Optimal Re-Roll Strategy for Maximizing Dice Product
You roll two fair six-sided dice and observe the results $D_1$ and $D_2$. Your payoff is the product $D_1 \cdot D_2$. Before collecting your payoff, you have one opportunity to re-roll exactly one of the two dice (or you can choose not to re-roll at all).
- What is the optimal re-roll strategy to maximize your expected payoff?
- What is the overall expected payoff under this optimal strategy?
Hints
- Compare the current value of each die to what you would expect from a fresh roll.
- A fair die has $E[D] = 3.5$. Re-rolling die $i$ replaces $d_i$ with a draw of expectation 3.5, so the expected product becomes $3.5 \cdot d_j$. When is this better than $d_1 \cdot d_2$?
- The re-roll condition simplifies to $d_i < 3.5$. Partition into three cases (0, 1, or 2 dice below 3.5) and compute each contribution to the overall expectation.
Worked Solution
How to Think About It: You are looking at two dice and deciding whether to gamble on improving one of them. The key insight is dead simple: re-rolling a die replaces its current value with a fresh draw that has expected value $E[D] = 3.5$. So re-rolling die $i$ is profitable exactly when the current value $d_i < 3.5$, i.e., when $d_i \in \{1, 2, 3\}$. If both dice are below 3.5, re-roll the lower one (or either if tied) since you can only re-roll once. If both are 4 or above, don't touch anything.
Quick Estimate: If both dice show 4+, the current product averages around $4 \times 5 = 20$, which beats $3.5 \times 5 = 17.5$, so holding is right. If one die shows 2 and the other shows 5, the current product is 10, but re-rolling the 2 gives expected product $3.5 \times 5 = 17.5$ -- a big improvement. The overall expected payoff should be meaningfully above the no-re-roll baseline of $3.5^2 = 12.25$.
Approach: Condition on whether 0, 1, or 2 dice show values $\leq 3$ and compute the expected product under the optimal action in each case.
Formal Solution:
Let $d_1, d_2$ be the observed values. If you re-roll die $i$, the expected product is:
$E[\text{product} \mid \text{re-roll } i] = 3.5 \cdot d_j$
where $d_j$ is the die you keep. The current product is $d_1 \cdot d_2$. Re-rolling die $i$ is better than holding when $3.5 \cdot d_j > d_1 \cdot d_2$, which simplifies to $3.5 > d_i$.
Optimal strategy: - If both $d_1, d_2 \geq 4$: hold (don't re-roll). - If exactly one die shows $\leq 3$: re-roll that die. - If both dice show $\leq 3$: re-roll the smaller one (if tied, re-roll either).
Now compute the expected payoff. Partition by the number of dice showing $\leq 3$:
Case 1: Both dice $\geq 4$ (hold). Each die is 4, 5, or 6 with probability
$E[d_1 \cdot d_2 \mid d_1, d_2 \geq 4] = E[d \mid d \geq 4]^2 = 5^2 = 25$
since $E[d \mid d \geq 4] = (4+5+6)/3 = 5$.
Case 2: Exactly one die $\leq 3$ (re-roll it). Probability:
$E[3.5 \cdot d_j \mid d_j \geq 4] = 3.5 \times 5 = 17.5$
Case 3: Both dice $\leq 3$ (re-roll the smaller). Probability: $(1/2)^2 = 1/4$. You keep the larger die (or either if tied) with value $d_{\max}$ and re-roll the other. We need $E[d_{\max} \mid d_1, d_2 \leq 3]$ where $d_1, d_2$ are uniform on $\{1,2,3\}$.
The distribution of $\max(d_1, d_2)$ for uniform dice on $\{1,2,3\}$: - $P(\max = 1) = 1/9$ - $P(\max = 2) = 3/9$ - $P(\max = 3) = 5/9$
$E[\max] = 1 \cdot \frac{1}{9} + 2 \cdot \frac{3}{9} + 3 \cdot \frac{5}{9} = \frac{1 + 6 + 15}{9} = \frac{22}{9}$
Expected product after re-rolling: $3.5 \times \frac{22}{9} = \frac{77}{9} \approx 8.56$.
Overall expected payoff:
$E = \frac{1}{4}(25) + \frac{1}{2}(17.5) + \frac{1}{4}\left(\frac{77}{9}\right) = \frac{25}{4} + \frac{35}{4} + \frac{77}{36}$
$= \frac{225}{36} + \frac{315}{36} + \frac{77}{36} = \frac{617}{36} \approx 17.14$
This is a substantial improvement over the no-re-roll baseline of
Answer: Re-roll any die showing 1, 2, or 3 (choosing the smaller if both qualify). The optimal expected payoff is $\frac{617}{36} \approx 17.14$, compared to
Intuition
The core principle is remarkably clean: re-rolling replaces one factor of a product with its expected value of 3.5, so you re-roll whenever the current factor is below that threshold. This is a direct application of conditional expectation and the independence of dice -- the decision for each die depends only on its own value, not the other die's value. The threshold $3.5$ is just the breakeven point where re-rolling is a zero-EV trade.
This type of reasoning -- comparing a known quantity to the expected value of a random replacement -- shows up constantly in trading. It is the same logic behind deciding whether to scratch a fill and re-enter, or whether to hold a position versus liquidating and re-deploying capital. The key habit is always having a "replacement value" benchmark in your head. If what you currently hold is worse than what the market offers in expectation, trade out of it.