The Dice Game Pig and Optimal Stopping

The game. On your turn in Pig, you repeatedly roll a single die. Each roll adds its value to your turn total — unless you roll a 1, in which case you lose everything accumulated this turn and your turn ends immediately. At any point before rolling a 1, you can choose to "hold," banking your turn total permanently. The question: what's the optimal stopping rule, and roughly why does it land where it does?

This is a genuinely rich optimal-stopping problem hiding behind a simple game, and it's a favorite interview vehicle because a clean, intuitive rule ("hold at 20") turns out to be a very good approximation to the mathematically optimal (and much more complicated, state-dependent) policy.

Setting up the one-roll decision

At any point in your turn with current turn total $t$, rolling once more has two outcomes: with probability $5/6$ you get a uniform value from $\{2,3,4,5,6\}$ added (average $4$ conditional on not rolling a 1), and with probability $1/6$ you lose the entire turn total $t$. The expected change in your turn total from one more roll is

$$E[\Delta] = \frac{5}{6}(4) - \frac{1}{6}(t) = \frac{20 - t}{6} .$$

In plain English: rolling again is worth it, in expectation, exactly when the expected gain from a good roll ($\frac{5}{6}\times 4 \approx 3.33$) outweighs the expected loss from busting ($\frac{1}{6} \times t$) — and those two forces balance exactly when $t = 20$.

Worked example: why the crossover is at 20

Set $E[\Delta] = 0$: $\frac{20-t}{6} = 0 \Rightarrow t = 20$. Below $t=20$, $E[\Delta] > 0$ — rolling again has positive expected value, so you should keep rolling. Above $t=20$, $E[\Delta] < 0$ — you have more to lose (a large banked total from this turn) than to gain from one more average roll, so you should hold. Check $t=15$: $E[\Delta] = (20-15)/6 = 0.83$, positive, keep rolling. Check $t=25$: $E[\Delta] = (20-25)/6 = -0.83$, negative, hold. This gives the well-known "hold at 20" heuristic: as soon as your turn total reaches 20, bank it.

Why the true optimum is more subtle (but close)

The one-roll analysis above only checks "is one more roll worth it," treating everything after as fixed. The actual optimal policy accounts for your opponent's current score too — if you're far behind, it's worth pushing your luck higher than 20 because a modest expected gain isn't enough if you need a big turn to catch up; if you're far ahead, holding earlier can lock in a win faster. Computed via dynamic programming (backward induction over game states), the fully optimal policy varies the hold threshold between roughly 20 and 27 depending on both players' scores — but "hold at 20," derived from the simple one-roll expected-value argument above, gets remarkably close to optimal play throughout the game.