Hello,
I have been reading MPT, and I had a quick question regarding optimal bet sizing in a heads-up GTO clairvoyant game where Hero has AA and QQ in his range and Villain always has KK (In position). Both players have $100 and there is $100 in the pot.

The book mentions that on the river (assuming all of the stated conditions), Hero should always go all in and that his EV will increase as his bet size increases, because he can bluff more often then. I understand that at a high level, but I am curious at to the math of why this works.

Furthermore, why is the EV equal to $100 times the total bet frequency for all of the rows here? I have attached a screenshot of the table that I am referencing below.

I should specify that this is on the river, and the 5 cards on the board don't affect anything (so AA beats KK and QQ loses to KK)

In the book The Theory of Poker Applied to No-Limit by David Sklansky in the chapter "Heads-Up on The End" and in the sub-chapter "Being Polarized and Betting Big" on page 201 this is explained.

Mason

Here's a different way to think about the EV which might make more sense. Since P1 is bluffing optimally, player 2s EV when he puts in $$ is 0, it is neither positive or negative b/c P1 is bluffing optimally. P1 will only check when he has QQ so P2 makes money whenever P1 checks. Since P1s only options are betting and checking, if his betting frequency increases, his checking frequency must decrease as both of these percentages must always total 100%. To maintain an optimal bluff strategy P1s betting and checking frequency's are dictated by the size of the bet relative to the pot. In order to maintain optimal bluffing frequency the bigger bets must be balanced by an increased frequency of betting. Since we know that P2s EV is directly related to P1 checking frequency then P1s EV must increase the more often he bets and P2s EV must decreased by the same amount.

Since the total pot is $100 P2s EV is $100 * % of times that P1 checks. P1s EV therefore must be $100 - P2s EV b/c all the EVs must add up to $100 (the size of the pot). If P1 checks 100% of the time (I.E. he always has QQ and can never bluff), P2s EV is $100, if P1 checks back 25% of the time, P2s EV is $25 ($100 *.25 + .75% * 0). P1s EV would $75 as shown in the table

What bet size would allow P1 to bet 100% of the time and still maintain optimal bluffing frequency? An infinite amount, therefore if P1 is perfectly balanced, they should always go all in b/c it minimizes the amount of times they need to check back for balance thus minimizing P2s EV.