N00b GTO question
Herf
Red Chipper Posts: 11 ✭✭
GTO solvers often recommend a mixed strategy. For example, in a given situation they may say to bet 95% of the time and check 5% of the time.
In NoLimit Hold 'em For Advanced Players, Matthew Janda says that a solver will never recommend more than one action unless the EV of both actions are equal (assuming an optimal opponent).
If checking or betting both have the same EV, the only reason I can think of to have a mixed strategy is to mask your plays (ranges). In that case, the percentage would seem to be necessarily arbitrary, since there would be no change in EV.
If not EV, then what criteria does a solver use to determine the frequencies of actions in a mixed strategy?
In NoLimit Hold 'em For Advanced Players, Matthew Janda says that a solver will never recommend more than one action unless the EV of both actions are equal (assuming an optimal opponent).
If checking or betting both have the same EV, the only reason I can think of to have a mixed strategy is to mask your plays (ranges). In that case, the percentage would seem to be necessarily arbitrary, since there would be no change in EV.
If not EV, then what criteria does a solver use to determine the frequencies of actions in a mixed strategy?
Best Answer

TheGameKat Las VegasPosts: 5,440 That's a fascinating question, and I honestly don't know the answer in any detail, but I suspect it goes something like this.
Let's suppose you get a funny solution where you have some largeish number of combos, and the silicon overlord claims that every one is 5% bet, 95% check. What that is saying is that if you bet very rarely, and check a lot, you've created a situation that is the least open to exploitation from your opponent.
So while the EV of each line might be the same, it's only the same if you take the actions at those frequencies. If you suddenly start betting a lot, there is something your opponent can do (call or raise), that reduces the overall EV of the mixed strategy.
Answers
I get it now. The lines are only equal EV at equilibrium. Thanks again!