charlesreid1

Baby Poker is a two-player dice game that fits on a napkin and still manages to be a real bluffing game. Each player rolls a single die concealed in a cup. There is one round of betting. That is it. And yet the optimal strategy involves mixed frequencies, a value range, a bluffing range, and a call/fold cutoff. It is the smallest game I know of that has all the moving parts of full poker.

Our working notes: Baby Poker.

The Problem

This one comes from the FiveThirtyEight Riddler column (now defunct, but happily still on the Wayback Machine): Can You Deal With These Card Game Puzzles?.

The rules:

Each player antes $1.
Both roll a single die in a cup and look only at their own die.
Player A acts first: call (both dice shown, higher wins the $2 pot) or raise (bet another dollar).
If A raises, Player B chooses call (match the dollar, higher wins the $4 pot) or fold (A takes the pot, B loses only the ante).
Ties in a called pot split equally.

Two questions:

What is the optimal strategy for each player?
How much is the game worth to Player A? (Equivalently, how much should A pay B up front to make it fair?)

Setting Up the Game

Let $x \in \{1,\dots,6\}$ be A's roll and $y \in \{1,\dots,6\}$ be B's. Let $p_x$ be the probability A raises given $x$, and $q_y$ the probability B calls a raise given $y$. The strategy for each player is a vector of six numbers in $[0,1]$.

Because this is a zero-sum game of imperfect information, we are looking for a Nash equilibrium: a pair of strategies $(p, q)$ such that neither player can improve their expected value by deviating, holding the other player's strategy fixed. The standard tool is the indifference principle: at equilibrium, if a player mixes between two actions with a given hand, they must be indifferent between those actions. Otherwise they would just pick the better one and stop mixing.

That gives us two levers to pull. B's calling frequencies $q_y$ must be tuned so that A is indifferent to bluffing with the worst hand. A's raising frequencies $p_x$ must be tuned so that B is indifferent to calling with the marginal hands.

Player B's Indifference

B only has to think when A has raised. Given a raise, folding costs B one dollar (the ante). Calling puts B in for two more dollars into a four-dollar pot, and the payoff depends on how A's raising range $p_x$ is distributed.

B's EV of calling with roll $y$, conditional on A having raised, is

$$ E_B(\text{Call} \mid y) = \frac{1}{\sum_x p_x} \left( \sum_{x<y}(+2)p_x + \sum_{x=y}(0)p_x + \sum_{x>y}(-2)p_x \right). $$

Setting this equal to $-1$ (the EV of folding) and clearing the denominator gives B's indifference condition on A's range:

$$ 3\sum_{x<y} p_x + p_y - \sum_{x>y} p_x = 0. $$

Player A's Range: Value and Bluffs

The shape of A's optimal range falls out of two observations.

Top of the range is trivial. A should always raise with 5 and 6. These are the hands most likely to win at showdown, and if you don't raise them you leave money on the table. So $p_5 = p_6 = 1$.

Bluffing balances the range. If A only ever raised with strong hands, B would just always fold and A would never get paid off. So A has to raise with some bad hands too, to keep B honest. The cleanest choice is to bluff with the worst hand ($x=1$). A middling hand like 3 has real showdown equity that gets wasted by turning it into a bluff, but a 1 has almost none.

So the trial strategy is: value-bet with 5 and 6, call with 2, 3, 4, and bluff some fraction $p_1$ of the time with 1.

Plug this into B's indifference formula at the boundary hand $y=3$ (the marginal hand B might call with):

$$ 3 p_1 + 0 - (p_5 + p_6) = 0 \;\implies\; 3 p_1 - 2 = 0 \;\implies\; p_1 = \tfrac{2}{3}. $$

A bluffs with 1 exactly two-thirds of the time.

Player B's Range

Same trick from the other side. A must be indifferent between calling and raising when holding the worst hand. Calling with $x=1$ has EV

$$ E_A(\text{Call} \mid x=1) = \tfrac{1}{6}(0 - 1 - 1 - 1 - 1 - 1) = -\tfrac{5}{6}. $$

Raising with $x=1$ wins one dollar when B folds and loses two when B calls (a 1 loses to anything but a tie, and there is no tie with a 1). Letting $S_q = \sum_y q_y$,

$$ E_A(\text{Raise} \mid x=1) = \left(1 - \tfrac{S_q}{6}\right)(1) + \tfrac{S_q}{6}(-2) = 1 - \tfrac{S_q}{2}. $$

Equating and solving: $S_q = \tfrac{11}{3}$.

B always folds 1 and always calls 5 and 6 (again, the extremes are easy), so $q_2 + q_3 + q_4 = \tfrac{11}{3} - 2 = \tfrac{5}{3}$. The monotone solution is $q_2 = 0$, $q_3 = \tfrac{2}{3}$, $q_4 = 1$.

The Answer

Player A. Always raise with 5 and 6. Always call with 2, 3, 4. Raise with 1 exactly $\tfrac{2}{3}$ of the time.

Player B (facing a raise). Always call with 4, 5, 6. Always fold with 1 and 2. Call with 3 exactly $\tfrac{2}{3}$ of the time.

Averaging A's conditional EV over all six rolls (the arithmetic is on the wiki), the value of the game is

$$ \text{EV}(A) = \tfrac{5}{54} \approx 0.093. $$

Acting first is worth about a dime a hand. A gets to pick the timing of the bluffs and force B into a mixing spot. To make the game fair, A pays B $\tfrac{5}{54}$ of a dollar per hand up front.

What It Teaches

The real point of Baby Poker is that it is a toy problem for game theory. It is small enough to solve by hand, and it exercises the full machinery: a two-player zero-sum game of imperfect information, solved by finding a Nash equilibrium in mixed strategies via the indifference principle. That is exactly the same math von Neumann and Morgenstern set up in Theory of Games and Economic Behavior, just shrunk to a six-sided die.

Baby Poker is a good teaching game because it has exactly the pieces of full poker and nothing else:

A value range (raise with 5, 6): hands you want called.
A bluff range (raise with 1 sometimes): hands you want to fold out better hands.
A calling range (call with 2, 3, 4): hands with showdown equity you don't want to inflate the pot with.
A mixed frequency ($\tfrac{2}{3}$) that ties the two players together via indifference.
A positional advantage: acting first is worth real money because you get to define the ranges the second player has to respond to.

Every one of those concepts is a game-theoretic construct that survives all the way up to no-limit Texas hold 'em. Baby Poker just strips the game down to the smallest version where the full solution concept (Nash equilibrium in mixed strategies) still has all its moving parts.

References

Our wiki notes: Baby Poker
Can You Deal With These Card Game Puzzles?, the original FiveThirtyEight Riddler column (archived)
Related: Enumeration Solitaire, the other puzzle from the same column