# 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

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)

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+1 vote

Once upon a time there were three rivals who got together in a dark wood on an open spot for composing a quarrel by using guns. It was a kind of duel but there were three people in it and therefore it is a truel. There were A, B and C. there was some rules to be followed in the game. The rules of the games were:
-They have to draw the lots in order to find who will fire first and who will be second and who will be third.
- The next rule is that, they will start firing on each other unless there is a last person alive.
- Each person will decide on his own about which person he will fire on.
-it is quite clear that A hits and kills in all the shots and it is 100%. Whereas 80% of the shots are hit by B and C hits in 50% of the shots he tries to fire in. the three persons have their own strategy for firing. Nobody is killed by the stray bullet.

Question: who will have the highest chances for surviving in the truel and how much is the chance?