Rerolling a Die: When to Take the Number

You roll a fair 6-sided die once. You may keep the number shown, or reroll it exactly once more and must keep whatever comes up the second time. Which numbers should you keep on the first roll, and what's your expected payoff under the optimal rule?

Work it out before reading on. The answer isn't "keep anything above 3" — it's close, but the exact cutoff takes a genuine comparison, not a guess.

Reframe: comparing to a fixed number, not a feeling

The rule you're building is a threshold strategy: keep the first roll if it's at least some number $t$, reroll otherwise. The right $t$ is whatever makes "keep" and "reroll" break even — anything strictly better than the expected value of rerolling should be kept, and anything strictly worse should be tossed back.

If you reroll, your payoff is a fresh fair die: $E[\text{reroll}] = \frac{1+2+3+4+5+6}{6} = 3.5$.

So the rule is immediate: keep the first roll if it's greater than 3.5, i.e., keep 4, 5, or 6; reroll on 1, 2, or 3. Anything above the average is worth keeping; anything below it, you're better off starting over.

Computing the payoff of the optimal rule

Under "keep 4/5/6, reroll 1/2/3": half the time (rolls 4, 5, 6) you keep that value; half the time (rolls 1, 2, 3) you reroll and get a fresh average of 3.5.

$E[\text{payoff}] = \frac{1}{6}(4+5+6) + \frac{1}{2}(3.5) = \frac{15}{6} + 1.75 = 2.5 + 1.75 = 4.25$

The optimal strategy averages 4.25, versus 3.5 for always keeping the first roll (no strategy) or 3.5 for always rerolling. The option to choose, used correctly, is worth $4.25 - 3.5 = 0.75$ of a pip — real value, purely from knowing when to stop.

A second example: unlimited rerolls, capped at 3 tries

Same die, but now you may reroll up to two times (three rolls total), keeping the last number you land on or stopping earlier. What's the optimal rule and its expected value?

Solve backward from the last roll, the standard technique for any multi-stage stopping problem. Roll 3 (forced keep): you must take whatever shows, expected value 3.5 — this is now playing the role of "the reroll option" for roll 2's decision.

Roll 2: compare the number shown to 3.5 (the value of falling through to roll 3). Same threshold as before: keep 4/5/6, reroll on 1/2/3. This gives, as computed above, expected value 4.25 — this becomes the fallback value for roll 1's decision.

Roll 1: now compare against 4.25, not 3.5. Keep the roll only if it beats 4.25 — that means keep only a 5 or a 6; reroll on 1, 2, 3, or 4 (even a 4 is now worse than the expected 4.25 you'd get by falling through to the roll-2/3 process).

$E[\text{payoff}] = \frac{1}{6}(5+6) + \frac{4}{6}(4.25) = \frac{11}{6} + \frac{17}{6} = \frac{28}{6} \approx 4.67$

Each extra reroll raises the threshold for keeping the current number, because the fallback option itself has gotten better — this is the general pattern in every multi-stage stopping problem.

Where this generalizes

This backward-induction, threshold-against-the-fallback logic is the exact machinery behind job-offer acceptance problems ("take this offer or keep searching"), the classic secretary problem, and — closer to a trading desk — deciding whether to lift an offer now versus waiting for a better print, where the "reroll" is the expected value of continuing to work an order. Any time a puzzle involves "take it or try again," the fix is the same: compute the expected value of trying again, and take anything that beats it.