Problem 203 Answer

Question

The game of one-card poker has a three-card deck, an ace, deuce, and trey. The ace is lowest and the trey is highest. Each of two players ante $1 into the pot. Then, each player gets one card. The order of betting is predetermined, with player 1 to act first. Player 1 may either bet $1 or check. If player 1 bets, player 2 may either call or fold. If player 1 checks, player 2 may either bet $1 or check. If player 1 checks, and player 2 bets, then player 1 may either call or fold. If both players check, or one bets and one calls, then the higher card wins the pot. Assuming both players are perfect logicians, what is the optimal strategy for each player?

Answer

The obvious plays are neither player should call with the ace, always call with the trey, and never bet with a deuce.

For the more difficult decisions, let's define the following probabilities:

• p₁ = Probability player 1 bets with an ace (bluffing).
• p₂ = Probability player 1 calls with a deuce after player 2 bets (hoping player 2 is bluffing).
• p₃ = Probability player 1 bets with a trey (hoping player 2 will call).
• q₁ = Probability player 2 bets with an ace, after player 1 checks (bluffing).
• q₂ = Probability player 2 calls with a deuce, after player 1 bets (hoping player 1 is bluffing).

Player 2 has probabilities q₁=1/3 and q₂=1/3. Player 1 has more flexibility. He may use any probabilites as long as these too equations are satisfied:

3p₂-p₃=1
3p₁=p₃

For example, p₁=1/5, p₂=8/15, p₃=3/5.

Acknowledgement: The idea of this problem came from the article Game Theory and Poker, by Jason Swanson (April 3, 2005). The paper works out the same solution. I kept the variables names the same for my own benefit, and yours.