Distribution of the minimum of discrete Uniform R.V.s

Question

What is the cumulative distribution function of the minimum of $n$ discrete uniform i.i.d R.V.s? In particular when $X_i\sim\text{Unif}\,\{1,6\}$.

For $n=2$ the problem becomes quite simple as listing all possible choices is not too bad. This becomes fairly hard when $n$ is large. Are there any closed forms?

Answer 1

Let's denote the cdf by $F_n$ when there are $n$ dice. Then

$F_n(k) = P(\min(X_1, X_2, \ldots, X_n) \le k)$

Now for the minimum to be at most $k$ is a complicated event, but its complement is simpler. So we'll rewrite this as

$F_n(k) = 1 - P(\min(X_1, X_2, \ldots, X_n) > k).$

For the minimum to be greater than $k$, all of the $X_i$ have to be greater than $k$, and then you can use the independence and identical distribution of the $X_i$. You can get a closed form in terms of $k$ and $n$, and check it against the results you can get by listing out all cases when $n = 2$.

Answer 2

Let $Y=min\{X_1,...,X_n\}$. The cdf of Y is $G(y)=P(Y\le y)$. But the probability that the minimum is less than y is 1 minus the probability that it is greater than y, $1-P(Y>y)$. Since the minimum is at least y, everything in the sample is at least y, i.e. $1-P(X_1>y)...P(X_n>y)$. But since the $X_i$ are discrete iid on $Unif[1,6]$, this is just $1-[P(X>y)]^n$. Again doing the trick that $P(A)=1-P(A^c)$, this is equivalent to $1-[1-P(X\le y)]^n=1-[1-F_X(y)]^n$.

The actual cdf is $G(Y)=\begin{cases}0&y<1\\1-(\frac 5 6)^n&1\le y<2\\1-(\frac 2 3)^n&2\le y<3\\1-(\frac 1 2)^n&3\le y<4\\1-(\frac 1 3)^n&4\le y<5\\1-(\frac 1 6)^n&5\le y<6\\1&y\ge6\\\end{cases}$