Survival probability of A:1-p (b+c) =1-[p (b)+p (c)-p (b) p (c)] = 0.12 (p stands for probability).
B's survival probability: 0 will die.
C probability of survival: 1 must survive.
There is another version of the same kind of logic analysis case:
Three musketeers A, B and C robbed a diamond in partnership and decided to fight with bullets.
The rules are as follows: draw lots to decide the shooting order. In order, everyone fired a shot until one person was alive. As we all know, A is a dead shot. B's shooting percentage is 80%; C's hit rate is 50% (that is, there is a 50% chance of hitting it). Now, who is most likely to get the diamond alive in the end?
There are three situations according to who shoots first.
If the lottery results in shooting first. Because he is a sharpshooter. He is sure to kill one, and the remaining one will shoot him. If A kills C, the probability of B shooting A is 80%. If A kills B, C has a 50% chance of shooting him. So a will kill b and wait for c to do it. The survival probability of him and C is 50%.
If b makes the first move. If he shoots A, he has an 80% chance of killing A, and then waits for C to judge himself; If he doesn't kill A, it's like A or C shooting first. If he shoots C, he has an 80% chance of killing C, and then 100% chance of being hit by A; If he doesn't kill C, it's like A or C shooting first. So b should shoot a.
If c makes the first move. If he shoots A, he has a 50% chance of killing A and 80% chance of being hit by B. If he shoots B, he has a 50% chance of killing B and a 100 chance of being hit by A. If C is empty. If you don't kill A or B, the first two situations will recur. In other words, the result of c shooting a or b is not as good as shooting empty. So c chose not to fight.
If everyone's survival probability is Pi(j), i= 1, 2,3. J= a，b，c。
In the first case,
p 1(A)= 50% p 1(B)= 0% p 1(C)= 50%。
The second situation
The premise of A's survival is that B misses, A hits B, and then C also misses. The premise of B's survival is to hit A, and then C can't hit C ... If C is not hit, C will keep shooting until one person falls. Suppose b and c play, and c takes the lead. The survival probability of B is L (B), and the survival probability is L (C). Then l (b) = 0.5 * (0.8+0.2 * l (b)); L(c)= 0.5+0.5 * 0.2 * L(c)； l(b)= 4/9； l(c)= 5/9；
P2(A)= 0.2 * 0.5 = 10% P2(C)= 0.2 * 0.5+0.8 * L(C)= 54% P2(B)= 0.8 * L(B)= 35.6%。
In the third case, C is not chosen correctly, so the result is still the first two.
P3 (A) =30% P3 (B) = 17.8% P3 (C) =52%.
Therefore, it is concluded that C, the worst marksman, is most likely to survive to get the diamond.