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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: To Make Oskar's Solution Work... | Comment 11 of 22 |
(In reply to To Make Oskar's Solution Work... by David Shin)

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

I agree that allowing meetings between dead birds or between live and dead birds does not change the outcome, as those meetings do not kill anyone and can be ignored.

However, the numbers do not agree with my calculations.  For example, if there are 2 hawks, the formula you give yields a probability of 1/3 that a given dove will survive, regardless of the number of doves.  Suppose there are 2 doves.  If the first meeting is between the two hawks, you will survive, and this probability is 1/6.  You can also survive if the first meeting is between one of the hawks and the other dove and then you go on to survive the remaining situation of 2 hawks and 1 dove.  This other way of surviving has the product of the following probabilities: 5/6 that it is indeed a hawk/dove meeting; 1/2 that it isn't you that's involved; 1/3 that you survive after it has been reduced to 2 hawks and you.  So the second way of surviving has probability (5/6)*(1/2)*(1/3), or 5/36.  Added to the 1/6 = 6/36 of the first way of surviving, that's 11/36 altogether, not 1/3.

The 1/(H+1) works only if there is only one dove.

Edited on September 13, 2004, 8:25 pm
  Posted by Charlie on 2004-09-13 12:09:43

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