Small knowledge, big challenge! This article is participating in the creation activity of “Essential Tips for Programmers”.

Today, I will use an example to help you understand the situation that you may encounter in life and work.

In the above table, there are two players in the game are A and B respectively. In the strategy set {A1,A2}\{A_1,A_2\}{A1,A2} represent the eagle and dove respectively, which represents toughness and compromise. The return matrix is as follows.


A 1 A_1
A 2 A_2

A 1 A_1 		A-C,A-C 2 a, 0

A 2 A_2 		0, 2 A, A,A

Therefore, when A−C> 0A-C >0A−C>0, both A and B choose A1A_1A1, which is their dominant strategy. In this case, no matter WHETHER B chooses eagle or dove, as long as A chooses eagle, it is the dominant strategy. Therefore, both sides choose eagle to achieve the equilibrium of dominant strategy. When A−C< 0a-C <0A −C<0, the strategy selection of B needs to consider the strategy selection of A. Assuming that the probability of A choosing A1A_1A1 is α\alphaα without knowing the choice of A, the result is obtained


E ( A 1 ) = Alpha. x ( A – C ) + ( 1 – Alpha. ) x 2 A \mathbb{E}(A_1) = \alpha \times (A-C) + (1-\alpha) \times 2A

E ( A 2 ) = Alpha. x 0 + ( 1 – Alpha. ) x A \mathbb{E}(A_2) = \alpha \times 0 + (1-\alpha) \times A

Above is the expectation of Party B to choose A1A_1A1(eagle) and A2A_2A2(dove), so when Party A chooses A1A_1A1, the profit of which strategy party B chooses is the same.


Alpha. x ( A – C ) + ( 1 – Alpha. ) x 2 A = Alpha. x 0 + ( 1 – Alpha. ) x A \alpha \times (A-C) + (1-\alpha) \times 2A = \alpha \times 0 + (1-\alpha) \times A

Alpha. = A C \alpha = \frac{A}{C}

When the probability of A1A_1A1 eagle adopted by Party A is AC\ FRAc {A}{C}CA, no matter whether Party B chooses eagle or pigeon, the benefits will not change. The probability that both sides select eagles is α∗=AC\alpha^* = \frac{A}{C}α∗=CA then an equilibrium point is reached.

Here, the bigger A is, the more likely it is to have A choice eagle, because there is A big gain in the dispute, and the bigger C is, the bigger the loss is, the less likely it is to have A dispute. In addition, the increase of C can make the selection of pigeon more profitable, because the pigeon’s profit is A(1−AC)A(1 – \frac{A}{C})A(1−CA), so the higher THE C is, the higher the selection of pigeon will be.