being killed always approaches 0 as N goes to infinity, regardless of M. -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.com/
