You are in a free-for-all shootout involving three people, and by some rotten luck, you are the worst shooter of the group. You, shooter A, hit your target 33% of the time. The other two, B and C, have 50% and 100% accuracy respectively.

At least you get to shoot first, with B going after you, and C going last. Everyone takes turns shooting until all but one are dead.

Who will you fire at in the first round to maximize your odds of survival?

(In reply to re: So close! by friedlinguini)

Friedlinguini-

You are still assuming that A should try to hit someone. Again I say, he stands a slightly better chance if he deliberately misses

He can't kill both of his opponants during the first round. if he kills one, the other has a "free" shot at him. If he misses, then the other two will shoot each other, and one will die. He will survive the round.

The chances of A being alive and not having killed his opponant after n rounds rapidly become vanishingly small (though not non-zero) for n > 2, so we can avoid all of the iterations I dreaded by just looking at his chances of surviving to the end of round 2.

These are the possible outcomes of the first round, the probability of arriving at that outcome, and the probability of then surviving Round 2

Case 1 Shoot at B

a) B dies (33%); C shoots A A does not survive to round 2 (0%)

b) A misses (67%) B shoots C and C Dies (50%) A and B go to round 2, where A's chances of surviving to round 3 are 67%

c) A misses (67%) B shoots C and misses(50%). C kills B. A and C go on to round 2, where A must try to kill C (33%)

Total chance to survive to the end of Round 2: 0% + 67%(50%)(67%) + (67%)(50%)(33%) = 33%

Case 2 Shoot at C

a) Hit C (33%); B shoots at A (50%); A and B go to Round 2 (50%)(67%)

b,c) Miss C (Same as b and c above)

Total chance of surviving to the end of round 2: 33%(50%)(50%)(67%) + 33% = 38.5%

Case Three, deliberately miss

b) A misses (100%) B shoots C and C Dies (50%) A and B go to round 2, where A's chances of surviving to round 3 are 67%

c) A misses (100%) B shoots C and misses (50%). C kills B. A and C go on to round 2, where A must try to kill C (33%)

Total chance to survive to the end of Round 2: (100%)(50%)(67%) + (100%)(50%)(33%) = 51%




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Posted by TomM on 2002-08-28 13:20:43