Game Theory Matrix Pure strategy and mixed strategy

Now two people are playing, B is selected, and A is selected.

B will get the previous benefit, for example, B will get chess A will choose the movie, then B will get 3 and A will get 2.

Pure strategy means that in a completely informational game, where A and B know all the possible gains, A makes a decision, "I'll choose chess."", so that if B boy chooses the movie, A boy can get the highest income 3, but if B boy chooses chess, the two of them will wait for zero income to kneel.

The mixed strategy means that B boy will not make his own decision explicitly, but will make a choice with a certain probability, such as tossing a coin, and choosing chess if he throws Grandpa Mao, otherwise he chooses a movie; Because of game theory.

The initial assumption was that people would be inclined to make definite choices with greater returns, while hybrid strategies randomized people's decisions and were generally used in more complex games.

To be honest, the landlord.,I don't know what you're going to ask.,And I don't know much about game theory.,I'm just very interested in your question.,Let's discuss it.。。。

Horizontal or vertical movies and chess are reversed.

First of all, the Nash equilibrium of pure strategy is (movie, movie) and (chess, chess).

This is a standard "sex war" game.

When you're done correcting the question, I'll help you calculate the hybrid strategy.

Answer]: Pure strategy refers to the course of action that participants can choose to adopt in the game. The mixed strategy is a kind of probability distribution in the pure strategy space, which means that the participants are randomly selected to implement it in the pure strategy according to this probability distribution when they actually make decisions.

In Game Theory and Information Economics, Zhang Weiying gives a definition of mixed strategic equilibrium, saying, "The expected utility of each participant is a linear function of his or her own mixed probability, which means that if i is an optimal mixed strategy relative to a given -i, then for all ik>0, the following equation holds:."

vi(s_ik,σ_i)>=vi(s'ik, i), to arbitrary s'_ik∈s_i

That is, if i is an optimal mixed strategy relative to a given mixed strategy of other people, and if this mixed strategy dictates that i chooses a pure strategy s ik with strictly positive probability, then s ik itself must be an optimal strategy relative to -i. ”

How is the above inequality deduced from the inequality of the Nash equilibrium of mixed strategy? I'm a beginner and don't really understand how this step is obtained, so please advise! Thank you!