Maximize the Product of Two Dice

Expectation · Medium · Free problem

You roll two fair 6-sided dice.

Part 1. What is the expected value of the product of the two dice?

Part 2. Now suppose that after seeing both dice, you may optionally re-roll exactly one of them (you must accept the new value). Under optimal play, what is the expected value of the product?

Hints

  1. For Part 1, remember that expectation of a product of independent random variables equals the product of their expectations.
  2. For Part 2, think about when a re-roll improves your expected product. Compare the current value of the smaller die to the expected value of a fresh roll ($3.5$).
  3. The optimal strategy is to re-roll the smaller die whenever $\min(d_1, d_2) \leq 3$. Enumerate all 36 outcomes, applying $E = M \cdot 3.5$ for re-roll cases and $E = M \cdot m$ for keep cases.

Worked Solution

How to Think About It: Part 1 is a warm-up -- the dice are independent, so the expected product is the product of the expected values. Part 2 is where it gets interesting. You see both dice and can re-roll one. Intuitively, you would only ever re-roll the smaller die (keeping the larger one gives you the best multiplier). The question is: when is the expected value of a fresh roll ($3.5$) better than the value you already have? That happens whenever the smaller die shows $3$ or less. If both dice show $4$ or higher, you are already beating the re-roll expectation on both, so you keep them.

Quick Estimate: Without the re-roll option, the expected product is $3.5 \times 3.5 = 12.25$. With the re-roll, you are boosting the low outcomes. Most rolls (27 out of 36) have a min of $3$ or less, so you are re-rolling often. The max die averages around $4$-$5$ in those cases, and the re-roll averages $3.5$, so a typical re-roll product is around $4.5 \times 3.5 \approx 16$. The keep cases (min $\geq 4$) average around

5$. Rough blend: $(27/36)(16) + (9/36)(25) \approx 12 + 6.25 = 18.25$. The true answer should be a bit below this since I rounded up -- expect something around
7$.

Approach: Direct enumeration over all 36 outcomes, applying the optimal re-roll rule.

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Part 1: No Re-roll

Since $D_1$ and $D_2$ are independent:

$E[D_1] = E[D_2] = \frac{1+2+3+4+5+6}{6} = \frac{7}{2}$

$E[D_1 \cdot D_2] = E[D_1] \cdot E[D_2] = \frac{7}{2} \cdot \frac{7}{2} = \frac{49}{4} = 12.25$

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Part 2: Optimal Re-roll

After rolling $(d_1, d_2)$, let $M = \max(d_1, d_2)$ and $m = \min(d_1, d_2)$. If you re-roll the smaller die, the expected product becomes $M \cdot 3.5 = 7M/2$. If you keep both, the product is $M \cdot m$. You re-roll when $7M/2 > M \cdot m$, i.e., when $m < 3.5$, i.e., when $m \leq 3$.

Optimal rule: Re-roll the smaller die if and only if $\min(d_1, d_2) \leq 3$.

Now compute the expected value by summing over all 36 equally likely outcomes:

Re-roll cases ($m \leq 3$, contributing $\frac{7M}{2}$ each):

Group by $M$: for each value of $M$, count how many outcomes have $\max = M$ and $\min \leq 3$.

$ and min $\leq 3$: $(1,2),(2,1),(2,2)$, count $= 3$. Contribution: $3 \times \frac{14}{2} = 21$
  • $M = 3$: outcomes with max $3$ and min $\leq 3$: $(1,3),(2,3),(3,1),(3,2),(3,3)$, count $= 5$. Contribution: $5 \times \frac{21}{2} = \frac{105}{2}$
  • $M = 4$: outcomes with max $4$ and min $\leq 3$: $(1,4),(2,4),(3,4),(4,1),(4,2),(4,3)$, count $= 6$. Contribution: $6 \times \frac{28}{2} = 84$
  • $M = 5$: outcomes with max $5$ and min $\leq 3$: $(1,5),(2,5),(3,5),(5,1),(5,2),(5,3)$, count $= 6$. Contribution: $6 \times \frac{35}{2} = 105$
  • $M = 6$: outcomes with max $6$ and min $\leq 3$: $(1,6),(2,6),(3,6),(6,1),(6,2),(6,3)$, count $= 6$. Contribution: $6 \times \frac{42}{2} = 126$

    • Re-roll total: $\frac{7}{2} + 21 + \frac{105}{2} + 84 + 105 + 126 = \frac{7 + 42 + 105 + 168 + 210 + 252}{2} = \frac{784}{2} = 392$

    Keep cases ($m \geq 4$):

    Keep total:

    6 + 40 + 48 + 25 + 60 + 36 = 225$

    Final answer:

    $E[\text{product with optimal re-roll}] = \frac{392 + 225}{36} = \frac{617}{36} \approx 17.14$

    Answer:

    Intuition

    The core idea here is a simple threshold rule: you should replace a random variable with a fresh draw whenever the current realization is below the fresh draw's expected value. For a fair die, that threshold is $3.5$, so you re-roll any die showing

    $,
    $, or $3$. This is the same logic behind "should I re-roll in Yahtzee?" and, more broadly, behind any decision to replace a known quantity with an uncertain one -- you do it when the expected value of the replacement exceeds what you currently hold.

    A common mistake is to compute $E[\max]$ and $E[\text{effective min}]$ separately and multiply them. This fails because the re-roll decision depends on the min, which is correlated with the max. When both dice are low, the max tends to be low too. You must sum over all 36 outcomes directly, applying the re-roll rule case by case. This is a good reminder that $E[XY] = E[X] \cdot E[Y]$ requires independence -- and conditioning on a strategy that depends on both variables destroys that independence.

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