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?

How about a different approach that isn’t about simulation?

<o:p> </o:p>

Everybody agrees that if you’re a hawk, survival chance is zero if an H is an even number and 1/H if H is an odd number.  Also that if you’re a dove, survival chance is zero if H is an odd number.  The contentious issue is what if you’re a dove with an even number of hawks?

<o:p> </o:p>

Use (H,D) to mean the state of the population of hawks and doves, and you’re a dove.  Each state has a probability of survival for you that is unchanging.  For instance P(0,D) = 1.

<o:p> </o:p>

Consider:

P(2,1)=(1/3)P(0,1)=1/3

P(2,2)=(2/6)P(2,1)+(1/6)P(0,2)+(1/6)P(2,2), a recursive expression.

Solving:

P(2,2)=(1/9)+(1/6)+(1/6)P(2,2)

6P(2,2)=(2/3)+1+P(2,2)

5P(2,2)=5/3

P(2,2)=1/3

<o:p> </o:p>

Similarly:

P(2,3)=(4/10)P(2,2)+(1/10)P(0,3)+(3/10)P(2,3) yields P(2,3)=1/5

Induction will show all P(2,D)=1/3

<o:p> </o:p>

Look at 4 Hawks:

P(4,1)=(6/10)P(2,1)=(6/10)(1/3)=1/5

P(4,2)=(4/15)P(4,1)+(6/15)P(2,2)+(1/15)P(4,2)

Solving:

P(4,2)=(4/15)(1/5)+(6/15)(1/3)+(1/15)P(4,2)

15P(4,2)=(4/5)+(6/3)+P(4,2)

14P(4,2)=14/5

P(4,2)=1/5

Induction will show all P(4,D)=1/5

<o:p> </o:p>

Induction will show that P(H,D)=1/(H+1)

Posted by bernie on 2004-09-24 09:39:59