Game-three gunmen fight

Three-person duel (gunners' game)

The question is as follows:

A, B, and C fight. A's shooting hit rate is 1/3, that is, if he tries his best, he can hit every three shots on average. B's shooting hit rate is 1/2; C's shooting hit rate is one (that is ). Given that a has the lowest hit rate, for the sake of fairness, they let a first launch, then B (if he is still alive), and then C (if he is still alive ). Then there is a, B, c, and so on until only one person is alive. You can only open a gun at each time, but you can either open it or empty it. Our question is: if all three ABC members follow the best choice, that is to say, to improve their survival rate as much as possible, who is most likely to survive? To be accurate, what is the probability that each person will survive?

In the first round of shooting, we can draw three obvious Inferences:

1) Since the hit rate of C is 100%, if a and B attack each other in the first round, they will not find death;
2) C in A and B, B is given priority because B's hit rate is 1/2 higher than a's hit rate of 1/3;
3) B will shot C and B will shot C before C. If it is inferred that C will shot C first, it will certainly shot C first, Because surrender will die, rather than a bo.

Next, we can first simplify the problem and consider the situation where only two people exist. Note that the hit rate of C is 100%. Therefore, whether only A and C are used, or only B and C are used, the situation is relatively simple. If it is C's turn, C will win.

For example, if only a and c fight now, if a starts the gun first, the survival rate of A is 1/3 (when a hits the gun). If C starts the gun first, a will die.

Now let's take a look at what will happen if only a and B fight. Obviously, when only two people are there, if both sides want to improve their survival rate as much as possible, no one will put a gun empty. Suppose it is the turn of a to open a shot at this time, then:

1) a shot, with 1/3 of the chances of hitting B to win, and 2/3 of the chances of not hitting B to enter the next round;
2) B shot, with 1/2 of the chances to win a, and 1/2 of the chances to win the next round;
3) Repeat the above steps until one person wins.

As shown in:

When only a and B fight, for example, if a is shot first, the survival probability P (A1) and P (B1) of A and B are:

P (A1)
= 1/3 + 1/3*(1/3 + 1/3 *...)
= 1/3*(1 + 1/3 + (1/3) ^ 2 + ...)
= 1/3*(1/(1-1/3 ))
= 1/2

P (B1) = 1-P (A1) = 1-1/2 = 1/2.

If B is shot first, the survival probability P (A2) and P (B2) of A and B are:

P (A2) = P (A1) * 1/2 = 1/4 (equivalent to B's first shot before a's first shot .)

P (B2) = 1-P (A2) = 1-1/4 = 3/4.

Obviously, between B and C, A does not want to fight against C. The above P (A1) and P (B2) are also used.

If a shot B in the first round, it would have a 1/3 probability to hit B directly, and then be killed by C. 2/3 probability to enter the next round of duel, that is, A has a probability of 1/3 in the first round. In the next round, he had a 1/2 probability of a duel with B, and a 1/2 probability of a duel with C. The situation does not look very good.

If A and B share the same enemy and enemy, what should I do if I start shooting C first? He will have a 1/3 probability of shooting C, and then be shot by B with a 1/2 probability, that is, if he first shot C, then there is a probability that 1/3*1/2 = 1/6 will go down in the first round.

What if the first round of a is empty? At this time, B will obviously shoot C. If a is shot (1/2 probability), it will be in the case of A and B dual. In addition, a is shot first, And A has a 1/2 probability of survival; if B does not shot C (1/2 probability), C will obviously kill B when it is his turn to C. In the case of a and c dual, a will first kill C, at this time, a has a 1/3 probability of survival.

Obviously, the first round of empty gun A is a better strategy. The following figure shows the possible process of A, B, and C DUAL battles:

Among them, the probability of survival of three persons A, B, and C is:

P (A) = 1/2*1/3 + 1/2 * P (A1) = 1/6 + 1/3 = 5/12

P (B) = 1/2 * P (B2) = 1/4

P (c) = (1-1/2) * (1-1/3) = 1/3

P (A1) and P (B2) are the intermediate variables calculated when only a and B are involved in the duel.

We can see that P (a)> P (c)> P (B), A has the highest survival probability (close to 1/2), and C has the second survival probability, B has the minimum survival probability.

The analysis method is worth learning, from simplicity to depth, and recurrence in sequence.