NLHE, for the sake of example both players have to flip their hands over after the turn.

So let's say a hand gets dealt to the turn, the players flip their cards, and Hero has 60% equity and Villain has 40%. Pot is $100 so Hero's EV if there was no more betting would be $60 and villain's would be $40.

We'll say there's an effective stack of $20 remaining.

Hero bets his $20 obviously because he has 60% equity and knows it, and villain is clearly better off calling than folding. Size of call / new pot = required equity, so 20/140 = only 14.5% equity required for a correct call.

Hero's EV is now 60% of $140 ($84), obviously greater than his old equity of $60, yet villain has played his hand correctly. Hero's EV has increased as a result of villain's correct decision, which is not what the fundamental theorem says should happen.

Did I make a mistake, is the all-in screwing with the math, is there an issue with the information no longer being incomplete, is the correct comparison between the $100 EV for hero if villain folds, or is the fundamental theorem not as robust as its calculus namesake?

https://en.wikipedia.org/wiki/Fundam...eorem_of_poker

I think this theorem need update.

Both players played perfectly, so in one game EV for favorite is higher then in other, but that's fine because it's two different games.
Also EV for second game is 0.6*140-20=64$.

If stacks were 10000$ player with 60% would jam and have EV of 100$. If the other guy folds, but if other guy calls(incorrectly) he would have EV of 2060$, so in that case EV is increased.

You need to compare different scenarios within the same game.

This isn't a contradiction. If villain had made a worse decision hero's EV would have increased more. The correct decision is sometimes the least bad one.

Just before villain calls, hero's EV is $100 if villain folds, and then after the villain chooses to call hero's EV falls to $64. That sounds like the villain playing his hand correctly to me.

I’d argue hero never has an EV of 100 because there is a zero chance of villain folding.

The way the wiki lists the theorem I feel is not accurate as it makes it sound like a villain making the correct decision will always reduce a hero’s EV, which is not true.

But I think it’s true that a villain making a decision differently than if they could see the cards and ran equilab and the math is always going to be worse for villain.

Reduce hero's EV relative to what?

Hero's EV is reduced relative to the "wrong" decisions when villain makes the "right" decision. Hero's EV in the event of river going check/check doesn't matter from villain's perspective after hero already bets. He can only optimize by folding, calling or raising.

True, but in this situation:
—Hero knows villain will play correctly
—Hero is still increasing his EV by betting the turn

"Every time you play a hand differently from the way you would have played it if you could see all your opponents' cards, they gain; and every time you play your hand the same way you would have played it if you could see all their cards, they lose."

We agree on the EV and math, but arguing semantics, the second half of the phrase above makes it sound like hero should LOSE when villain plays his hand correctly.

Use of the word “lose” is misleading because generally we use the word to mean something a person already had. In this sense the only thing 60% equity dude is “losing” is the EV he’d have gotten if 40% dude blundered and folded.

I would re-word to something like

Every time a poker player makes the same decision (heads up) they would have if they could see their opponent’s cards and then correctly computed their pot odds, it will be their least bad (highest EV) play and the play that will give their opponent the least good (lowest EV) expected result.

But the hero does lose EV when villain plays his hand correctly.

When facing the 20bb all-in bet, the villain cannot raise, he can only call or fold. If he folds, (playing his hand incorrectly), then hero wins $100. If villain calls, (playing his hand correctly), then hero wins $64, and this is a relative loss of $36 for hero.

I think it is more of a generalization. For an even simpler example than the one provided in OP, imagine you are playing HU and you are both dealt AA and get it in preflop with reasonable raise sizes. In this situation you both played perfectly.

HeroÂ’s EV after betting and being called is 60 + 0.2x where x is the bet amount. VillainÂ’s EV is 40 - 0.2x. After a fold heroÂ’s EV is 100, villainÂ’s is 0.

Now letÂ’s consider different values for x. For x=20, hero has EV of $64, villain has EV of $36. Hero increases EV by $4 vs not betting so a bet is correct. Villain has a correct call since folding is EV 0. Once hero bets, his EV is either $100 or $64 depending on villainÂ’s action. Villain correctly calls reducing heroÂ’s EV by $36.

Now let x=$200. Notice that villain now risks 200 to win 300, so he must win 200200+300)= 40% of the time to break even. Now heroÂ’s EV after a fold is still $100. After a call itÂ’s 60 + 0.2(200) = $100. VillainÂ’s EV after calling is 40 - 0.2(200) = 0. Calling or folding are equivalent options as expected and hero is indifferent to villainÂ’s choice.

Finally let x=300. If villain calls heroÂ’s EV is $120. VillainÂ’s EV would be -$20. Villain should fold, and by doing so he costs hero $20 in EV.

In all cases the Fundamental Theorem holds - heroÂ’s EV is lower when villain acts correctly vs what it would be when villain does not make the correct choice. ThatÂ’s all the FT really says.

You have to look at winning/losing in the long term, not over the EV of a single hand.
Play situations like these a million times, where often the equity is closer to the pot odds and people will start making mistakes.
The one who makes the least mistakes will win over the long term.

Say villain needs 20% equity to continue but only has 19%, he will sometimes make bad calls.
Or villain needs 20% equity has 21% but makes a bad fold.

If you would make the exact same mistakes as villain in the exact same situations over an infinite sample size, you would both break even (assuming no rake)
If however you make more correct decisions (or less mistakes) you will end up winning because you have more often played your hand in a correct way.

Knowing your opponents cards isn't the entire story either.
You also need to be able to exactly calculate your equity and pot odds, assuming it's the final decision point in the hand.
It gets much more complicated when there can be more betting or raising.