Approach through Assurance Game: Who's Being a Fool?

Today we talked about the assurance game. To symbolize it, assuming red is player 1, blue is player 2.

Before that, one important assumption has to be made, namely that each player believes that the other is in the same situation and will have the same probability as the thinker have of making their decision.

Let P(S)=the probability that the Agent would stand.
      P(R)=the probability that he would run.
and E(S) and E(R) are the expected utilities from the actions two players act.
Now, P1 wants to calculate his final utility from how P2 act.
1. E(S)=P(S)*A+P(R)*C
2. E(R)=P(S)*B+P(R)*D
Then compare 1 and 2, yield E(S)-E(R)=P(S)*(A-B)+P(R)*(C-D)

Here comes the objection #1: Crisis of being demanding, and practically less frequently than prisoner's dilemma.

Assurance game requires that P(S)*(A-B)+P(R)*(C-D)=0, therefore if we assume that P(S)=P(R)=.5, we must have A-B+C-D=0. And if the result is not zero, there will be a dominant strategy. It seems to me that A-B+C-D=0 is too demanding, because the result of A-B+C-D is perfectly contingent, it follows that we have to make up four numbers to make the equation established, and it is more plausible for us to make up another four to destroy the equation.

Moreover, in the fool's objection, "maximized utility by breaking the covenant" means C>>A>D>B, this is definitely a prisoner's dilemma. it follows that A+C nevertheless does not equal to B+D. Therefore by making up numbers that satisfied the assurance game is simply escaping from the fool's attack. To me, prisoner's dilemma happens much more frequently than assurance game.

Objection #2: Crisis of methodology

P(S)=P(R)=.5 if and only if E(S)=E(R). That is to say, by assuming P(S)=P(R), we are meanwhile assuming that E(S)=E(R). It is true that the outcome of the decision would affect the probability of each action, because if I know by calculating the probability that E(R) definitely outweighs E(S), my probability of running away is 1 and that of staying decrease to 0. Therefore in assurance game, neither probability of each action nor expectation of the utility should be assumed.

Actually this method of calculating expectation is not very practical, because we always have to assume something when doing the calculation, otherwise P(S) and P(R) changes with the changes of E(S) and E(R), and E(S) and E(R) forwardly changes P(S) and P(R), we come up with nothing but infinite regression. Therefore we either abandon the method, or we continuously making assumption.

To my knowledge, so far the expectation seems to be the best tool we can use in prediction, and it follows that we should continuously making assumptions. To acknowledge this, let's go back to the equation, then assume that A-B=1 and C-D=-1, then E(S)-E(R)=P(S)-P(R). And Hobbes would say that this ends up with the same conclusion: do whatever the other one does. But as I said from above, it is not wise to assume that A-B=D-C.