Distribution of minimum of independent dice rolls.

Question

If $\{X_i\}_{1\le i \le n} $ are $n$ independent fair-die rolls, what is the distribution of $\min(X_1, \dots, X_n)$?

Let $X := \min(X_1, \dots, X_n), n=$ number of rolls, $P(X = x) = \frac{1}{6}$

Is it correct to say that $X \sim \text{binom}(n, p = \frac{1}{6})$?

Well yes, I'm trying to say $p(X = x) = \frac{1}{6}$ where $x = 1, 2,...6$ — user3904534, Dec 01 '15 at 05:23
Hint: What's the probability that the minimum is $>1$, i.e., that $1$ is not rolled in the $n$ throws. Similarly, what's the probability it's $>2$? Once you have these two numbers, what can you conclude? — stochasticboy321, Dec 01 '15 at 05:24

André Nicolas · Accepted Answer · 2015-12-01T05:44:06.040

The random variable $X$ can take on values $1$, $2$, $3$, $4$, $5$, or $6$. For all $i$ with $1\le i\le 6$, we will find $\Pr(X=i)$. It is perhaps useful to deal with them backwards.

$\Pr(X=6)$: This is the easiest one. The probability that the minimum is $6$ is the probability all the rolls are $6$. Thus $\Pr(X=6)=\left(\frac{1}{6}\right)^n$.

$\Pr(X=5)$: Next we find the probability the minimum is $5$. This happens if (i) all the throws are $\ge 5$ but (ii) not all the throws are $\ge 6$. The probability all the throws are $\ge 5$ is $\left(\frac{2}{6}\right)^n$. It follows that $\Pr(X=5)=\left(\frac{2}{6}\right)^n-\left(\frac{1}{6}\right)^n$.

$\Pr(X=4)$: Next we find the probability the minimum is $4$. This happens if (i) all the throws are $\ge 4$ but (ii) not all the throws are $\ge 5$. The probability all the throws are $\ge 4$ is $\left(\frac{3}{6}\right)^n$. It follows that $\Pr(X=4)=\left(\frac{3}{6}\right)^n-\left(\frac{2}{6}\right)^n$.

Halfway done! It's your turn.

Distribution of minimum of independent dice rolls.

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