Optimal Guess in a Dice Payout Game
You pick a number $n$ from $1$ to $6$. I roll a fair die. If the die shows $n$ or higher, you get paid $\$n$. If the die shows less than $n$, you get nothing.
What number should you pick, and what is the fair price to play this game?
Hints
- Write out the expected payoff as a function of your chosen number $n$. What is the probability of winning for each choice?
- The expected payoff is $E_n = n \cdot (7 - n) / 6$. This is a quadratic in $n$ -- where is it maximized?
- Evaluate $E_n$ at $n = 3$ and $n = 4$. Both give $\$2$, which is the maximum. The parabola peaks at $n = 3.5$, so the two nearest integers tie.
Worked Solution
How to Think About It: This is a classic expected value optimization. You are choosing a threshold $n$ that controls a trade-off: higher $n$ means a bigger payout when you win, but a lower probability of winning. Your gut should say that the extremes are bad -- picking $n = 1$ guarantees a win but only pays $\$1$, while picking $n = 6$ pays $\$6$ but you only have a $1/6$ chance. The sweet spot is somewhere in the middle. The fair price is whatever the optimal expected payoff turns out to be.
Quick Estimate: At the extremes, $n = 1$ gives $\$1 \cdot 1 = \$1$ and $n = 6$ gives $\$6 \cdot 1/6 = \$1$. Both are $\$1$. In the middle, try $n = 3$: you win with probability $4/6$ and get $\$3$, so the expected payoff is $3 \times 4/6 = \$2$. Try $n = 4$: you win with probability $3/6$ and get $\$4$, so $4 \times 3/6 = \$2$. The optimal expected payoff looks like $\$2$.
Approach: Compute $E_n = n \cdot P(\text{die} \ge n)$ for each $n$ and maximize.
Formal Solution:
If you choose $n$, the die shows $n$ or higher with probability $(7 - n)/6$, so your expected payoff is:
$$E_n = n \cdot \frac{7 - n}{6}$$
Evaluating for each choice:
- $n = 1$: $1 \cdot 6/6 = 1.00$
- $n = 2$: $2 \cdot 5/6 \approx 1.67$
- $n = 3$: $3 \cdot 4/6 = 2.00$
- $n = 4$: $4 \cdot 3/6 = 2.00$
- $n = 5$: $5 \cdot 2/6 \approx 1.67$
- $n = 6$: $6 \cdot 1/6 = 1.00$
The expected payoff is symmetric around $n = 3.5$ (since $E_n = n(7-n)/6$ is a downward-opening parabola with vertex at $n = 3.5$). The two integer values flanking the vertex, $n = 3$ and $n = 4$, both achieve the maximum of $\$2$.
Answer: Pick $n = 3$ or $n = 4$. The maximum expected payoff is $\$2$, so the fair price to play is $\$2$.
Intuition
This problem illustrates the fundamental risk-reward trade-off that shows up everywhere in quant finance: increasing your payout per event comes at the cost of reducing the probability of that event occurring. The expected payoff $n(7-n)/6$ is a quadratic that peaks in the middle, which is a simple instance of the general principle that optimal strategies balance aggressiveness against likelihood of success. The symmetry around $n = 3.5$ is a nice feature -- it tells you the game treats "conservative with high probability" and "aggressive with low probability" equivalently at the optimum.
In practice, this kind of threshold optimization appears whenever you set limit orders (higher limit price means bigger profit per fill but fewer fills), choose strike prices for options, or set take-profit levels. The key lesson: always write out expected payoff as a function of your decision variable and optimize, rather than guessing based on intuition alone. And notice that the fair price ($\$2$) is exactly double what you get from the naive extremes ($\$1$) -- the middle ground matters.