Maximum of Multiple Dice Rolls
You roll $k \geq 1$ fair $n$-sided dice, each showing a value in $\{1, 2, \ldots, n\}$. Let $M_k$ denote the maximum value among all $k$ rolls.
Find a closed-form expression for $P(M_k = r)$, the probability that the maximum equals $r$, for any $1 \leq r \leq n$.
Then compute the answer explicitly for $k = 3$, $r = 4$, and $n = 6$.
Hints
- The maximum is hard to pin down directly, but easy to bound from above. Try computing $P(M_k \leq r)$ first.
- By independence, $P(M_k \leq r) = P(X_1 \leq r)^k$. A fair $n$-sided die gives $P(X_i \leq r) = r/n$.
- Use the discrete CDF difference: $P(M_k = r) = P(M_k \leq r) - P(M_k \leq r-1) = \left(\frac{r}{n}\right)^k - \left(\frac{r-1}{n}\right)^k$.
Worked Solution
How to Think About It: The maximum is one of those quantities that is awkward to handle directly but clean to handle via its CDF. If you ask "what is the chance the max equals exactly 4?", it is hard to count directly. But if you ask "what is the chance the max is at most 4?", that is just the chance every single die shows 4 or less -- which factors immediately by independence. From there, $P(M_k = r) = P(M_k \leq r) - P(M_k \leq r-1)$ is a one-line subtraction. This CDF-then-difference trick is the standard tool for any discrete order statistic problem.
Quick Estimate: With $k = 3$, $r = 4$, $n = 6$: the max is at most 4 roughly $(4/6)^3 \approx 0.296$ of the time. The max is at most 3 roughly $(3/6)^3 = 0.125$ of the time. So the max equals exactly 4 about $0.296 - 0.125 = 0.171$. Sanity check: roughly 1-in-6, which feels about right -- out of six possible max values on a 6-sided die, equal weighting would be 1/6 $\approx 0.167$, so 0.171 is in the right ballpark.
Approach: Compute $P(M_k \leq r)$ by independence, then use the discrete difference $P(M_k = r) = P(M_k \leq r) - P(M_k \leq r-1)$.
Formal Solution:
Let $X_1, \ldots, X_k$ be the $k$ independent rolls. The CDF of the maximum is:
$$P(M_k \leq r) = P(X_1 \leq r, \ldots, X_k \leq r)$$
Since $M_k \leq r$ if and only if every die shows at most $r$, and the dice are independent:
$$P(M_k \leq r) = \prod_{i=1}^{k} P(X_i \leq r) = \left(\frac{r}{n}\right)^k$$
where $P(X_i \leq r) = r/n$ because the die is fair and there are $r$ values in $\{1, \ldots, r\}$ out of $n$ equally likely outcomes.
For the PMF, use the discrete difference:
$$P(M_k = r) = P(M_k \leq r) - P(M_k \leq r-1) = \left(\frac{r}{n}\right)^k - \left(\frac{r-1}{n}\right)^k$$
For $k = 3$, $r = 4$, $n = 6$:
$$P(M_3 = 4) = \frac{4^3 - 3^3}{6^3} = \frac{64 - 27}{216} = \frac{37}{216}$$
Answer: $P(M_k = r) = \left(\dfrac{r}{n}\right)^k - \left(\dfrac{r-1}{n}\right)^k$. For $k=3$, $r=4$, $n=6$: $\dfrac{37}{216} \approx 0.171$.
Intuition
The key principle here is that the CDF of the maximum of independent random variables factors cleanly: $P(\max \leq r) = \prod P(X_i \leq r)$. This is one of those identities that looks almost too simple, but it follows directly from the equivalence "the max is at most $r$" $\Leftrightarrow$ "every variable is at most $r$". Once you have the CDF, the PMF is just a discrete difference. This trick generalizes far beyond dice -- it is the foundation of order statistic theory, used whenever you care about extremes of a sample.
In practice, max and min distributions show up constantly in quant work: the maximum drawdown of a strategy, the first passage time of a price process, the worst-case loss across a portfolio of positions. The discrete version here is a clean entry point, but the continuous analog -- where $P(M_k \leq x) = F(x)^k$ for any CDF $F$ -- is what you will use in modeling. The common mistake is trying to count outcomes directly ("how many ways can the max be exactly 4?") rather than reaching for the CDF, which collapses the problem to a one-liner.