Equalized Expectations
For an \(n\times n\) matrix game with no PSS, we may use the idea of equalized expectations to solve for optimal strategy mixtures algebraically. Recall that, in the \(3\times 3\) case, we have
\[\begin{split}\vec r = \left[\begin{array}{r}r_1\\r_2\\r_3\end{array}\right]\hspace{5mm}\text{and}\hspace{5mm}\vec c = \left[\begin{array}{r}c_1\\c_2\\c_3\end{array}\right]\end{split}\]
where
\[\begin{split}\begin{align}0\leq r_1,r_2,r_3 &\leq 1\\[3mm]r_1 + r_2 + r_3 &= 1\end{align}\end{split}\]
Also, the same is true for \(c_1,c_2,\) and \(c_3\).
Example
Given the following matrix game, first verify there is no reduction by dominance. Then solve for the optimal strategy mixtures using equalized expectations for Rose and for Colin. What is the value of the game?
\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rrr}&A&B&C\\\hline A&1&-1&6&\\B&9&3&-2&\\C&0&6&-2&\end{array}\end{array}\end{split}\]
We consider the two values \(c_1\) and \(c_2\) since
\[c_3 = 1 - c_1 - c_2\]
Remember that Colin is trying to choose \(\vec c = (c_1,c_2,c_3)\) in such a way so that Rose’s expected values are equal. If he can do so, then Rose will have no incentive to change her own strategy mixture. We will set \(s=c_1\) and \(t=c_2\) to simplify the algebra, and we are thus trying to solve for expected values in the example game:
\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{r|rrc}&As&Bt&C(1-s-t)\\\hline A&1&-1&6&\\B&9&3&-2&\\C&0&6&-2&\end{array}\end{array}\end{split}\]
where Colin ensures that:
Strategy A is played \(s\) proportion of the time,
Strategy B is played \(t\) proportion of the time, and
Strategy C is played \(1-s-t\) proportion of the time.
Then, we have the three expected value calculations:
\[\begin{split}\begin{align}E(\text{Rose} A) = s - t + 6(1-s-t) &= -5s -7t + 6\hspace{11mm}(1)\\[2mm]E(\text{Rose} B) = 9s + 3t - 2(1-s-t)&=11s + 5t -2\hspace{16mm}(2)\\[2mm]E(\text{Rose} C) = (0) + 6t - 2(1-s-t)&=2s+8t-2\hspace{14mm}(3)\end{align}\end{split}\]
We can create a linear system of two equations and two unknowns by setting the first and second equations equal to one another and the second and third equations equal to each other.
\[\begin{split}\begin{align}(1)=(2)\implies -5s -7t + 6&=11s + 5t -2\\[2mm](2)=(3)\implies 11s + 5t -2&=2s+8t-2\end{align}\end{split}\]
Moving terms to the same side, we find the following:
\[\begin{split}\begin{align}16s + 12t -8&=0\\[2mm]9s - 3t&=0\end{align}\end{split}\]
We can solve the second equation above for \(s\) in terms of \(t\) as follows:
\[9s=3t \hspace{3mm}\implies\hspace{3mm} s=\frac{1}{3}t\]
We subsitute this into the first equation for \(t\), and we find the following:
\[\begin{split}\begin{align}16\left(\frac{1}{3}t\right)+12t&=8\\\frac{16}{3}t+\frac{36}{3}t&=8\\\frac{52}{3}t&= 8\\t = \frac{24}{52}=\frac{12}{26}&=\frac{6}{13} \end{align}\end{split}\]
Since \(s=\frac{t}{3}\), the above calculation demonstrate that
\[s=\frac{\frac{6}{13}}{3}=\frac{6}{13}\frac{1}{3} =\frac{2}{13}\]
and therefore that
\[1-s-t = \frac{13}{13} - \frac{2}{13} - \frac{6}{13}= \frac{5}{13}\]
Thus, Colin’s optimal strategy mixture is given as follows:
\[\begin{split}\vec{c} = \left[\begin{array}{r}\frac{2}{13}\\\frac{6}{13}\\\frac{5}{13}\end{array}\right]\end{split}\]
Rose must choose \(\vec r = (r_1,r_2,r_3)\) in such a way so that Colin’s expected values are equal. If she can do so, then Colin will have no incentive to change his strategy mixture. We will set \(v=r_1\) and \(w=r_2\) to simplify the algebra, and we are thus trying to solve for expected values in the example game:
\[\begin{split}\begin{array}{cc}&\text{Colin}\\\text{Rose}&\begin{array}{c|rrr}&A&B&C\\\hline Av&1&-1&6&\\Bw&9&3&-2&\\C(1-v-w)&0&6&-2&\end{array}\end{array}\end{split}\]
\[\begin{split}\begin{align}E(\text{Colin} A) = v + 9w + 0(1-v-w) &= v + 9w\hspace{21mm}(4)\\[2mm]E(\text{Colin} B) = -v + 3w + 6(1-v-w)&=-7v - 3w + 6\hspace{9mm}(5)\\[2mm]E(\text{Colin} C) = 6v - 2w -2(1-v-w)&=6v-2\hspace{22mm}(6)\end{align}\end{split}\]
To create the solvable linear system, we do the following:
\[\begin{split}\begin{align}(4)=(5)\implies v + 9w& = -7v - 3w + 6\\(5)=(6)\implies 6v-2 &= -7v - 3w + 6\end{align}\end{split}\]
Bring all the terms to the right-hand side:
\[\begin{split}\begin{align}8v + 12w - 6& = 0\\13v + 3w - 8 &= 0\end{align}\end{split}\]
The top equation can be solved for \(w\) in terms of \(v\) as follows:
\[12w = 6-8v \implies w=\frac{1}{2}-\frac{2}{3}v\]
We subsitute into the second equation as follows:
\[\begin{split}\begin{align}
13v + 3\left(\frac{1}{2}-\frac{2}{3}v\right) - 8 &= 0\\13v + \frac{3}{2} - 2v - 8 &= 0\\11v &=8 - \frac{3}{2}\\11v &= \frac{13}{2}\\v&=\frac{13}{22}\end{align}\end{split}\]
Thus, Rose’s optimal strategy mixture is given as follows:
\[\begin{split}\vec{r} = \left[\begin{array}{r}\frac{2}{13}\\\frac{6}{13}\\\frac{5}{13}\end{array}\right]\end{split}\]
\(\text{Rose} = (1/2, 1/6, 1/3)\)
