Is this well-known claim actually true?

Interesting, can you show the working to prove that shoving is optimal sizing for equilibrium opener?

I don’t have anything beyond what I posted. Shoving maximizes the positive EV you get when an opponent erroneously calls you with a hand that isn’t AA. Shoving does not affect your EV against an opponent playing a proper strategy - EV is zero regardless of bet size. I concede , depending on the elasticity of your opponent’s making errors and calling, there may be a smaller sizing that will generate more calls, but that’s getting quite a bit beyond the theory, which is telling us that we should never get anything other than zero EV regardless of what size we use. I will amend - shoving is most definitely a profitable action, but against some strategies there might be a better sizing.

But there are literally an infinite number of non-equilibrium strategies that an opponent might use, so I don’t see how we could possibly come up with a single optimal strategy to account for that. For example, an opponent might play a strategy of call an open with KK or AA if and only if the open sizing is not greater than X bb. Our optimal strategy would then be open only with AA using a size of X BB. Again, though, it’s impossible to make an optimal strategy for ALL possible non-equilibrium strategies. We know that shoving is

1. Profitable or zero EV against any opposing strategy
2. Avoids loss of EV from postflop errors

Therefore without claiming it to be optimal against all possible opposing strategies, shoving would be perfectly viable as a general strategy.

Sure, I get what you're saying. I just have trouble accepting that the opponent can simplify his strategy to only calling with AA.

But I do remember in Modern Poker Theory reading about how important betsize is and for example shoving with AA in regular NLHE is a much larger EV mistake than opening with 72o to 2bbs. So perhaps there is something to be said for using a smaller RFI side in the example in this thread too.

It isn’t simplification. Calling with AA is the full-blown GTO strategy. There exists a strategy that is zero EV regardless of opponent’s action, namely folding all hands. By definition, that means folding all hands is a GTO strategy, nothing any opponent does can cause it to lose EV. But there is also another strategy that has the same property, calling with only AA. No opening strategy can make that strategy -EV. This has the advantage of exploiting a non-GTO opening strategy since it will have +EV if someone erroneously opens a hand other than AA.

If you call with AA plus any other combo, call it XY, you can now be exploited. A strategy of opening only AA will be +EV against a calling range of AA+XY. Opener’s AA is zero EV against caller’s AA and +EV versus XY, so it is +EV versus that calling range. Since opener has +EV using an open range of AA only, caller must be -EV when calling with AA + XY, so AA+XY cannot be a GTO calling range.

Note the same argument applies to an open range of AA+XY. Calling with AA only would be +EV against that open range. Hence AA+XY cannot be a GTO opening range either.

Sure, I do understnad what you're saying.

So why don't people play NLHE this way, only calling pre with AA? Because the blinds give an incentive for the opener to open wider, and therefore the other players can defend wider as a result?

I think what you mean is that once you know the opener is only going to open with AA, then the GTO strategy is to only call with AA, right?

I'd like to see actual mathematical proof of how opening with only AA is the GTO approach.

You've been shown proof that it's the GTO approach, you simply refuse to accept it.

Try an analogous example. BTN jams (3bb), SB and BB both call (100bb). So it's a dry side pot.

Action checks down to the river. SB checks. BTN (who's all in) gets excited and tables a royal flush out of turn. Action on you in the BB.

To recap: The side pot is 0. The all-in player has a Royal flush. What is your strategy? Should you ever bluff? Should you bet the 3rd nuts?

Sure thing, I get it. Thanks for entertaining the discussion, it's interesting to hear what people think.

For me what really seals this is that if you know any opponent is only playing AA, then opening any other hand quite simply becomes minus EV (thanks Chat GPT, lol).

Having read the thread from start to finish it's safe to say no ''mathematical proof'' will convince you. You've gotten 2 really good explanations(that include math). The reason you're struggling to accept it is because you're confusing terms.

Did you read my last post?

I get it, and I agree with what has been posted. I was playing devil's advocate to some extent.

Isn't it quite easy, I just play AA. Show how you exploit that -> playing only AA is GTO.

IÂ’ll try again:

Suppose for the sake of simplicity itÂ’s heads up and you open all in for 100bb. (That way we donÂ’t have to worry about postflop equities in the EV calculation.)

LetÂ’s say OP opens a range of AA plus 22 as a bluff. What is the optimal calling range? We can call with AA only. For simplicity again, I assume AA vs 22 has 80% equity. Given that, caller has an EV of zero when opener has AA and an EV of 100x.8 - 100x.2=60bb when opener has 22. Note that caller is blocking openerÂ’s AA, so when called, he will have AA with probability of 1/7 and 22 with probability 6/7 so the overall EV for caller is 6/7x60=51.4 bb

Now letÂ’s add a hand to callerÂ’s range. LetÂ’s try AA plus KK. Opener still opens AA plus 22. The EV calculation for each possible matchup is identical - AA vs AA is zero. AA vs KK, AA vs 22 or KK vs 22 all are 60 for the higher pair. Opener with AA vs AA will happen 1/14 of the time (opener will have AA with probability 1/2, caller will have AA with probability 1/7 given that opener has AA). AA for opener vs KK happens with probability 6/7. Caller will have AA or KK with equal probability if opener has 22, so the 22 vs AA and 22 vs KK each happen with probability 1/4. From here it just is a matter of adding up the weighted EVs. Caller has +60EV with probability 1/2 and -60 with probability 3/7, so his overall EV is 60x1/2-60x3/7=4.3BB, far lower than the alternative of only calling with AA.

This is not a general proof, but it is suggestive. Here is a general proof. Let the opener bluff with probability b and open AA with probability 1-b. Let caller bluff with probability c, and therefore call with AA with probability 1-c. Let x be the equity of AA vs openers bluff, y the equity of AA vs callers bluff and z the equity of callers bluff vs openers bluff. The caller will in general have an EV of xb(1-c) + zbc - yc(1-b).

IÂ’m going to rewrite that expression in two algebraically equivalent ways. First
bx + bc(z-x) + cy(b-1). From this form it is clear that if caller only calls AA (c=0) then his EV is bx. If c is not zero, then b is a probability so the third term is clearly not positive (either zero if opener only opens AA and negative otherwise). The second term is likewise negative since z

The second form is -cy +bx(1-c) +bc(z+y). In this case if we set b=0 implying that opener only opens AA then caller has EV of -cy (this is zero when c=0 as expected). Again, c is a probability so the second term is nonnegative. The third term is obviously positive, so the overall formula will be greater than -cy if b is not zero. Thus a nonzero b increases callers EV, implying that openers EV is decreased. Therefore b=0 is an optimal strategy for opener , I.e. raising only AA.