# Optimal Strategies in a One-Card Poker Game with Perfect Logicians

I’m analyzing a one-card poker game using a deck with three cards: an ace (lowest), a deuce, and a trey (highest). Each player antes \$1 into the pot, and each player is dealt one card. The game proceeds as follows: • Player 1 acts first and can either bet \$1 or check.
• If Player 1 bets, Player 2 can either call the bet or fold.
• If Player 1 checks, Player 2 can either bet $1 or check. • If Player 2 bets after Player 1 checks, Player 1 can either call the bet or fold. • The winner is determined by the highest card if both players check, or if one player bets and the other calls. If one player bets and the other folds, the betting player wins by default. Given that both players are perfect logicians and aim to maximize their chances of winning, I need to figure out the optimal strategy for each player: • How should Player 1 decide whether to bet or check? • How should Player 2 respond to Player 1’s actions to play optimally? • What factors should each player consider in their decision-making process based on their own card and the potential actions of the opponent? • For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. Commented Jul 24 at 6:10 ## 1 Answer Finally, I found a solution. The obvious plays are: neither player should call with the ace, always call with the trey, and never bet with a deuce. By raising with a deuce, you will win \$1 against an ace, because your opponent will fold, and lose \$2 against a trey, because he will raise. A superior strategy with a deuce is to always check, and then to always call if the other player bets. By following the check and call strategy, you'll win \$2 instead of \$1 against an ace, and stay equal against a trey at a \$2 loss. If your opponent checks with the ace, you'll still win \\$1, and be no worse off. You can only gain by checking and calling with a deuce.

There are five other, more difficult, decision points to be faced by both players, as follows:

$$\textbf{1.}$$ $$p_1$$: Probability player 1 bets with an ace (bluffing).

$$\textbf{2.}$$ $$p_2$$: Probability player 1 calls with a deuce after player 2 bets (hoping player 2 is bluffing).

$$\textbf{3.}$$ $$p_3$$: Probability player 1 bets with a trey (hoping player 2 will call).

$$\textbf{4.}$$ $$q_1$$: Probability player 2 bets with an ace, after player 1 checks (bluffing).

$$\textbf{5.}$$ $$q_2$$: Probability player 2 calls with a deuce, after player 1 bets (hoping player 1 is bluffing).

The key to game theory problems like this is to put your opponent at an indifference point at non-obvious decisions. If the odds favor one way or the other, the person making the decision will always go with the odds and gain an advantage. To do that, let's express player 1's expected value.

\text{EV}_1 = \begin{aligned} &\text{pr(player 1 has A)} \times \text{pr(player 2 has 2)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 checks)} \times (-1) \\ &+ \text{pr(player 1 has A)} \times \text{pr(player 2 has 2)} \times \text{pr(player 1 bets)} \times \text{pr(player 2 folds)} \times (+1) \\ &+ \text{pr(player 1 has A)} \times \text{pr(player 2 has 2)} \times \text{pr(player 1 bets)} \times \text{pr(player 2 calls)} \times (-2) \\ &+ \text{pr(player 1 has A)} \times \text{pr(player 2 has 3)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 bets)} \times \text{pr(player 1 folds)} \times (-1) \\ &+ \text{pr(player 1 has A)} \times \text{pr(player 2 has 3)} \times \text{pr(player 1 bets)} \times \text{pr(player 2 calls)} \times (-2) \\ &+ \text{pr(player 1 has 2)} \times \text{pr(player 2 has A)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 checks)} \times (+1) \\ &+ \text{pr(player 1 has 2)} \times \text{pr(player 2 has A)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 bets)} \times \text{pr(player 1 folds)} \times (-1) \\ &+ \text{pr(player 1 has 2)} \times \text{pr(player 2 has A)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 bets)} \times \text{pr(player 1 calls)} \times (+2) \\ &+ \text{pr(player 1 has 2)} \times \text{pr(player 2 has 3)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 bets)} \times \text{pr(player 1 folds)} \times (-1) \\ &+ \text{pr(player 1 has 2)} \times \text{pr(player 2 has 3)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 bets)} \times \text{pr(player 1 calls)} \times (-2) \\ &+ \text{pr(player 1 has 3)} \times \text{pr(player 2 has A)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 checks)} \times (+1) \\ &+ \text{pr(player 1 has 3)} \times \text{pr(player 2 has A)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 bets)} \times \text{pr(player 1 calls)} \times (+2) \\ &+ \text{pr(player 1 has 3)} \times \text{pr(player 2 has A)} \times \text{pr(player 1 bets)} \times \text{pr(player 2 folds)} \times (+1) \\ &+ \text{pr(player 1 has 3)} \times \text{pr(player 2 has 2)} \times \text{pr(player 1 checks)} \times \text{pr(player 2 checks)} \times (+1) \\ &+ \text{pr(player 1 has 3)} \times \text{pr(player 2 has 2)} \times \text{pr(player 1 bets)} \times \text{pr(player 2 folds)} \times (+1) \\ &+ \text{pr(player 1 has 3)} \times \text{pr(player 2 has 2)} \times \text{pr(player 1 bets)} \times \text{pr(player 2 calls)} \times (+2) \end{aligned}

