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I have to confess I don't really understand the answer but does it mean that I can't include rake when trying to compare strategies ?
Yes. The reason is that if the game is not zero-sum (that is EVs of all players doesn't add up to the initial pot which is the case with rake as some money disappears in every pot) there might be more than one solution. Those solutions might have completely different EVs for both players. Simplifying it means that optimal play doesn't exist in games which are not zero-sum (rake, ICM). The reason we include an option to calculate results with both rake and ICM is that the hope is that if those disturb the results only a bit (small rake or relatively flat ICM structure) no matter what solution the solver finds it's still going to be useful and all the solutions are close enough to each other in overall EV. Unfortunately this is not guaranteed and there is nothing we can do about it as that's just math of it.
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Is the result I get here, where the "check whole range" strategy has higher EV than a mixed one, relevant or not?
It's not relevant. It just means that having 3 options there is more than one equilibrium. It also doesn't mean it's better to check having 3 options because the opponent may play differently there as well and then checking would have lower EV.
It might be difficult to visualize when you hear about it for the first time so here is an example to ponder:
Game:
-blinds are 1/2
-stacks are 1000
-it's HU and we play push or fold.
-rake is 5$ if it's all-in, 0$ if it ends without showdown
-both players have AA only in range
Consider the following pairs of strategies:
1)OOP pushes 100%, IP folds 100% if pushed into
2)OOP folds 100%, IP calls 100% if pushed into
In case 1) can OOP improve? They can't because pushing wins the pot (+2$ from the blind) and folding is obviously -1$. Can IP improve? They can't because calling has EV of -2.5$ (997.5$ - 1000$) (pot 2000$ -5$ rake and eq is 50%) while folding has EV of -2$. As neither player can improve it's equilibrium by definition.
In case 2) can OOP improve? They can't because a push will get called so pushing has EV of -2.5$ while folding has EV of -1$. Can IP improve? They can't because they are never pushed into and if they stop calling in case of being pushed into then OOP might start pushing.
In case 1) OOP wins 2$ every hand and IP losses 2$ every hand (blinds). In case 2) OOP losses 1$ every hand and IP wins 1$ every hand even though both are equilibriums and in both cases there is nothing any player can do about it. Two solutions, two completely different payoffs for both players.
Now, this to lesser extent happens in thousands of hundred of thousands nodes in the tree, especially on the river. The player who pushes first or already is commited to calling a lot forces the other to be more passive or they both lose to rake. That's why there might be (and almost always is) more than one equilibrium in games which are not zero sum. This makes comparing trees like that unreliable. Even if you run the same tree you could in theory get a different solution every time you run it (this won't happen as the solver is somewhat deterministic but if we change the algorithm a bit or tune some parameters it's already possible).
This fact about non-zero sum games is the reason we haven't implemented rake and ICM for a long time - we thought it's not that useful to calculate equilibria with those. There was a lot of demand though so we reluctantly added it. You have to be careful using those, especially if the rake is big or ICM close to the bubble/final table as then results really aren't very reliable.
We recommend solving without rake and then making reasonable adjustments for rake (that is avoiding playing marginal EV hands for example).
That helps a lot 