"The EV of a strategy profile is nondecreasing with respect to the number of options in the profile"

In other words, if I consider Strategy profile P1 that has n options and strategy profile P2 that has n+z options (z is obviously a positive integer), is the EV of P2 always greater than or equal to P1?

For a concrete example--is the EV of a strategy profile for playing the SB in a single raised pot that contains the options 3b-flat-or-fold at least as high at one which only contains 3b-or-fold?

In general the answer is no, check simple toy games like e.g. game of chicken to see why.

Only for 2-player zero sum games the answer is yes.

I think this is what we could call a "trivially true" statement. Let me explain why.

Suppose you are considering two strategies: P1 and P2 are otherwise identical, but strategy P1 always jams on the river in position with the nuts when checked to. Strategy P2 jams X% of the time and checks back (1-X)% of the time.

It is clear that strategy P1 has EV greater than or equal to P2, since checking back with the nuts is always incorrect. But strategy P2 has more options than P1. Nevertheless, we can claim that P2 has EV greater than or equal to P1 if we allow ourselves to set X = 0. If you are forcing X to be positive, your statement becomes false.

It intuitively seems true, but actually it isn't. You make the mistake to assume that the other players' GTO strategies remain unchanged when adding/removing options. That isn't the case.

Let's consider the game of chicken i mentioned above.

"The name "chicken" has its origins in a game in which two drivers drive towards each other on a collision course: one must swerve, or both may die in the crash, but if one driver swerves and the other does not, the one who swerved will be called a "chicken", meaning a coward; this terminology is most prevalent in political science and economics. "


Payoffs:
One swerves, one straight: -1, +1
Both swerve: 0, 0
Both straight: -1000, -1000

The GTO strategy for the game above is to swerve 99.9% and go straight 0.1% of the time, giving both actions an equal EV of -0.001.

If you take the option to swerve away from one of the players, the new GTO strategy for the other player is to swerve 100% of the time. The player with both options ends up with an EV of -1 and the player with only one option ends up with an EV of +1.

Sorry, should have been more clear. My example was only meant to talk about the 2-player zero-sum case. I figured you were right about the non-zero-sum case.

^^Cool stuff, thanks.

Here's another one. Assume a zero-sum game



"Someone playing a perfectly GTO strategy is guaranteed to extract at least some minimal amount of EV from the pot, call it EV_min, regardless of the opponent's strategy"

If true, is it also true that EV_min>=0? And would the value of EV_min depend entirely on opponent's strategy?

If this is true, would you prove it by contradiction?

Of course there exists some EV_min, but unless it's a two player zero sum game, the EV in GTO state is not necessarily equal to that minimum guaranteed EV. (For 2 player zero sum it's trivially the case: If an opponent could decrease your EV that means he is increasing his own, and if that's possible you were not in a GTO state in the first place.)

No, that's not true - not even for 2 player zero sum.