Lest say river spot 3 players in one player has nut/air range and the rest of players has same bluff catchers(they split the pot if they call and polarized range is bluffing).And bluff catching ranges dont block any part of polarized range.

How should each player play?What if bluff catching ranges must make call or fold at the same time dose that change the strategy?

P.S.Everybody has one pot size bet stack.

This is a really cool game. I've been thinking about it for a few hours and have a lot written down but not a definite solution yet.

My initial thought is that there's a simple equilibrium where the first bluffcatcher always folds to the nuts/air player's bet, and the second bluffcatcher calls 50%, which would be the case if the first bluffcatcher can never have positive (specifically greater than zero) EV in an equilibrium and is only ever indifferent between calling and folding.

But if the first bluffcatcher does have 0 EV then there might also be another complicated equilibrium where the first bluffcatcher calls sometimes which then prevents the second bluffcatcher from calling because it would require 1:1 nuts:air ratio from the bettor which is strictly dominated.

What's for sure is nuts/air player has to bluff sometimes and bluffs have to work at least 50% of the time, but I don't think 2:1 nuts:air is the right bluffing frequency.

This game is also interesting as the second bluffcatcher can seriously threaten the first bluffcatcher with irrational play, and the first bluffcatcher and the nuts/air player can bilaterally deviate to seriously disadvantage the second bluffcatcher.

I hope to work on this more later after some sleep or see someone else attack this.

Also I've been working on this assuming nuts/air player's entire range consists of more than 1/3 air. If that's not the case then one equilibrium is nuts/air player bets 100%, and both bluffcatchers fold.

The fun part of the problem is obviously finding equilibria when nuts/air player has more air than that and has the potential to overbluff.

OK, there are infinitely many equilibria of the following form as long as nuts/air player's range isn't more than 2/3 nuts. The first "simple equilibrium" I found was of this form.

Nuts/air player bets every nuts and and 1/2 as much air, checks the rest of the air.

Where x+(1-x)y=.5

First bluffcatcher calls with frequency x and folds with frequency 1-x.

If the first bluffcatcher calls, the second bluffcatcher always folds. If the first bluffcatcher folds, the second bluffcatcher calls at frequency y and folds at frequency 1-y.

I don't think any player stands to gain by unilaterally deviating from here. Nuts/air player could start overbluffing, but their EV would not change as all bluffs are 0EV. Each bluffcatcher could call more or less often, but each call for them is 0EV and folding is 0EV.

Sorry for the triple post.

Its seems to me like you said,second one should call 50 %.I dont get one thing,if that is solution second bluff catcher cant play that strategy and expect to at least break even regardless what other players doing.Is break even vs GTO opponent and wining vs rest of them,only property of equilibrium in 2 players games?If that is true for 3 player games then obv this would not be an equilibrium.
I dont get what x and y represents?

In your example none of the bluffs or bluff-catchers "win" anything, they are all exactly break-even. The first caller playing 50% is no better off than the other caller with 0%.

In any case, GTO does not guarantee that you break even from every single position or spot. e.g.: The blind positions are expected to lose money on average in the Nash Equilibrium. One player being worse off than another in a given spot is no contradiction of NE play, not even in 2-player games.

Bluffcatcher #1 always folding and Bluffcatcher #2 calling 50% is an equilibrium which satisfies the equation I provided.

Guaranteeing at least breakeven (non-loss) is not a property of Nash Equilibrium in games with 3 or more players (and only a property of 2 player games that are symmetric, e.g. a NLHE HU hand where either player is equally likely to be the BB). For example, in this equilibrium, after bluffcatcher #1 calls, bluffcatcher #2 can call every time, which drastically reduces the EV of both bluffcatchers at the benefit of the nuts/air player, and gives both bluffcatchers negative EV.

x and y are just variables that describe an equation that make nuts/air player indifferent to bluffing. It was derived as follows:

Bluffcatcher #1 will call some percent of the time (x).

x

Once bluffcatcher #1 folds, bluffcatcher #2 will call some percent of the time (y).

x+(1-x)y

The collective probability of bluffcatcher #1 and #2 calling needs to be 50% to make nuts/air player's bluffs 0EV.

x+(1-x)y=.5