Optimal Calling Strategy in a Simplified Poker Game

Game Theory · Hard · Free problem

Players A and B each ante $\

$, creating a $\$4$ pot. They each independently draw a random number from $U(0, 100)$ -- this is their "hand." The higher number wins at showdown.

The game proceeds as follows:

  1. Player A acts first and can either check or raise by betting an additional $\ $.
  2. If A checks, B must also check, and they go to showdown (reveal numbers, higher wins the pot).
  3. If A raises, B can either fold (losing their $\ $ ante) or call (matching A's $\ $ bet, making the pot $\$8$, then showdown).

Describe Player B's optimal strategy. Specifically, when A raises, for which values of B's number $b$ should B call vs. fold? Derive the equilibrium threshold.

Hints

  1. Think about pot odds. When A raises, how much does B risk relative to what's in the pot? What win probability does B need for calling to break even?
  2. A's raising range has two components: value bets (high numbers) and bluffs (low numbers). B's calling threshold must make A indifferent about bluffing -- but note that a bluff's payoff does not depend on the exact low number A holds, so the indifferent bluff is the boundary of the bluffing region, not the worst possible hand.
  3. Set up three equations: A's bluff boundary indifference ( x = 3t - 100$), B's pot-odds win probability ($3x = 100 - y$), and A's value-bet indifference ( y = t + 100$). Solve the system for the calling threshold $t$.

Worked Solution

How to Think About It: This is a toy poker game, and it teaches you the core mechanic of calling strategy: pot odds vs. the opponent's range. B doesn't know A's number, but B knows A is rational. A will raise with very strong numbers (value betting) and very weak numbers (bluffing), and check the middle. B needs to call just often enough to keep A honest -- call too rarely and A bluffs freely, call too often and A only raises strong hands and prints money. The equilibrium threshold is pinned down by pot odds.

Quick Estimate: When A raises, the pot is $\$6$ and B must risk $\ $ to call. That gives pot odds of $3:1$, meaning B needs to win at least

/4$ of the time when calling for it to be profitable. So in equilibrium A must bluff exactly
/4$ of the time when raising. A naive shortcut -- "set A's worst bluff indifferent to checking and read off the threshold" -- is tempting but wrong, because a bluff's payoff does not depend on which low number A holds. The exact cutoff falls out only from solving the full equilibrium; the careful derivation below gives $t = 40$. Let's derive it.

Approach: We solve for the Nash equilibrium of the full game, focusing on B's calling threshold.

Formal Solution:

Assume A plays a threshold strategy: raise with $a \in [0, x]$ (bluffs) and $a \in [y, 100]$ (value bets), check with $a \in (x, y)$. B plays a threshold strategy: call if $b > t$, fold if $b < t$.

Step 1 -- B's threshold from A's bluffing indifference.

A bluff is a number so low it always loses at showdown when called. So the EV of raising as a bluff does NOT depend on the exact value of $a$: A wins the pot only when B folds ($b < t$, net $+\

$) and loses when B calls ($b > t$, net $-\$4$):

$\text{EV(raise as bluff)} = \frac{t}{100}(+2) + \frac{100 - t}{100}(-4) = \frac{6t - 400}{100},$

which is constant across the whole bluff interval. Checking, by contrast, is a losing hand going to showdown, with EV that rises in $a$:

$\text{EV(check, } a) = \frac{a}{100}(+2) + \frac{100 - a}{100}(-2) = \frac{4a - 200}{100}.$

Since EV(check) increases with $a$ while EV(raise) is flat, A's indifferent bluff is the largest one -- the boundary $a = x$ -- not $a \approx 0$. Setting them equal at $a = x$:

$6t - 400 = 4x - 200 \implies 2x = 3t - 100.$

Step 2 -- verify via pot odds.

At $b = t$, B's probability of winning against A's raising range must equal

/4$ (the pot odds). A raises with $[0, x] \cup [y, 100]$, total length $x + (100 - y)$. B with $b = t$ beats all bluffs in $[0, x]$ (since $x < t$) and loses to all value bets in $[y, 100]$ (since $y > t$). So:

$P(\text{B wins} \mid \text{A raised},\, b = t) = \frac{x}{x + (100 - y)} = \frac{1}{4}.$

This gives the constraint $3x = 100 - y$, i.e., $y = 100 - 3x$.

Step 3 -- A's value-bet indifference pins down $x$, $y$, and $t$.

A with $a = y$ is indifferent between raising and checking:

$\text{EV(check, } a = y) = \frac{4y - 200}{100},$ $\text{EV(raise, } a = y) = \frac{2t + 4(y - t) - 4(100 - y)}{100} = \frac{8y - 2t - 400}{100}.$

Setting them equal:

$8y - 2t - 400 = 4y - 200 \implies 4y = 2t + 200 \implies 2y = t + 100.$

We now have three equations:

x = 3t - 100$, $3x = 100 - y$, and
y = t + 100$. From the last, $y = (t+100)/2$; substituting into $3x = 100 - y$ gives $3x = 100 - (t+100)/2 = (100 - t)/2$, so $x = (100 - t)/6$. Combining with x = 3t - 100$:

$2 \cdot \frac{100 - t}{6} = 3t - 100 \implies \frac{100 - t}{3} = 3t - 100 \implies 100 - t = 9t - 300 \implies 10t = 400 \implies t = 40.$

Back-substituting: $x = (100 - 40)/6 = 10$ and $y = (40 + 100)/2 = 70$.

Step 4 -- the full equilibrium.

A's raising range has total length

0 + 30 = 40$. Of that,
0$ is bluffs and $30$ is value bets. The bluff fraction is
0/40 = 1/4$, exactly matching the pot odds. This confirms the equilibrium.

Answer: B's optimal strategy is:

  • If A checks: B checks (forced by the rules). They go to showdown.
  • If A raises: B calls if $b > 40$ and folds otherwise.

The threshold comes directly from pot odds: B risks $\

$ to win a $\$6$ pot, so B needs to win at least
5\%$ of the time. In equilibrium, A bluffs exactly
/4$ of the time when raising, making B indifferent at the threshold.

Intuition

This is the fundamental mechanic behind every calling decision in poker and, more broadly, in any adversarial game with incomplete information. The defender (B) doesn't need to figure out whether the opponent is bluffing on any specific hand -- that's impossible with symmetric information structure. Instead, B commits to a calling frequency that makes the opponent's bluffs break even. The threshold is entirely determined by pot odds: if B must risk $\ $ to contest a $\$6$ pot, B needs to win 5\%$ of showdowns, so B calls with the top /3$ of hands, forcing A's bluff frequency to exactly

/4$ of raises.

The deeper lesson is that in equilibrium, the caller's strategy is pinned down by the bettor's indifference, and vice versa. This circular logic -- each player's strategy makes the other indifferent -- is the hallmark of mixed-strategy Nash equilibria. In quant interviews, this same structure appears in market-making problems (how much to shade your quote to deter informed traders), auction theory (how aggressively to bid), and adverse selection models (when to trade against a potentially informed counterparty).

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