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Where should you shoot first for the highest chance of survival?

+1 vote
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Amar, Akbar and Anthony plans for gun fighting and eachone got a gun to shoot each other until only one person is left.

Shooting History of all three
Amar hits his shot 1/3 of the time, gets to shoot first.
Akbar, hits his shot 2/3 of the time, gets to shoot next if still living.
Anthony having perfect record at shooting(100% accuracy) shoots last , if alive.

If you are Amar, where should you shoot first for the highest chance of survival?

posted Jul 2, 2014 by Pardeep Kohli

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1 Answer

0 votes

Amar should shoot Anthony first. This increases his change for survival from 50/189 to 59/189.
.
Suppose A shoots B first.
1/3 A hits B and 100% that C then kills A. (= 0% that A wins)
.
2/3 A misses B
followed by 2/3 B kills C, leaving just A and B standing.
A and B then have a shootout with A having a 3 in 7 chance of winning and B having a 4 in 7 change of winning (= 2/3 * 2/3 * 3/7 = 36/189 that A wins
.
OR after A misses B, 1/3 B misses C, followed by 100% C kills B, leaving just A and C standing.
A then has a 1/3 of killing C. If A misses, C will kill him (= 2/3 * 1/3 * 1/3 = 14/189 that A wins
.
Combining the chances that A wins gives (14 + 36 = 50/189)
.
[The shootout between A and B is 1/3 (=3/9) that A kills B immediately. If A misses, B has a 2/3 chance of killing B, so 2/3 * 2/3 = 4/9 that B wins. There is also a 2/9 chance that both miss. This series converges into 3 in 7 for A winning and 4 in 7 for B winning.
...
Alternatively, A can try to shoot C.
1/3 A kills C, followed by 2/3 that B kills A.
That leaves 1/9 that only A and B are left. A then has a 3/7 chance of winning (= 1/9 * 3/7 = 9/189 that A wins)
.
If A misses (2/3 chance), there is a 2/3 chance that B kills C, leaving only A and B. The total chance of A winning this series = 2/3 * 2/3 * 3/7 = 36/189 that A wins
.
OR B misses C (1/3 chance), leaving C to kill B. At this point only A and C remain and A has 1/3 of killing C. Total chance of A winning this series =2/3 * 1/3 * 1/3 = 14/189
.
Combining the chances that A wins gives (9 + 36 + 14 = 59/189)

answer Mar 31, 2017 by Jcm



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