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Hawks and Doves (Posted on 2004-09-13) Difficulty: 3 of 5
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  83404532/1003377375
10 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 0.0831238
10 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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