Value of Zero sum game

Question

In part iii) I am unsure as to why we subtract 1 from the value of the game (underlined in green)

Answer 1

To restate what I already said in the comments above:

The original game for part (iii) had payoff matrix: $M=\begin{bmatrix}x&-1&-1\\-1&y&-1\\-1&-1&z\end{bmatrix}$

We choose to consider a different game which is closely related to the original game but where instead all entries in the payoff matrix are non-negative. To do so, we can choose to add one to every entry. That is:

$M' = M+Ones_{3\times 3} = \begin{bmatrix}x+1&0&0\\0&y+1&0\\0&0&z+1\end{bmatrix}$

Worded without math, whatever the result of the game, we simply pay $A$ an additional dollar compared to what it would have originally been. The strategy for $A$ and $B$ will not have changed, however in every instance, $A$ will end up with one more dollar than otherwise. Therefore, if the original value of the game was $v$, the new value of the modified game is $v+1$.

Now, this new payoff matrix is very convenient for us because it very closely resembles the payoff matrix that you saw in part (ii) of the problem. In part (ii), you saw that with payoff matrix $\begin{bmatrix}\color{green}{x} & 0&0\\0&\color{red}{y}&0\\0&0&\color{blue}{z}\end{bmatrix}$, the value of the game would be $\dfrac{\color{green}{x}\color{red}{y}\color{blue}{z}}{D}$ where $D$ is written as above in the image.

This implies that the value of the game with matrix $M' = \begin{bmatrix}\color{green}{x+1} & 0&0\\0&\color{red}{y+1}&0\\0&0&\color{blue}{z+1}\end{bmatrix}$ would be $v=\dfrac{\color{green}{(x+1)}\color{red}{(y+1)}\color{blue}{(z+1)}}{D'}$ where $D'$ is written as above in the image.

However, this is the value for the game with payoff matrix $M'$, i.e. the game where player $A$ got an extra dollar than they would have otherwise in every case. It follows then that the value for the game with payoff matrix $M = M' - Ones_{3\times 3}$ will be $v-1$

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For additional justification for why adding the $Ones$ matrix does not change the strategies of the game but changes the value of the game:

Suppose we have a game with payoff matrix $M$ and optimal strategies $a^*$ and $b^*$.

Then $a^* M b \geq a M b$ for all strategies $a$ and $b$ (including $b^*$). Similarly $a M b^* \leq a M b$ for all strategies $a$ and $b$ (including $b^*$).

Now, note that $a (M+\lambda Ones)b = aMb + \lambda(a(Ones)b)$. Note further what $(Ones)b$ is however. As $b$ is a distribution vector, the sum of the entries of $b$ is one. It follows that $(Ones)b$ is simply a vector of all ones. It then follows that $a(Ones)b = a(Ones)=1$.

From here, we see that for all strategies $a$ and $b$, the value of the game with matrix $M+\lambda Ones$ is simply $\lambda$ plus the value of the game with matrix $M$ with the same strategies. It follows then that $a^*$ and $b^*$ are still the optimal strategies.