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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?

Riccardo Bucco
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2 Answers2

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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.

angryavian
  • 89,882
  • More compactly, one can say that the (complementary) CDF of the (minimum) maximum roll is the pointwise product of the individual (complementary) CDFs of the dice. As far as dice probability problems go, this is quite clean; compare the next step up of taking the $k$th lowest die. – HighDiceRoller Jan 27 '23 at 01:56
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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 $0 \times(1-\frac27)+1 \times(\frac27-0) + 2 \times(0-0)=\frac27$.

Henry
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