Game Theory (Prisoner's Dilemma): Why dominant strategy yields lower payoff than other possible actions?

Question

From class notes:

Dominated Strategy

Action $i$ is a dominated strategy for player $j$ if it yields a lower payoff than at least one other actions available to player $j$ for every possible actions by all other players. So action $a^{j}_{i}$ is a dominated strategy for player $j$ if:

$V^{j}(a^{1}_{j},...,a^{j}_{i},..., a^{M}_{l}) < V^{j}(a^{1}_{j},...,a^{j}_{k},..., a^{M}_{l})$

for at least one $k \neq i$ and all $l$

In the image above, the dominated strategy for both player is $({W, W})$. I understand that the payoff value for picking W is $(6, 6)$ for both players as the other highest payoff would be to pick $NW$ which gives a $(7,3)$ for player 1 or $(3, 7)$ for player 2.

But I don't understand the dominant strategy.

Dominant Strategy

{...}

$V^{j}(a^{1}_{j},...,a^{j}_{i},..., a^{M}_{l}) > V^{j}(a^{1}_{j},...,a^{j}_{k},..., a^{M}_{l})$

From the same image above, now playing $(NW, NW)$ is the dominant strategy for both players. Given that picking $NW$ yields $(4, 4)$. The payoff value is much lower than other available actions. Yet, the dominant strategy states that:

action $i$ is a dominant strategy for player $j$ if it yields a higher payoff than any other actions available to player $j$ for every possible actions by all other players.

Answer 1

By definition,

action $i$ is a dominant strategy for player $j$ if it yields a higher payoff than any other actions available to player $j$ for every possible actions by all other players.

There are multiple quantifiers there; it might be clearer this way:

Action $i$ is a dominant strategy for player $j$ if for every possible selection of actions by all other players, action $i$ it yields a higher payoff to player $j$ than any other action available to player $j$.

So consider action $NW$ for player $1.$ I'll write $NW_1$ for that action to distinguish it from the action $NW_2$ that can be taken by player $2.$ There is only one other action available to player $1,$ namely $W_1,$ so $NW_1$ is dominant if it is true that no matter which action player $2$ takes, $NW_1$ yields a higher payoff to player $1$ than $W_1$ does.

If player $2$ takes action $W_2$, then $W_1$ pays $6$ to player $1$ and $NW_1$ pays $7$ to player $1.$ So $NW_1$ yields the higher payoff.

If player $2$ takes action $NW_2$, then $W_1$ pays $3$ to player $1$ and $NW_1$ pays $4$ to player $1.$ So $NW_1$ yields the higher payoff.

And now we've run out of things the other player can do, and in every case $NW_1$ yields a higher payoff to player $1$ than $W_1$ does. So $NW_1$ is dominant.

Player $2$ faces a similar situation; in each case $NW_2$ pays better than $W_2.$

Note that it is not sufficient to observe that the sum of payoffs of $NW_j$ is greater than the sum of payoffs of $W_j$ for player $j$. Suppose we had another game with this payoff matrix:

$$ \begin{array}{cc} & \text{Player $2$} \\ \text{Player $1$} & \begin{array}{c|c|c|} & A & B \\ \hline A & (6,6) & (5,5) \\ \hline B & (5,5) & (17,17) \\ \hline \end{array} \end{array} $$

Here, for each player the sum of payoffs for action $B$ is $22$ while the sum of payoffs for action $A$ is only $11,$ but $B$ does not dominate $A$; the payoff for $B$ is less than $A$ when the other player chooses $A.$

Answer 2

I now understand the reason why the dominated strategy is $W$ for both players.

If player 2 chooses $W$, then the payoff value would be $6+3=9$.

Comparably, choosing $NW$ would yield $7+4=11$.

Thus, $V^{2}(W) < V^{2}(NW)$. This is the definition of a dominated strategy.

Vice versa, $V^{2}(NW) > V^{2}(W)$ would be the dominant strategy, picking $NW$.