Learn Poker Theory 1: Finding the Mathematically Optimal Value Bet Sizing in the AKQ Game

1. Introduction

In poker, choosing the right value bet size is crucial for maximizing expected value (EV). In this article, we’ll use a simple AKQ game to take a deep dive into the optimal bet size for value hands.

2. What Is the AKQ Game?

The AKQ game is a simplified model designed to teach the fundamentals of poker strategy. There are several variations, but in this article we’ll use the following rules.

• Players
  This game is played heads-up between Hero (you) and Villain (your opponent).
• Cards
  The deck contains only three cards: As, Ks, Qs. Hand strength is A > K > Q. Each player is dealt one card at random, and no two players can receive the same rank (ignore suits).
• Position
  Hero is always in position (IP), and Villain is always out of position (OOP).
• Action
  1. The game starts directly on the river, and Villain (OOP) acts first. Villain always checks.
  2. Then Hero (IP) can choose any bet size or check back.
  3. If Hero checks back, the hand goes straight to showdown.
  4. If Hero bets, Villain can only call or fold. Raising is not allowed.
• Pot size
  The initial pot is 1.

This simple model helps you learn core ideas like betting strategy and range construction.

3. Building Strategies by Hand and Pure Strategies

First, let’s summarize the optimal play for each hand for both Hero and Villain.

Hero’s Optimal Strategy

• 1. As
  Since this is the strongest hand, you should always value bet to target Villain’s K.
• 2. Ks
  A always calls and beats you, and Q just folds, so betting loses EV. Always check.
• 3. Qs
  This is the weakest hand, but you should mix it in as a bluff alongside your A value bets, bluffing at a frequency that makes Villain’s K indifferent.

Villain’s Optimal Response After Facing a Bet

• 4. As
  Villain is always ahead, so they always call.
• 5. Ks
  Villain loses to Hero’s A, but to avoid folding too often versus Hero’s Q bluffs, Villain should bluff-catch at the correct frequency (call often enough that Hero is indifferent about bluffing with Q).
• 6. Qs
  Villain is always behind, so they always fold.

Here, actions 1, 3, 4, and 6 are fixed at 100% frequency. If you deviate, you clearly lose EV for the reasons above.

A strategy where an action is taken with 100% frequency is called a pure strategy. Getting a pure strategy wrong causes a large EV loss, so you need to be careful.

Since you can’t change the pure strategies, the key to solving this game is determining:

• Hero’s Q bluff frequency (3), and
• Villain’s K calling frequency (5).

For 3 and 5, the optimal approach is to balance both actions. This is called a mixed strategy. If you don’t balance properly and lean too far in one direction, your opponent can exploit you. Here’s what that looks like.

If Hero bluffs too often with Q, Villain exploits by calling with K 100% of the time. If Hero bluffs too rarely with Q, Hero becomes too value-heavy with A, and Villain exploits by folding K 100% of the time.

If Villain calls too often with K, Villain becomes too bluff-catch heavy, and Hero exploits by bluffing with Q 0% of the time. If Villain calls too rarely with K, Villain overfolds, and Hero exploits by bluffing with Q 100% of the time.

So how do we find the correct frequencies for 3 and 5? We’ll cover that in the next section.

4. Mixed Strategy Frequencies and Indifference

To find optimal mixed strategy frequencies, we need to define what it really means to “make the opponent unsure.” The key concept here is indifference.

Indifference means a hand has the same EV between multiple actions.

That definition can feel abstract, so here’s a concrete example. Suppose the pot is 100, and your opponent bets 100 (a pot-sized bet).

You must call 100. If you call, the final pot becomes 100 (pot) + 100 (opponent bet) + 100 (your call) = 300. In other words, you pay 100 to compete for 300, so when your equity is exactly 100/300 = 1/3, calling and folding have the same EV (0).

This situation is called call-fold indifference. In general, when a hand is indifferent, it tends to remain in a mixed strategy to maintain balance.

In this AKQ game, the key to determining the mixed strategies in 3 and 5 is to adjust frequencies so that a specific opponent hand becomes indifferent.

Specifically:

• 3. Hero’s Q: choose a bluff frequency that makes Villain’s K call-fold indifferent.
• 5. Villain’s K: choose a calling frequency that makes Hero’s Q bet-check indifferent.

5. Solving the AKQ Game When the Only Bet Size Is 50% Pot

To keep things simple, in this section we’ll restrict Hero’s bet size to 50% pot (0.5).

Hero’s Strategy

Hero value bets A at frequency 1, bluffs with Q at an optimal frequency ($f_Q$), and chooses $f_Q$ so that Villain’s K has the same EV for calling and folding. If Villain holds K and calls, the payoff is:

• Hero has A (1 combo): -0.5 (Villain loses the 0.5 call)
• Hero has Q ($f_Q$ combos): 1 + 0.5 (Villain wins the pot 1 plus Hero’s bet 0.5)

If Villain folds K, their stack doesn’t change, so the payoff is 0.

Setting EV(call) = EV(fold), we get:

$$\frac{1 ・ (-0.5) +f_Q ・(1 + 0.5)}{1+f_Q} = 0$$

Solving gives $f_Q =\frac{1}{3}$.

