you're right up until that sentence. the odds will fluctuate with the decrease in population to favor either side, but will eventually end up at 50% each again. as there will never be an odd number of killers, there will not be a murderous survivor. they'll kill each other
The only way for a killer to die is if he meets another killer, in which case they both die. This means that if we start with ten killers they will eventually all kill each other off (regardless of the ratio to pacifists); however, were we to start with 11 killers and 11 pacifists (as posed by Bonanova) there would be one killer left. No matter how the population decreases affect the odds, killers can only be killed when they meet another killer. So if we have an odd number of killers in the initial population we will always be left with 1. After we're down to 1 killer it really doesn't matter how many pacifists are left (0, 3, 11), because they will eventually meet the lone killer until he is all that's left. I stand by my theory of if there is an even number of both groups (i.e. 10 to 10) the odds are everyone dies (but it would be possible for one or more Pacifists to survive if the killers killed themselves off first), but if there is an odd number of both groups (i.e. 11 to 11) then a lone killer will survive.
Edited by Aunt Minta, 26 March 2008 - 06:11 PM.