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I am given the following game to find nash equilibria in pure and mixed strategies:

$\begin{pmatrix}& & Litte John &\\ & & c & w \\Big John & c & (5,3) & (4,4) \\ & w & (9,1) & (0,0) \end{pmatrix} $

The pure-strategy equilibria will be if one plays $w$ and the other one $c$, no matter which one.

Now for the mixed strategy, I assigned the probabilities $p,1-p$ to strategies $c,w$ for big John, and $q,1-q$ for $c,w$ for little John, i.e.:

$\begin{pmatrix}& & Litte John &\\ & & q & 1-q \\Big John & p & (5,3) & (4,4) \\ & 1-p & (9,1) & (0,0) \end{pmatrix}$

Then I would create functions that show the Expectational value of bigJohn w.r.t. $p$, because he can only change his own strategy, hence

$B(p)=5pq+4p(1-q)+9(1-p)q$, and similar for Little John, $L(q)=4pq+4p(1-q)+1(1-p)q$

Then we find the derivatives, $B'(p)=-8q+4$ and $L'(q)=-2p+1$. Then these are set equal to zero to maximize the payoff for both Johns, and then? Is that the end? I am suspicious to my solution because now everyone's optimal strategy mix will depend on the other's mix. Isn't it the purpose to maximize the payoff independent of the enemy player? Can somebody clear this up for me please?

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    $\begingroup$ ...or should this rather go to mathematics? $\endgroup$
    – Marie. P.
    Commented Feb 27, 2012 at 15:59

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Your problem was introduced when you took the derivatives. There is no value of p that makes B'(p) = -8q + 4 = 0. That is, there are no local maxima. As such, you will be looking for a boundary solution. If q > 1/2, B' < 0, and will therefore be maximized when p is as small as possible (which is 0, since p is bounded between 0 and 1). If q < 1/2, B' > 0, and therefore p should be as large as possible, which, in this case is 1. If q=1/2, all values of p produce the same payoff. Therefore, the best response function for p should be:

p=0, if q > 1/2. p=1, if q < 1/2. p in [0,1] if q = 1/2.

Now go through and do the same thing for q. The Nash equilibrium will be the set of probability weightings that simultaneously satisfy both best-response functions.

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  • $\begingroup$ Thank you very much! That's exactly what I needed! marie $\endgroup$
    – Marie. P.
    Commented Feb 27, 2012 at 21:23

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