Ten-Sided Die Game: Expected Payoff with Multiple Rolls

Expectation · Medium · Free problem

You are offered a game involving a fair

0$-sided die with faces $\{1, 2, \ldots, 10\}$. Each play costs $\$5$ and pays you $\$N$, where $N$ is the outcome of the roll. Answer the following parts:

  1. Single roll: What is your expected profit?
  1. Two rolls (keep the maximum): You roll the die twice and receive the higher of the two outcomes. What is your expected profit?
  1. Many rolls: What happens to the expected profit as the number of rolls grows?

Hints

  1. For a single roll, the expected value of a fair
    0$-sided die is the average of
    $ through
    0$.
  2. For the maximum of two rolls, use $P(\max = k) = P(\max \leq k) - P(\max \leq k-1)$ and expand using the CDF $(k/10)^2$.
  3. You will need $\sum_{k=1}^{10} k^2 = 385$ and $\sum_{k=1}^{10} k = 55$ to evaluate $E[\max(N_1, N_2)]$.

Worked Solution

How to Think About It: The expected value of a single fair $d_{10}$ roll is $5.5$, which exceeds the $\$5$ cost, so the game always has positive expected value. The interesting part is how the expected payoff grows when you take the maximum of $n$ rolls -- it approaches $\

0$ as $n \to \infty$, giving a limiting profit of $\$5$. The key to the multi-roll case is computing $E[\max(N_1, \ldots, N_n)]$ efficiently.

Quick Estimate: For two rolls: the max of two uniform draws on $\{1, \ldots, 10\}$ should be somewhere around $7$-$8$ (much larger than the mean of a single draw). So expected profit $\approx 7$ or $8$ minus $5 = $ roughly $\

$ to $\$3$. The exact calculation below gives $7.15$.

Approach: For a single roll, apply the definition of expected value. For multiple rolls, use the CDF of the maximum.

Formal Solution:

Part 1: Single Roll

$E[N] = \frac{1+2+\cdots+10}{10} = \frac{55}{10} = 5.5$

Expected profit $= 5.5 - 5 = \$0.50$.

Part 2: Two Rolls (Keep the Maximum)

Let $M = \max(N_1, N_2)$. For a discrete uniform on $\{1, \ldots, 10\}$: $P(M = k) = P(\max \leq k) - P(\max \leq k-1) = \left(\frac{k}{10}\right)^2 - \left(\frac{k-1}{10}\right)^2 = \frac{2k-1}{100}$

$E[M] = \sum_{k=1}^{10} k \cdot \frac{2k-1}{100} = \frac{1}{100} \sum_{k=1}^{10} (2k^2 - k) = \frac{1}{100}\left(2 \cdot 385 - 55\right) = \frac{715}{100} = 7.15$

(using $\sum_{k=1}^{10} k^2 = 385$ and $\sum_{k=1}^{10} k = 55$).

Expected profit $= 7.15 - 5 = \ .15$.

Part 3: Many Rolls

For $n$ rolls, $E[\max(N_1, \ldots, N_n)] = \frac{1}{10^n} \sum_{k=1}^{10} k \left(k^n - (k-1)^n\right)$. As $n \to \infty$, the maximum converges to

0$ almost surely, so: $E[\max] \to 10 \quad \text{as } n \to \infty$

Expected profit approaches $\

0 - \$5 = \$5$. The game always has positive expected value, and the profit grows toward $\$5$ as you roll more times.

Answer: (1) $\$0.50$ expected profit. (2) $\

.15$ expected profit. (3) Profit approaches $\$5$ as the number of rolls grows.

Intuition

The expected value of the maximum of $n$ i.i.d. draws is a classic order statistic calculation. For a discrete uniform on $\{1, \ldots, m\}$, the maximum concentrates near $m$ as $n$ grows -- specifically, $E[M_n] \approx m - m/(n+1)$ for large $n$, which approaches $m$ from below. In this problem that means profit approaches $\$5$ asymptotically.

Order statistics like this come up in options pricing (barrier options pay off based on extremes), in market microstructure (the best bid/ask across multiple venues), and in risk management (tail loss estimates). The technique -- computing the PMF of the max via the CDF of each individual draw -- is one you should be able to apply immediately. It generalizes to continuous distributions as well, where $f_{M_n}(x) = n [F(x)]^{n-1} f(x)$.

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