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Assume you have k dice, each with some random number of sides, 1...n. I'm trying to figure out, after rolling all k dice, what is the probability of obtaining at least one roll with a given value x - say, perhaps, 1? What is the probability of obtaining no 1s at all? I can write this by hand for 2-3 dice, and I'm trying to see if there's a way to write it for k dice.

And can this be generalized to a point where, say, some of your dice have no 1s on them to begin with? So the probability of rolling a 1 for that particular die = 0.

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up vote 3 down vote accepted

If you have $k$ dice with $n$ sides then the probability of a particular die showing on each dice is $p=1/n$. The probability that the die doesn't show that particular number is $1-p$. Since the dice are independent, you multiply to find the probability that none of the dice shows that particular number:

$${\rm Pr}(\textrm{No 1s on $k$ dice}) = (1-p)^k = \left( 1 - \frac{1}{n}\right)^k $$

and the case of showing at least one 1 is the complement of this, so it's probability is 1 minus the probability of showing no 1s:

$${\rm Pr}(\textrm{At least one 1 on $k$ dice}) =1- (1-p)^k = 1-\left( 1 - \frac{1}{n}\right)^k $$

If we generalize to the situation where the probability of dice $i$ showing a 1 is $p_i$ (i.e. different probabilities for different dice) then the probability of that die not showing a 1 is $1-p_i$, and to find the probability of no dice showing a 1 we again multiply all the individual probabilities together, and again the probability of at least one 1 is found by subtracting the probability of zero 1s from 1:

$${\rm Pr}(\textrm{No 1s on $k$ dice}) = \prod_{i=1}^k (1-p_i) $$

$${\rm Pr}(\textrm{At least one 1 on $k$ dice}) = 1- \prod_{i=1}^k (1-p_i) $$

where the $\prod$ symbol (a capital Greek letter 'pi') stands for 'product' and it means "multiply together all the numbers after this symbol". For example,

$$\prod_{i=1}^3 (1-p_i) = (1-p_1)(1-p_2)(1-p_3)$$

Now notice that if a particular die has no chance of showing a 1 then $p_i=0$ for that die, so $1-p_i=1$. In the expressions above that means we are multiplying by 1, which has no effect on the result - so any dice that can't possibly show a 1 can just be ignored!

Explained differently - if you have $k$ dice where $r$ of them can show a 1 and $s$ of them can't, the probabilities of getting no 1s and getting at least one 1 are exactly the same as if you only had $r$ dice to begin with.

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