考察Learn Poker Theory 2: Finding the Mathematically Optimal Bet Sizing Split in the AKQJT9 Game
IntroductionThis article is a continuation of the previous one. If you haven’t read it yet, please start there first: https://pokerqz.com/blog/theoretical_poker_1This time, the topic is bet size splitting. In GTO, stronger value hands generally tend to use larger bet sizes. But why does that happen? In this article, we’ll explore that question from a mathematical perspective using a toy game called the AKQJT9 game.1. What is the AKQJT9 game?The AKQJT9 game is a toy game I created for this topic, and it’s a variation of the AKQ game introduced last time. Compared to AKQ, the only change is that the number of cards increases from 3 to 6, but let’s quickly review the rules.PlayersThis game is played heads-up between Hero and Villain.CardsThe deck consists of only six cards: As, Ks, Qs, Js, Ts, 9s. Hand strength is A > K > Q > J > T > 9. Each player is dealt one random card, and no two players can receive the same rank.PositionHero is always in position, and Villain is always out of position.ActionsThe game starts directly on the river, and Villain (OOP) acts first. Villain always checks.Then Hero (IP) can choose any bet size or check back.If Hero checks back in step 2, the hand goes straight to showdown.If Hero bets in step 2, Villain can only call or fold. Raising is not allowed.Pot sizeThe initial pot is 1.2. The GTO solution computed by an algorithmNext, we want to compute the optimal solution with formulas. But to make the intuition easier to grasp, let’s first look at a GTO solution computed using the CFR algorithm. (I’d like to cover CFR in more detail another time.)Figure 1 GTO solution on the river when the board is 2s 2h 2d 3s 3hFigure 1 visualizes the GTO solution using a tool I built. The range chart is styled similarly to GTO Wizard. This chart shows Hero’s full-range strategy in a real poker spot where the board is 2s 2h 2d 3s 3h (a board unrelated to A through 9 in this toy game) after Villain checks.However, both players’ hand ranges are restricted to six combos: AsAh, KsKh, ... , 9s9h (only the spade-heart pocket pair combos are used).Hand strength is AsAh > KsKh > ... > 9s9h, and since the players can’t hold the same hand, this situation is equivalent to the AKQJT9 game.This time, I included many bet size options from 15% to 160%. Red indicates large bets, orange indicates small bets, and green indicates checks.You can see that the nuts A uses a larger value bet size than the second nuts K, meaning the strategy splits bet sizes.Figure 2 Strategy for KsKh in Figure 1 (left column: frequency, right column: EV; EV is shown multiplied by 100)Figure 3 Strategy for AsAh in Figure 1 (left column: frequency, right column: EV; EV is shown multiplied by 100)Figures 2 and 3 show the detailed strategies when Hero holds KsKh and AsAh. For each hand, the best approach is to use the bet size that maximizes EV, and you can see that actions with higher EV get higher frequency.Here’s what we can summarize about the AKQJT9 game from Figures 1 through 3.Summary so farThe nuts A uses a pure large bet. The optimal size seems to be between 120% and 130% pot, slightly closer to 120%.The second nuts K uses a pure small bet. The optimal size seems to be between 25% and 30% pot, slightly closer to 30%.Marginal hands Q through T pure check.The weakest hand 9 uses the same sizes as A and K for value betting, and bluffs at the appropriate frequencies.Additional notesQ cannot value bet. If Hero bets Q, Villain will always call with A and K and always fold 9. Even if Villain bluff-catches with J and T at 100% frequency, Hero’s equity when called becomes 50%, so the bet does not function as a value bet.T never bluffs. From a combo-count perspective, 9 alone provides enough bluff combos (explained later).3. The optimal solution when the AKQJT9 game allows two arbitrary bet sizesNow let’s move into the mathematical analysis.Since this game features bet size splitting, let’s define the value bet size used with A as $b_{A}$, and the value bet size used with K as $b_{K}$.From the previous article, when you pure value bet a strong hand at size $b$, the bluffing frequency that makes Villain’s marginal hands indifferent between calling and folding is $\frac{b}{1+b}$. Because we use two bet sizes this time, Hero’s optimal strategy becomes the following.Hero’s strategyHero (IP) handFrequency of bet $b_{A}$Frequency of bet $b_{K}$Frequency of checkA100K010Q , J , T0019$\frac{b_{A}}{1+b_{A}}$$\frac{b_{K}}{1+b_{K}}$$1 - \frac{b_{A}}{1+b_{A}} -\frac{b_{K}}{1+b_{K}}$Against this strategy, Villain responds as follows. For now, let $f_{1}$ be the bluff-catch frequency of Villain’s marginal hands (K through T) versus bet $b_{A}$, and let $f_{2}$ be the bluff-catch frequency of Villain’s marginal hands (Q through T) versus bet $b_{K}$. (We will compute the exact values next.)Villain’s strategy versus Hero’s bet $b_{A}$Villain handFrequency of callFrequency of foldA (1 combo)10K , Q , J , T (4 combos)$f_{1}$$1 - f_{1}$9 (1 combo)01Villain’s strategy versus Hero’s bet $b_{K}$Villain handFrequency of callFrequency of foldA , K (2 combos)10Q , J , T (3 combos)$f_{2}$$1 - f_{2}$9 (1 combo)01From the previous article, Villain’s optimal bluff-catch frequency makes Hero’s weak hand indifferent between bluffing and giving up by checking. In other words, we want(EV of Hero’s 9 betting $b_{A}$) = (EV of Hero’s 9 betting $b_{K}$) = (EV of Hero’s 9 checking)So the following equation must hold:$$\frac{1 ・ (-b_{A}) + 4f_1 ・(-b_{A}) + 4(1-f_1) ・1 }{5} = \frac{2 ・ (-b_{K}) + 3f_2 ・(-b_{K}) + 3(1-f_2) ・1 }{5} = 0$$Solving this gives $f_1 = \frac{4-b_{A}}{4(1 + b_{A})}$ and $f_2 = \frac{3-2b_{K}}{3(1 + b_{K})}$.Also, $1 - f_1 = \frac{5b_{A}}{4(1 + b_{A})}$ and $1 - f_2 = \frac{5b_{K}}{3(1 + b_{K})}$. (We’ll use these later.)4. Which $b_{A}$ and $b_{K}$ maximize Hero’s overall range EV?Now that we know both players’ optimal strategies, let’s find $b_{A}$ and $b_{K}$ that maximize Hero’s overall range EV. In the previous article, we noted that the EV of marginal hands and weak hands does not depend on Hero’s bet size. Therefore, the optimal $b_{A}$ and $b_{K}$ can be found by focusing only on the EV of A and K (the value hands).Let’s compute the EV for A and K. We can compute A’s EV the same way as before, but K’s EV requires one important caution: K can value-own itself against Villain’s A. Keeping that in mind, let Hero’s EV with A be $E_{A}(b_{A})$ and with K be $E_{K}(b_{K})$. Then:$$E_{A}(b_{A}) = \frac{4f_1・(1 + b_{A}) + (4(1-f_1) + 1)・1}{5} = \frac{1}{5} ((4-b_{A}) + \frac{5b_{A}}{1 + b_{A}} + 1) = \frac{1}{5} (5 + \frac{5b_{A}}{1 + b_{A}} - b_{A})$$$$E_{K}(b_{K}) = \frac{1・(-b_{K}) + 3f_2・(1 + b_{K}) + (3(1-f_2) + 1)・1}{5} = \frac{1}{5} (-b_{K} + (3-2b_{K}) + \frac{5b_{K}}{1 + b_{K}} + 1) = \frac{1}{5} (4 + \frac{5b_{K}}{1 + b_{K}} - 3b_{K})$$Next, we find the $b_{A}$ that maximizes $E_{A}(b_{A})$ and the $b_{K}$ that maximizes $E_{K}(b_{K})$ by differentiating each function.Using the standard derivative rule for rational functions:$$\frac{d}{db_{A}} E_{A}(b_{A}) = \frac{1}{(1 + b_{A} )^2} - \frac{1}{5}$$$$\frac{d}{db_{K}} E_{K}(b_{K}) = \frac{1}{(1 + b_{K} )^2} - \frac{3}{5}$$Both $\frac{d}{db_{A}} E_{A}(b_{A})$ and $\frac{d}{db_{K}} E_{K}(b_{K})$ are monotonically decreasing, so the optimal bet sizes are the values of $b_{A}$ and $b_{K}$ that make these derivatives equal to 0.Solving gives $b_{A} =\sqrt{5} -1 \approx 1.236$ and $b_{K} = \sqrt{\frac{5}{3}} -1 \approx 0.291$.This means Hero’s optimal bet size is 123.6% pot with A, and 29.1% pot with K. Earlier, based on the algorithmic results, we predicted:The nuts A uses a pure large bet. The optimal size seems to be between 120% and 130% pot, slightly closer to 120%.The second nuts K uses a pure small bet. The optimal size seems to be between 25% and 30% pot, slightly closer to 30%.And the math matches perfectly. Also, the bluffing frequencies for 9 are:For 123.6% pot: $\frac{b_{A}}{1 +b_{A}} \approx \frac{1.236}{1 + 1.236} \approx 0.553$For 29.1% pot: $\frac{b_{K}}{1 +b_{K}} \approx \frac{0.291}{1 + 0.291} \approx 0.225$Even combined, they do not exceed 1. In other words, 9 alone provides enough bluff combos, so there’s no need to turn T into a bluff.5. A key warning when applying bet size splitting in real gamesSo far, we’ve discussed how bet size splitting can increase the EV of your range. But you need to be careful when applying this concept in real poker.In this toy game, Villain had no option to raise. But in real poker, your opponent can raise.If your small-bet range contains zero nut hands, and your opponent realizes it, they can exploit you by responding with a wide, polarized raising strategy.That can significantly reduce your range EV.So if your opponent is strong enough to raise properly, you should include some nut hands in your small-bet range to stay balanced (especially when you are out of position).ConclusionThis ended up being another advanced article, but thank you for reading all the way through. In this article, I explained that GTO bet size splitting is an effective strategy with solid mathematical justification.To summarize the key takeaways:When you split bet sizes, using larger bet sizes with stronger value hands increases EV.Select bluff hands from the weakest, lowest-equity hands, and bluff at the correct frequency for each bet size (in practice, hands with better blockers tend to choose larger bluff sizes; this toy game had no blocker effects).In real poker, opponents can raise, so even if you split bet sizes, you need to distribute some nut hands across sizes to avoid being exploited.