Average of rolling $N$ $K$-sided dice and only taking the value of the lowest die roll

Question

This question is related, but mine is about a much more general case.

I want to roll $N$ dice $D_1, D_2, ..., D_{N}$. Each die $D_i$ has $K_i$ sides (so different dice possibly have different number of sides). Once I've rolled them, I want to keep the lowest value.

The sides of die $D_i$ are labeled $0, 1, ..., K_i-1$.

Example:

1. There are 4 dice (with, respectively, 4 sides, 7 sides, 2 sides, 9 sides)
2. I roll them, and get the values 2, 3, 1, 0
3. 0 is the lowest values, so that's the final result of the experiment.

What is the expected value of the result? On average, what is my final result?

Answer 1

Let $X$ be the minimum of the $N$ rolls. It takes values in $0,1,\ldots, \min_i(K_i) - 1$.

$$E[X] = \sum_{m=1}^{\min_i(K_i) - 1} P(X \ge m) = P(X \ge 1) + P(X \ge 2) + \cdots + P(X \ge \min_i(K_i) - 1)$$

If you can compute each probability $P(X \ge m)$ you can get the expectation. Note that $$P(X \ge m) = P(\text{every dice roll is $\ge m$}) = P(\text{1st roll is $\ge m$}) P(\text{2nd roll is $\ge m$}) \cdots P(\text{$N$th roll is $\ge m$}).$$

The probability that the $i$th roll is $\ge m$ is $\frac{\max(K_i - m, 0)}{K_i}$.

In the end you get a pretty gnarly expression due to the generality of the $K_i$ possibly being all different.

Answer 2

Your actual example is not too difficult to calculate, since one of the dice can only take the value $0$ or $1$, but it illustrates the general method:

• The probability the lowest value is $2$ or more is the the probability all the dice are $2$ or more, which is $0$ since one of them cannot exceed $1$ but we could write it as $\frac{4-2}{4}\times \frac{7-2}{7}\times \frac{2-2}{2}\times \frac{9-2}{9}=0$

• The probability the lowest value is $1$ or more is the the probability all the dice are $1$ or more, which is $\frac{4-1}{4}\times \frac{7-1}{7}\times \frac{2-1}{2}\times \frac{9-1}{9} =\frac27$

• The probability the lowest value is $0$ or more is the the probability all the dice are $0$ or more, which is $1$ since thaey are all at least $0$ but we could write it as $\frac{4-0}{4}\times \frac{7-0}{7}\times \frac{2-0}{2}\times \frac{9-0}{9}=1$

From this we could say the expectation is $\frac27 +0=\frac27$ though some people might prefer the equivalent $0 \times(1-\frac27)+1 \times(\frac27-0) + 2 \times(0-0)=\frac27$.