*The painter ChatGPT “A dramatic scene depicting a small, desperate mouse biting a large, surprised cat.” (from the Japanese proverb “A cornered mouse bits a cat.” which means despair makes cowards courageous.)
Preflop All-In: Practical Example
The game I play has a lot of beginners, so there are a lot of players who go all-in pre-flop. Therefore, it feels like whoever wins the pre-flop all-in will win Texas Hold’em. I feel like I want to think seriously about this issue for a while. However, if we were to mathematically formulate this problem and solve it rigorously, it would become a matter of game theory (stochastic dynamic optimization problems or reinforcement learning), and if there are multiple players, there would be an infinite number of Nash equilibria, so a model of the opponent would be necessary. Since it is currently difficult to deal with this issue strictly, the purpose of this article is to consider it starting with a simple example.
To consider an optimal preflop all-in strategy, let us consider a simplified problem. Assume there are two players and you only have a 9bb stack. Let us assume your opponent has 100+bb and is betting preflop an amount equal to your stack every game with an “you’re out, give up” mentality. Since you are in a cornered situation, you will not think about risk aversion and will make the decision to go all-in or fold to maximize your expected value. In this case, what hand should you use to make the best all-in move? For the first game we will assume you are the SB.
In this case, if you keep folding you will lose your entire stack in the 12th game, so you need to go all-in before that happens.
| Game | Blind | Stack |
|---|---|---|
| 1 | 0.5 | 8.5 |
| 2 | 1.0 | 7.5 |
| 3 | 0.5 | 7.0 |
| 4 | 1.0 | 6.0 |
| 5 | 0.5 | 5.5 |
| 6 | 1.0 | 4.5 |
| 7 | 0.5 | 4.0 |
| 8 | 1.0 | 3.0 |
| 9 | 0.5 | 2.5 |
| 10 | 1.0 | 1.5 |
| 11 | 0.5 | 1.0 |
| 12 | 1.0 | 0.0 |
The probability of winning for each hand in a two-player game is given in the following table.
It is important to note that the 169 squares are not equally likely to occur, and the number of each square in the \(\small {}_{52}C_2=1326\) possible combinations is as follows:
If the probability of winning for each square is \(\small p_{ij}\) and the probability of being dealt that hand is \(\small d_{ij}\) (the number in the table above divided by 1326),
\[ \small q_{ij} = p_{ij} \times d_{ij} \]
is the probability of winning given that hand on each square. If we express the range of entering hands as (\small R), the expected probability of winning at that time can be calculated as:
\[ \small p_R = \sum_{(i,j)\in R} q_{ij}. \]
The problem in this article is to find the optimal preflop all-in range for each game in this case.
Optimal Stopping Problem
The problem of deciding the optimal timing when a state (in the case of the problem in the previous section, the hand) is given probabilistically, like the problem in the previous section, is called the optimal stopping problem. This is a problem that often appears in financial engineering, and a typical example is the pricing problem of American options (the problem of determining the optimal timing to exercise an option). Generally, the optimal timing for the optimal stopping problem is determined by inductively finding the optimal solution from later problems on the time axis (this type of solution method is called backward induction).
We will use symbols to denote the entry hand range in the \(\small n\)-th game as \(\small R^{(n)}\), the winning probability of the hand as \(\small p_R^{(n)}\), the remaining stack size as \(\small s^{(n)}\), the expected value of chips if you fold as \(\small V_{\text{Fold}}^{(n)}\), and the expected value of chips if you go all-in as \(\small V_{\text{All-in}}^{(n)}\). It is easy to understand that the expected value of your chips when you go all-in is:
\[ \small V_{\text{All-in}}^{(n)} = 2 \times s^{(n-1)} \times p_R^{(n)}. \]
The expected value of folding must be calculated inductively from the last game. In the 12th game, if you fold you will lose all your chips, so \(\small V_{\text{Fold}}^{(12)}=0\). Therefore, on the 12th game, you will go all-in no matter what the hand is (in this case, your probability of winning is 0.5). If you fold in the 11th game, you will go all-in in the 12th game, which can be represented as:
\[ \small V_{\text{Fold}}^{(11)}=E\left[V_{\text{All-in}}^{(12)}\right] = 2 \times 0.5 = 1.0. \]
The 10th game can be calculated as:
\[ \small V_{\text{Fold}}^{(10)}= E\left[\max\left\{ V_{\text{All-in}}^{(11)}, V_{\text{Fold}}^{(11)} \right\} \right] \]
because you can choose to go all-in or fold in the 11th game. Note that since the hand that will be dealt in the 11th game is unknown at the time of the 10th game, it is necessary to calculate the expected value. By performing a recursive calculation, we can express
\[ \small V_{\text{Fold}}^{(n)}= E\left[\max\left\{ V_{\text{All-in}}^{(n+1)}, V_{\text{Fold}}^{(n+1)} \right\} \right] \]
as for all \(\small n\).
From the above considerations, it follows that you should go all-in with a hand range that has an expected value that exceeds the expected value of your chips if you fold, out of the 169 squares. Therefore, recursively from subsequent games, you just need to go all-in with a hand whose winning probability \(\small p_{ij}^{(n)}\) satisfies
\[ \small R^{(n)} = \left\{(i,j)| 2 \times s^{(n-1)} \times p_{ij}^{(n)} > V_{\text{Fold}}^{(n)} \right\}. \]
Solution
The calculations are quite difficult, so I’ll just show you the solution. The skeptical reader can do the calculations for himself. The expected value of chips when folding \(\small V_{\text{Fold}}^{(n)}\) and the probability of winning \(\small p_{ij}^{\ast(n)}\) such that \(\small 2s^{(n-1)}p_{ij}^{\ast(n)} = V_{\text{Fold}}^{(n)}\) can be estimated for each game as follows.
| Game | Stack | Fold Value | Equilibrium Prob |
|---|---|---|---|
| 1 | 8.50 | 8.9532 | 49.74% |
| 2 | 7.50 | 7.9944 | 47.03% |
| 3 | 6.50 | 7.2710 | 48.47% |
| 4 | 6.00 | 6.3122 | 45.09% |
| 5 | 5.00 | 5.6240 | 46.87% |
| 6 | 4.50 | 4.6640 | 42.40% |
| 7 | 4.00 | 4.0274 | 44.75% |
| 8 | 3.00 | 3.0570 | 38.21% |
| 9 | 2.50 | 2.5010 | 41.68% |
| 10 | 1.50 | 1.5010 | 30.02% |
| 11 | 1.00 | 1.0000 | 33.35% |
| 12 | 0.00 | 0.0000 | 00.00% |
Even if a player folds, he or she has the right to choose their hand in the subsequent game, so it can be seen that the expected value of chips at the time of folding is estimated to be higher than the stack at the time of folding. You can also see that if you are in the big blind position, you tend to be looser because you will lose more chips if you fold.
The optimal preflop all-in hand range is one that is played with hands that have a higher probability of winning than equilibrium probability. It may seem very loose, but when you’re in a corner, if you don’t gamble boldly, you’ll just end up in a tight spot. The hand ranges for each of the 12 rounds are as follows. This is a simplified example, but if you find yourself in a tight spot in a tournament, it might help you decide what hand range to go with depending on how much of your stack you have left. However, in tournaments, it’s important to survive as many games as possible in order to improve your ranking, so things may get a little tighter. Considering an all-in strategy to maximize the number of games you survive is a topic for another time.
- 1st game
- 2nd game
- 3rd game
- 4th game
- 5th game
- 6th game
- 7th game
- 8th game
- 9th game
- 10th game
- 11th game
- 12th game
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