Villain’s Strategy

Villain calls with K at an optimal frequency ($f_K$) and chooses $f_K$ so that Hero’s Q has the same EV for betting and checking.

If Hero holds Q and bets, the payoff is:

• Villain has A (1 combo): -0.5 (Hero gets called and loses)
• Villain has K and calls ($f_K$ combos): -0.5 (Hero gets called and loses)
• Villain has K and folds ($1 - f_K$ combos): 1 (Hero wins the pot 1)

If Hero checks Q, Hero always loses at showdown, so the payoff is 0.

Setting EV(bet) = EV(check), we get:

$$\frac{1 ・ (-0.5) +f_K ・(-0.5) + (1-f_K) ・1 }{2} = 0$$

Solving gives $f_K =\frac{1}{3}$.

Putting it all together, the optimal strategies are:

Hero’s Strategy

Villain’s Strategy (vs Hero bet 0.5)

6. Solving the AKQ Game With a Single Arbitrary Bet Size (b)

In the previous section, we solved the game with a fixed bet size of 0.5. We can solve the same way for a single bet size of $b$ (a constant). Replacing 0.5 with $b$ in the two equations gives:

$$\frac{1 ・ (-b) +f_Q ・(1 + b)}{1+f_Q} = 0$$

$$\frac{1 ・ (-b) + f_K ・(-b) + (1-f_K) ・1 }{2} = 0$$

Solving yields $f_Q = \frac{b}{1+b}$ and $f_K = \frac{1-b}{1+b}$.

Summarizing the optimal strategies:

Hero’s Strategy

Villain’s Strategy (vs Hero bet $b$)

Now look at $f_K = \frac{1-b}{1+b}$. When $b$ is greater than 1 (a pot-sized overbet), $f_K$ becomes negative, which clearly makes no sense. What’s happening?

In general, an overbet bluff requires a bluff success rate above 50%. When Hero bluffs with Q in this game, Villain holds either A or K. But Villain never folds A, so the bluff can never succeed more than 50% of the time.

That leads to an important conclusion: in the AKQ game, Hero should not use overbet bluffs. And if bluffs don’t exist at that size, value bets can’t be balanced either, so Hero won’t use overbets in this game. Therefore, from here on we assume $0 < b \leq 1$.

7. What Bet Size Maximizes EV for Hero’s Entire Range?

So far, we’ve solved the game for any single bet size $b$. Now we get to the main point.

In fact, Hero’s EV with K and Q does not change with the bet size $b$. Here’s why:

• Hero’s K: you always check, then lose to A and beat Q at showdown. You win the pot 1 with 50% probability, so EV is always 0.5.
• Hero’s Q: Villain’s K calling frequency makes Q indifferent between bluffing and giving up, so EV is always 0.

That means only A’s EV changes with bet size. So Hero’s optimal bet size is the one that maximizes EV when Hero holds A. Let’s compute the EV of betting A with size $b$.

$$(EV\text{ of Hero’s A}) = (P(\text{call}))・(1+b) + (P(\text{fold}))・1$$

Villain calls when they hold K ($\frac{1}{2}$) and choose to call ($\frac{1-b}{1+b}$), so:

$$P(\text{call}) = \frac{1}{2}\frac{1-b}{1+b}$$

Let this be $P(b)$. Then:

$$(EV\text{ of Hero’s A}) = P(b)・(1+b) + (1-P(b))・1 = bP(b) + 1 = \frac{1}{2}\frac{b(1-b)}{1+b} +1$$

Ignoring the constant term, we just need to maximize $\frac{b(1-b)}{1+b}$.

$$\frac{b(1-b)}{1+b} = \frac{-b(1+b) + 2(1+b) -2 }{1+b} = 2 -(b+ \frac{2}{1+b}) = 3 -((1+b) + \frac{2}{1+b})$$

By the arithmetic-geometric mean inequality, $(1+b) + \frac{2}{1+b}$ is minimized when $1+b = \sqrt{2}$, meaning $b = \sqrt{2} -1 \approx 0.414$. At that point, $\frac{b(1-b)}{1+b}$ reaches its maximum value of $3-2\sqrt{2}$.

This means Hero’s optimal bet size is 41.4% pot, because it maximizes the EV of A. We started with poker bet sizing, and somehow a square root appeared. Interesting.

If we plug this maximum back into the original EV expression:

$$(\text{Maximum EV of Hero’s A}) = \frac{1}{2} ・(3-2\sqrt{2}) +1 = \frac{5}{2} - \sqrt{2} \approx 1.086$$

If Hero couldn’t bet, A’s EV would be 1. So being in position clearly increases the EV of Hero’s overall range.

8. Summary

Thanks for reading through a long and challenging article. Here, we analyzed how Hero’s bet size affects expected value (EV). The key takeaways are:

• The AKQ game is a simplified poker model, and studying it in detail helps you learn many theories used in real poker.
• When the opponent cannot raise, value hands can have an optimal bet size that maximizes EV.

By analyzing the AKQ game, you can build a deeper understanding of bet sizing and GTO (Game Theory Optimal) strategy. In the next article, we’ll explore optimal bet sizing in more complex situations and how to apply these ideas in real games.

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