Consider this:

Hero is playing 6max NLH and opens from EP with range A. The villain is in CO and can be one of two types of villains each with a flatting frequency equal to one another--one with a middling to weak range (range B) and another with a middling to strong range (range C) with varying frequenches such that the cumulative flatting frequencis of all hands in ranges B and C are equal.

The question I pose is this:

Given these circumstances knowing that range A is vs. Either range B or range C and that either scenario is equally as likely does one maximize ev by taking a pseudo gto approach vs a range (range D) that is the average of ranges B and C or does one do something more extreme?

From hero's perspective, the villain's flatting range is the average of B and C, right? I don't think the hero has any incentive to deviate from a strategy that would be optimal against range D.

yeah this, for you it makes no difference where he randomizes. take a spot where villain wants to bluff his air 50% of the time. if he takes the minute mark of his watch, then technically he has two ranges too, but unless you figure out the pattern you cannot do better than playing against the average of the two ranges.

Is there a simple way to prove this?

Yes

This is true, but if I've understood OP, one of the players has a weaker range than the other. The two strats are not "balanced" or "equally viable", as such.
This makes me think that hero should focus on exploiting the wearker range/strat, if exploiting it doesn't lose too much vs the alternative (stronger flatting range) strat.

Just to be sure, when OP says Range D is the "average" of Range B and Range C, I take it to mean the frequency-weighted average of the two ranges. If Aces make up 10% of Range B and 4% of Range C, then Aces would make up 7% of Range D.

Under that assumption, then I think the answer to OP is clearly Yes. If the optimal strategy against Range B is HeroB and if the optimal strategy against Range C is HeroC, then it seems to me that the optimal strategy against a 50/50 "frequency mixture" of Range B and Range C is itself a 50/50 "frequency mixture" of HeroB and HeroC. I think this is a direct result of EV's being "linear in probabilities".

So, if Arty is saying that Hero should lean towards the counter-strategy that gives him more chips (more exploitative), I don't think this is correct. Or, saying it better, that is already taken into account by the linear-in-probability mixtures.

This is my assumption as well, I'm asking if I can prove this?

Isn't it just an implication of how EV's are linear and additive?

If X is Hero's strategy that maximizes EV[B], where B is Villain's range, and if Y is Hero's strategy that maximizes EV[C] then p*X + (1-p)*Y must be the strategy that maximizes EV[p*B + (1-p)*C] since EV[p*B + (1-p)*C] = p*EV[B] + (1-p)*EV[C].

Seems that “ranges of equal probability “ implies that each range has an equal number of combos as the other. If not, then I don’t think that you can use the term equal probability.

Why not plug the hands into a solver like this:

Hands that are included in both ranges get plugged in at 100% frequency.

Hands that aren’t included in both ranges are plugged in at 50% frequency.

Then solve.

You can also solve it by using the probabilities in the equations using both ranges and solving for minimax. These are done when a single player has non-uniform range with point masses. Yes, absolutely solvable and provable. Post it in the forum when completed BorkenStars!

Hmm. Thinking about this more, I'm not sure anymore. The bolded part could be significant. Since it's two different villains playing within their own range, I think some postflop decisions could end up being different than a single villain playing the combined range.

Let's take a dumb example. Flatting range B is QQ-JJ. Flatting range C is 100% of QQ and 50% of AA and JJ. So range D would be 25% of AA, 100% of QQ and 75% of JJ.

The flop comes 222. Hero goes all-in for 15x pot with KK+ and some bluffs.

A single villain playing range D can just call AA and fold everything else correctly.

Now assume two equally likely villains, one with range B and one with range C. The latter can still play correctly by just calling AA. However, the villain playing range B can't just assume that, from hero's perspective, there could be a player in the same seat who has AA in their range. From the villain's perspective, they never have KK+ here and hence need to bluffcatch with QQ some of the time. So in this case the villains' combined calling range would end up being AA with some percentage of QQ, instead of just AA.

I've changed my mind and gone back to agreeing with the first answers. :/

Suppose we assign villain B the arbitrary range of JJ, and villain C a range of 44.

What is the EV of hero's max exploit strat vs JJ (range B)? Let's call it EVx.
What is the EV of hero's max exploit strat vs 44 (range C)? Let's call it EVy.
What is the EV of hero's strat vs JJ/44 (range D)? Let's call it EVz.

I'm guessing that EVz is not exactly the average of EVx and EVy. However, it doesn't seem possible to find a strategy that does better than the one that treats both hands as being in a composite range. EVy (vs 44) is likely to be considerably higher than EVx (vs jacks), but since both hands are equally likely, hero can't use a strategy that focuses on 'punishing' one hand at the expense of the other.
The GTO strat of playing as if villain has JJ/44 will automatically take into consideration that villain has a good hand (JJ) 50% of the time and a bad hand (44) 50% of the time. Whatever it 'loses' vs jacks, the GTO strat will more than make up vs 44. If hero deviated from GTO and focused solely on punishing the weak range (44), he'd lose too much vs JJ.

Absolutely agree. If hero had no prior to exploit, then standard GTO* would be the most EV, but with a 50/50 mix of A/B ranges, then the modified GTOz will do the same, with slightly better EV.

This question is in regards to anon zoom tables im guessing?