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duel - a simpler solution

(The original puzzle was posted on Friday, August 03, 2001 and was solved by Thomas W. Millett on Monday, August 06, 2001)

Let us call the three guys A (you, the 33% guy), B (the 50% guy) and C (the 100% guy), and let q(BCA) be the probability of YOUR SURVIVAL if three gunmen are left and it's B's turn to shoot.  The symbols q(CAB), q(AC), q(BA) and q(AB) are defined similarly.  Now, consider the following choices:

(i) You shoot B.  Your survival prob. is
(1/3) (0) + (2/3) q(BCA).

(ii) You shoot C.  Your survival prob. is
(1/3) q(BA) + (2/3) q(BCA).

(iii) You shoot the air.  Your survival prob. is q(BCA), which is equal to:
(1/3) q(BCA) + (2/3) q(BCA).

By comparing the three probabilities above, it is clear that choice (i) is inferior to choice (ii), and

(ii) is worse than (iii)  when  q(BA) < q(BCA)

Clearly, q(BA) = (1/2) (0) + (1/2) q(AB).  Hence

(ii) is worse than (iii)  when  (1/2) q(AB) < q(BCA) ......... (*)

How large is q(BCA)?  If all three gunmen are still there and it's B's turn to shoot, who or where will B shoot?  Well, B must shoot C.  For otherwise, it is clear that in the next round, C (the perfect gunman) will kill B in order to increase his/her chance of survival.  Therefore,

= (1/2) q(AB) [if B kills C] + (1/2) q(CAB) [if B fails to kill C]
= (1/2) q(AB) + (1/2) q(AC)  [because C kills B]
> (1/2) q(AB).

So we conclude by (*) that choice (iii), i.e. shooting the air, is the optimal decision.

Hankel O'Fung
Friday, February 20, 2004

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