Dice Bets: Expected Profit and Kelly Criterion
Two independent fair six-sided dice are rolled simultaneously. A bookmaker offers the following six bets, each costing $\
| Bet | Event | Net profit if win | |-----|-------|-------------------| | (i) | Sum is 2 or 3 | $\$5$ | | (ii) | Sum is exactly 4 | $\$4$ | | (iii) | Sum is exactly 10 | $\$4$ | | (iv) | Sum is 11 or 12 | $\$5$ | | (v) | Sum is odd | $\$0.30$ | | (vi) | Sum is even | $\$0.30$ |
- For each bet, compute its expected profit.
- Suppose you must choose exactly one bet each round, staking a fixed fraction $f$ of your current wealth. Which bet maximizes expected wealth after a single round?
- Which bet maximizes expected log-wealth over many rounds (i.e., which is optimal under the Kelly criterion)? Justify quantitatively.
Hints
- Start by computing the probability of each event using the 36 equally likely dice outcomes. Notice several bets share the same probability.
- For expected wealth in one round, note that $E[W_1] = W(1 + f \cdot E[\text{profit}])$, so the bet with the least negative expected profit wins.
- For Kelly: compute $f^{*} = (p(1+g) - 1)/g$ for each bet. If $f^{*} < 0$, the edge is negative and you shouldn't bet. Compare which bet is "least bad" by evaluating the log-growth rate.
Worked Solution
How to Think About It: This is a classic "expected value vs. growth rate" problem. First, compute the probabilities and expected profits for each bet. Then recognize that maximizing single-round expected wealth is different from maximizing long-run growth. The Kelly criterion tells you that log-wealth growth rate, not arithmetic expectation, determines which bet compounds best over many rounds. A bet with the highest expected profit might still be worse for long-run growth if it loses too often.
Quick Estimate: The tail bets (sums 2-3 and 11-12) have very low probabilities (about $3/36$ each) but pay $5:1$. The even/odd bets are essentially coin flips paying $0.30:1$. Gut says the even/odd bets should have negative expected value since $0.30$ is far below the fair payout of
Formal Solution:
Step 1: Compute probabilities.
With two fair dice, there are 36 equally likely outcomes. The number of ways to get each sum:
- Sum 2: 1 way. Sum 3: 2 ways. So $P(2 \text{ or } 3) = 3/36 = 1/12$.
- Sum 4: 3 ways. So $P(4) = 3/36 = 1/12$.
- Sum 10: 3 ways. So $P(10) = 3/36 = 1/12$.
- Sum 11: 2 ways. Sum 12: 1 way. So $P(11 \text{ or } 12) = 3/36 = 1/12$.
- Sum odd: 18 ways. So $P(\text{odd}) = 18/36 = 1/2$.
- Sum even: 18 ways. So $P(\text{even}) = 18/36 = 1/2$.
Step 2: Expected profit for each bet.
$E[\text{profit}_i] = p_i \cdot (\text{net win}) - (1 - p_i) \cdot 1$
| Bet | $p$ | Win | $E[\text{profit}]$ | |-----|-----|-----|--------------------| | (i) |
All bets have negative expected profit. Bets (v) and (vi) have the least negative expected value at $-\$0.35$.
Step 3: Maximizing expected wealth after one round.
If you stake fraction $f$ of wealth $W$ on bet $i$, your expected wealth after one round is:
$E[W_1] = W \cdot \bigl[p_i(1 + f \cdot g_i) + (1 - p_i)(1 - f)\bigr]$
where $g_i$ is the net profit per dollar staked. This simplifies to:
$E[W_1] = W \cdot \bigl[1 + f \cdot (p_i \cdot g_i - (1-p_i))\bigr] = W \cdot [1 + f \cdot E[\text{profit}_i]]$
Since all expected profits are negative, the maximum expected wealth is achieved by setting $f = 0$ (don't bet). But if you *must* bet ($f > 0$ is fixed), you want the least negative expected profit.
Best bet for single-round expected wealth: (v) or (vi), with $E[\text{profit}] = -0.35$, the smallest loss in expectation.
Step 4: Maximizing long-run growth (Kelly criterion).
The expected log-growth rate for bet $i$ with stake fraction $f$ is:
$G_i(f) = p_i \ln(1 + f \cdot g_i) + (1 - p_i) \ln(1 - f)$
The Kelly fraction maximizes this. Taking the derivative and setting to zero:
$\frac{p_i \cdot g_i}{1 + f^{*} \cdot g_i} = \frac{1 - p_i}{1 - f^{*}}$
For bets (v)/(vi): $p = 1/2$, $g = 0.3$.
$f^{*} = \frac{p \cdot g - (1-p)}{g} = \frac{0.5 \times 0.3 - 0.5}{0.3} = \frac{-0.35}{0.3} < 0$
For bets (i)/(iv): $p = 1/12$, $g = 5$.
$f^{*} = \frac{p(1+g) - 1}{g} = \frac{(1/12)(6) - 1}{5} = \frac{0.5 - 1}{5} = \frac{-0.5}{5} = -0.10$
For bets (ii)/(iii): $p = 1/12$, $g = 4$.
$f^{*} = \frac{(1/12)(5) - 1}{4} = \frac{-7/12}{4} < 0$
All Kelly fractions are negative, meaning the optimal strategy for every bet is to not bet at all (or ideally, take the other side). Since all bets have negative edge, the Kelly criterion says don't wager on any of them.
If forced to choose, we compare $G_i(f)$ at the forced stake $f$ for each bet. For small $f$, the growth rate is approximately $G_i(f) \approx f \cdot E[\text{profit}_i] - \frac{f^2}{2}\text{Var}(\text{return}_i)$. Since bets (v)/(vi) have the least negative EV *and* the lowest variance (outcomes are $+0.3$ or $-1$ with equal probability, vs. rare large wins for the other bets), they dominate on both terms.
Best bet for long-run growth: (v) or (vi), for the same reason -- least negative edge and lowest variance.
Answer: All six bets have negative expected profit. Bets (v) and (vi) are the least bad at $E = -\$0.35$ per dollar. For both single-round expected wealth and long-run Kelly growth, bets (v) or (vi) are optimal (or equivalently, least destructive). The Kelly fraction is negative for all bets, meaning the theoretically optimal action is to not bet at all.
Intuition
This problem illustrates the fundamental distinction between expected value and growth rate in betting. Even when all bets are losing propositions, the Kelly criterion still ranks them -- it penalizes high-variance bets more heavily than low-variance ones. The tail bets (sums 2-3 or 11-12) might seem attractive because they pay $5:1$, but they hit so rarely (