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We are given a bimatrix game (A,B) = $\begin{bmatrix}0,4 & 4,0 & 5,3\\4,0 & 0,4 & 5,3\\ 3,5 & 3,5 & 6,6 \end{bmatrix}$. Suppose (p, q) is a Nash equilibrium in (A,B). Prove that {1,2} $\not\subseteq$ C(p), where C(p) denotes the carrier of the strategy p.

To solve this problem I was planning on using the following Corollary:

"A strategy pair (p, q) is a Nash equilibrium in a bimatrix game (A,B) if and only if C(p) $\subseteq$ P$B_1$(q) and C(q) $\subseteq$ P$B_2$(p)".

Now, suppose that {1,2} $\subseteq$ C(p). Since C(p) $\subseteq$ P$B_1$(q) we also have {1,2} $\subseteq$ P$B_1$(q) i.e. e$^{1}$*Aq* = max k e$^{k}$*Aq* and e$^{2}$*Aq* = max k e$^{k}$*Aq*. We calculate e$^{k}$*Aq* for k = 1,2,3. Suppose q = (q$_1$, q$_2$, 1 - q$_1$ - q$_2$).

e$^{1}$*Aq* = 5 - 5q$_1$ - q$_2$
e$^{2}$*Aq* = 5 - 5q$_2$ - q$_1$
e$^{2}$*Aq* = 6 - 3q$_2$ - 3q$_1$

We see that {1,2} $\subseteq$ C(p) if 5 - 5q$_1$ - q$_2$ = 5 - 5q$_2$ - q$_1$ (so if q$_1$ = q$_2$) and 5 - 5q$_1$ - q$_2$ $\geq$ 6 - 3q$_2$ - 3q$_1$ and 5 - 5q$_2$ - q$_1$ $\geq$ 6 - 3q$_2$ - 3q$_1$ . Substitute q$_1$ = q$_2$ into the second or third inequality to obtain 5 $\geq$ 6 which is ofcourse impossible. We conclude that {1,2} $\not\subseteq$ C(p).

Can anyone tell me if my solution is right or not? Thanks in advance! :)

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