Recall that they have the following payoff matrix:
Player 2
Krump
Flitz
Player 1
Krump
(W,W)
(X,Y)
Flitz
(Y,X)
(Z,Z)
where W>Z.[1] We can apply a positive affine transformation (that is, n↦an+b where a>0) to all of W,X,Y,Z without changing the game. So let's pick the function n↦n−ZW−Z. This sends W to 1 and Z to 0, leaving us with just two parameters: R=X−ZW−Z and S=Y−ZW−Z.
So what happens if we plot the space of these games on a graph? The lines {X,Y}={W,Z} become {R,S}={0,1}, i.e. vertical and horizontal lines. The lines X+Y=2W and X+Y=2Z become the diagonals R+S=2 and R+S=0; and X=Y becomes the diagonal R=S. Drawing those lines, and relabelling in terms of W,X,Y,Z, it looks like this:
Note: I tried to limit the vertical size of that but couldn't figure out how. Sorry!
Note that Cake Eating (my favorite game) is the only one with a finite boundary; the other boxes extend to infinity. There are also finite components in the Farmer's Dilemma (with X+Y<2W), and Stag Hunt and Studying For a Test (with X+Y>2Z). As drawn, Prisoner's Dilemma occupies almost all of the box it shares with Too Many Cooks; but Too Many Cooks (above the line X+Y=2W) is also infinite. (I initially got those the wrong way around, so the drawing isn't very clear there.)
I don't know if we learn much from this, but here it is.
In the previous post I mostly ignored equalities because it was mildly convenient to do so. But the analysis here completely fails if we allow W=Z. So now I'm ignoring them because it's considerably more convenient to do so. ↩︎
Loosely and non-rigorously, x/0 is infinite, and so all games with W=Z are extreme forms of the corner games (unless X=W=Z or Y=W=Z).
X~Y>W=Z gets you an anti-coordination game and W=Z>X~Y gets you a pure or relatively pure coordination game (Let's Party).
Y>W=Z>X and X>W=Z>Y are interesting, because they equate games as different as the PD (or Too Many Cooks) and Abundant Commons. I would describe this game as more similar to the Abundant Commons than the PD, as Flitz/Flitz is a perfectly acceptable equilibrium. The value transfer here is neither hyperefficient nor inefficient, but merely efficient.
The triple equalities here are equivalent under name change, so, WLOG, let's take X=W=Z. Then, there are two games: Y>X and Y<X. Looking at the diagram, X=W=Z>Y should resemble Studying for a Test, while Y>X=W=Z should resemble the Farmer's Dilemma.
The former game has a primary theme of avoiding Y, and so, while Flitz/Flitz is an equilibrium, I would expect to see more Krump/Krump, as it is never beneficial to play Flitz when there's any risk of Krump.
The latter game is more complex, but the equilibrium you actually see is Flitz/Flitz, because the only way to get Y is if you play Flitz.
Finally, with all four equal, there is no longer much of a game. All strategies are equilibria, the payoff is identical in each case. This is the trivial game.
Expanding on the Y>W=Z>X and X>W=Z>Y, I would split Abundant Commons at Y=Z, into Abundant Commons above the line and Deadlock below it. Then, the games equated are Deadlock and the PD, and those form a natural continuum.
I'm not familiar with all these games by name, and feel like the post would be better with some explanations. I'm particularly confused by how Y very negative and X very positive becomes 'abundant commons'. Attempting to Google 'abundant commons game' doesn't immediately clarify.
Some time after writing Classifying games like the prisoner's dilemma, I read a paper (I forget which) which pointed out that these games can be specified with just two numbers.
Recall that they have the following payoff matrix:
where W>Z.[1] We can apply a positive affine transformation (that is, n↦an+b where a>0) to all of W,X,Y,Z without changing the game. So let's pick the function n↦n−ZW−Z. This sends W to 1 and Z to 0, leaving us with just two parameters: R=X−ZW−Z and S=Y−ZW−Z.
So what happens if we plot the space of these games on a graph? The lines {X,Y}={W,Z} become {R,S}={0,1}, i.e. vertical and horizontal lines. The lines X+Y=2W and X+Y=2Z become the diagonals R+S=2 and R+S=0; and X=Y becomes the diagonal R=S. Drawing those lines, and relabelling in terms of W,X,Y,Z, it looks like this:
Note: I tried to limit the vertical size of that but couldn't figure out how. Sorry!
Note that Cake Eating (my favorite game) is the only one with a finite boundary; the other boxes extend to infinity. There are also finite components in the Farmer's Dilemma (with X+Y<2W), and Stag Hunt and Studying For a Test (with X+Y>2Z). As drawn, Prisoner's Dilemma occupies almost all of the box it shares with Too Many Cooks; but Too Many Cooks (above the line X+Y=2W) is also infinite. (I initially got those the wrong way around, so the drawing isn't very clear there.)
I don't know if we learn much from this, but here it is.
In the previous post I mostly ignored equalities because it was mildly convenient to do so. But the analysis here completely fails if we allow W=Z. So now I'm ignoring them because it's considerably more convenient to do so. ↩︎