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 Hawks and Doves (Posted on 2004-09-13)
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?

 See The Solution Submitted by Federico Kereki Rating: 3.5000 (2 votes)

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 re: Bird's Eye View ! | Comment 8 of 22 |
(In reply to Bird's Eye View ! by Syzygy)

"If 2 H... Then H = 0% ; D = 50% ( As the 2 hawks will eventually meet and die leaving the remaining doves alive )"

The below is incorrect, as pointed out in later discussion:

Suppose there are 2 hawks and 1 dove. That dove survives only if the first meeting is between the two hawks, which has probability 1/3, so that dove's chance of survival is 1/3.

If there are 2 hawks and 2 doves, a given dove survives if either the first meeting is between the two hawks, with probability 1/6, or the first meeting is between one of the hawks and the other dove and then the dove survives the 2H-1D stage.  The latter has probability (5/6)(1/2)(1/3)=5/36, making the total probability 11/36.

For each even number of hawks and number of doves, the probability of a given dove's survival is the sum of (1) the probability that the first meeting is between two hawks and the dove survives in the ensuing H-2,D round, and (2) the probability that the first meeting is between a hawk and a dove, but that dove is a different one and then the dove survives the ensuing H,D-1 round.

This could be done on a spreadsheet, but to get rational numbers I used UBASIC:

5   Num=5
10   dim P(2,Num)
20   for D=1 to Num:P(1,D)=1:next D:' zero hawks
30   for H=2 to Num*2 step 2
40     P2h=H*(H-1)//((H+1)*H)
50     P(2,1)=P2h*P(1,1)
55     print P(2,1);
60     for D=2 to Num
70       P2h=H*(H-1)//((H+D)*(H+D-1))
80       P(2,D)=P2h*P(1,D)+(1-P2h)*(D-1)//D*P(2,D-1)
85       print P(2,D);
90     next D
95     print
97   for D=1 to Num:P(1,D)=P(2,D):next D
100   next H

which gave, for up to 10 hawks and 5 doves:

` 2 hawks 1/3  11/36  17/60  53/200  131/525 4 hawks 1/5  41/225  317/1890  6863/44100  4807/33075 6 hawks 1/7  769/5880  5477/45360  356407/3175200  572722//5457375 8 hawks 1/9  1451/14175  296167/3118500  2280244/25727625  83404532/100337737510 hawks 1/11  38537/457380  1273871/16216200  53105341/721440720  14759033447/213050462625`

where each line shows the probability of a given dove surviving if there are 1, 2, 3, 4 or 5 doves altogether.

In decimal, these come out to

` 2 hawks 0.3333333 0.3055556 0.2833333 0.2650000 0.2495238 4 hawks 0.2000000 0.1822222 0.1677249 0.1556236 0.1453364 6 hawks 0.1428571 0.1307823 0.1207451 0.1122471 0.1049446 8 hawks 0.1111111 0.1023633 0.0949710 0.0886302 0.083123810 hawks 0.0909091 0.0842560 0.0785555 0.0736101 0.0692748`

so with larger numbers of either hawks or doves, each dove's chances of survival goes down.

Edited on September 13, 2004, 8:23 pm
 Posted by Charlie on 2004-09-13 10:34:07

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