\text{EV}_1 = \begin{aligned} &\frac{1}{3} \times \frac{1}{2} \times (1 - p_1) \times 1 \times (-1) \\ &+ \frac{1}{3} \times \frac{1}{2} \times p_1 \times (1 - q_2) \times 1 \\ &+ \frac{1}{3} \times \frac{1}{2} \times p_1 \times q_2 \times (-2) \\ &+ \frac{1}{3} \times \frac{1}{2} \times (1 - p_1) \times 1 \times (-1) \\ &+ \frac{1}{3} \times \frac{1}{2} \times p_1 \times 1 \times (-2) \\ &+ \frac{1}{3} \times \frac{1}{2} \times 1 \times (1 - q_1) \times 1 \\ &+ \frac{1}{3} \times \frac{1}{2} \times 1 \times q_1 \times (1 - p_2) \times (-1) \\ &+ \frac{1}{3} \times \frac{1}{2} \times 1 \times q_1 \times p_2 \times 2 \\ &+ \frac{1}{3} \times \frac{1}{2} \times 1 \times 1 \times (1 - p_2) \times (-1) \\ &+ \frac{1}{3} \times \frac{1}{2} \times 1 \times 1 \times p_2 \times (-2) \\ &+ \frac{1}{3} \times \frac{1}{2} \times (1 - p_3) \times (1 - q_1) \times 1 \\ &+ \frac{1}{3} \times \frac{1}{2} \times (1 - p_3) \times q_1 \times 1 \times 2 \\ &+ \frac{1}{3} \times \frac{1}{2} \times p_3 \times 1 \times 1 \\ &+ \frac{1}{3} \times \frac{1}{2} \times (1 - p_3) \times 1 \times 1 \\ &+ \frac{1}{3} \times \frac{1}{2} \times p_3 \times (1 - q_2) \times 1 \\ &+ \frac{1}{3} \times \frac{1}{2} \times p_3 \times q_2 \times 2 \end{aligned}

$$\text{EV}_1 = \frac{1}{6} \times \left( -1 + p_1 + p_1 - p_1 q_2 - 2 p_1 q_2 - 1 + p_1 - 2 p_1 + 1 - q_1 - q_1 + q_1 p_2 + 2 q_1 p_2 - 1 + p_2 - 2 p_2 + 1 - q_1 - p_3 + p_3 q_1 + 2 q_1 - 2 p_3 q_1 + p_3 + 1 - p_3 + p_3 - p_3 q_2 + 2 p_3 q_2 \right)$$

$$\text{EV}_1 = \frac{1}{6} \times (p_1 - q_1 - p_2 - 3 p_1 q_2 + 3 q_1 p_2 - p_3 q_1 + p_3 q_2) \quad \text{(Equation 1)}$$

$$\text{EV}_1 = \frac{1}{6} \times \left[q_1 \times (-1 + 3 p_2 - p_3) + q_2 \times (-3 p_1 + p_3) + p_1 - p_2 \right]$$

What values for $$q_1$$ and $$q_2$$ will make $$\text{EV}_1$$ equal to a constant, causing player 1 to be indifferent to all values of $$p_1$$, $$p_2$$, and $$p_3$$?

The answer is $$q_1 = \frac{1}{3}$$, and $$q_2 = \frac{1}{3}$$, so...

$$\text{EV}_1 = \frac{1}{6} \times \left[ \frac{1}{3} \times (-1 + 3 p_2 - p_3) + \frac{1}{3} \times (-3 p_1 + p_3) + p_1 - p_2 \right]$$

$$= \frac{1}{6} \times \left[ -\frac{1}{3} + p_2 - \frac{p_3}{3} - p_1 + \frac{p_3}{3} + p_1 - p_2 \right]$$

$$= \frac{1}{6} \times \left(-\frac{1}{3}\right)$$

$$= -\frac{1}{18}$$

So, player 2 should bluff with an ace after a check and call with a deuce, each with probability $$\frac{1}{3}$$. If player 2 does that, it won't make any difference what player 1 does at his three difficult decision points; his overall expected value will be $$-\frac{1}{18}$$.

Next, what should player 1's optimal strategy be so that player 2 can't exploit a bad strategy? Let's go back to Equation 1. The expected value of player 2 will be the opposite of the expected value of player 1. So...

$$\text{EV}_2 = \left(-\frac{1}{6}\right) \times \left( p_1 - q_1 - p_2 - 3 p_1 q_2 + 3 q_1 p_2 - p_3 q_1 + p_3 q_2 \right)$$

$$\text{EV}_2 = \left(-\frac{1}{6}\right) \times \left[ p_1 \times (1 - 3 q_2) + p_2 \times (3 q_1 - 1) + p_3 \times (q_2 - q_1) - q_1 \right]$$

Next, determine values for $$p_1$$, $$p_2$$, and $$p_3$$ so that the $$q_1$$ and $$q_2$$ terms cancel out, leaving a constant. This will result in a strategy where player 2 is indifferent at the difficult decision points.

If $$3 p_2 - p_3 = 1$$, then the $$q_1$$ terms will cancel out. If $$3 p_1 = p_3$$, then the $$q_2$$ terms will cancel out.

So, two equations must be satisfied. Examples of values that work are $$p_1 = \frac{1}{5}$$, $$p_2 = \frac{8}{15}$$, and $$p_3 = \frac{3}{5}$$.