Whenever a hawk meets a dove, the dove is killed. Whenever two hawks meet, they fight to death, and both are killed. And if two doves meet, nothing bad happens.
There are H hawks and D doves, and you are either a hawk or a dove. Assuming that meetings are random, what are your chances of survival?
The only remaining problem is the expectation for the number of doves surviving the last two hawks in case of an even number of hawks. Lets call this number M, then the chances of survival for doves are zero in case of an odd number of hawks and (M/D) in case of an even number of hawks. I agree to the earlier mentioned chances of survival for hawks ((1/H) in case of an odd number of hawks and zero in case of an even number of hawks. These probabilities are already the good answers.
So we only need to calculate M for the final solution. The answer maybe found by simulation in an Excel sheet, however, we can also calculate a direct guess for M.
First it is important to understand that the answer does not depend on the ratio H/D at the start. In case of a large ratio (lots of hawks, little doves) the hawks start killing each other until a certain point where the total population is somehow in balance. Offcourse the hawks will also kill some doves during the period directly after start, but these events will be rare. On an average the number of dead hawks will exceed the number of dead doves in the start period. The same principle goes for the case of a small ratio (little hawks, lots of doves): the hawks start killing the doves until the same point where the total population is somehow in balance.
This balance is found when the probability of a hawkhawk meeting equals half of the probabillity of a hawkdove meeting. In that case after three meetings there will be on an average two dead hawks and two dead doves. Suppose this balanced situation occurs with H hawks and D doves. The probabilities of the meetings are:
Pr(hawkhawk) = H*(H1)/((H+D)*(H+D1))
Pr(hawkdove) = 2*H*D/((H+D)*(H+D1))
(factor 2 in last probability because you have both the hawkdove and dovehawk meeting)
in case Pr(hawkhawk) = 0,5 * Pr(hawkdove) then
H*(H1) = H*D D = H1
It is easy to see that the underlined equation really describes the balanced situation. In case (D = H1) the expected number of dead hawks equals the expected number of dead doves, so after n meetings the number of hawks and doves will be (H(n/2)) and (D(n/2)). The new condition for the new situation is than (D(n/2)) = (H(n/2)) 1 which condition is always valid given the initial condition (D = H1). So the population really tends to maintain a mix of H hawks and (H1) doves during the balanced situation.
Given this insight it is easy to see that the case with the even number of hawks on an average will end with H = 2 and D = 1. This is the balanced situation for H = 2. The last dove has a chance of 1/3 to survive this last situation. So on an average M = 1/3 dove survives this case. With that number for M we find a (1/(3*D)) chance of survival for a dove in case of an even number of hawks.
Offcourse the actual number of surviving doves maybe 1, 2 or even more, but lots of simulations end in zero doves left. On an average 1/3 dove survives, independent from the ratio (H/D) at start. In case of simulation with very large ratio (H/D) or (D/H)use large numbers of birds to avoid that the balanced situation will not occur before the end.
Matthijs

Posted by Matthijs
on 20040916 11:51:55 