Note: the math and the picture didn't transfer. I may try to fix it in future, but for now you might want to just read it at the original site. [Mod/Edit note: Should all be fixed now!]

Consider games with the following payoff matrix:

Player 2

Krump

Flitz

Player 1

Krump

(W, W)

(X, Y)

Flitz

(Y, X)

(Z, Z)

One such game is the Prisoner's Dilemma (in which strategy "Krump" is usually called "Cooperate", and "Flitz" is usually called "Defect"). But the Prisoner's Dilemma has additional structure. Specifically, to qualify as a PD, we must have $Y > W > Z > X$ . $Y > W$ gives the motivation to defect if the other player cooperates, and $Z > X$ gives that motivation if the other player defects. With these two constraints, the Nash equilibrium is always going to be Flitz/Flitz for a payoff of $(Z, Z)$ . $W > Z$ is what gives the dilemma its teeth; if instead $Z > W$ , then that equilibrium is a perfectly fine outcome, possibly the optimal one.

I usually think of a Prisoner's Dilemma as also having $2 W > X + Y > 2 Z$ . That specifies that mutual cooperation has the highest total return - it's "socially optimal" in a meaningful sense¹ - while mutual defection has the lowest. It also means you can model the "defect" action as "take some value for yourself, but destroy value in the process". (Alternatively, "cooperate" as "give some of your value to your playmate², adding to that value in the process".) We might consider instead:

If $2 W < X + Y$ , then defecting while your playmate cooperates creates value (relative to cooperating). From a social perspective, Krump/Flitz or Flitz/Krump is preferable to Krump/Krump; and in an iterated game of this sort, you'd prefer to alternate $X$ with $Y$ than to get a constant $W$ . Wikipedia still classes this as a Prisoner's Dilemma, but I think that's dubious terminology, and I don't think it's standard. I might offhand suggest calling it the Too Many Cooks game. (This name assumes that you'd rather go hungry than cook, and that spoiled broth is better than no broth.)
If $2 Z > X + Y$ , then defecting while your playmate defects creates value. I have no issue thinking of this as a Prisoner's Dilemma; my instinct is that most analyses of the central case will also apply to this.

By assigning different value

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