Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For example, If I roll 4 dice (set size 4, range of possible values 1-6), what is the probability of getting at least 2 of 6s.

share|cite|improve this question
Ideally, the body of the question should be self-contained without relying on the title; in the present case, the two seem to contradict each other. Rolling $4$ dice doesn't generate a permutation. Also, the concept of "value of an object" isn't explained in the body. If, as the example seems to suggest, your question is what the probability is of getting a particular value $V$ at least $X$ times drawn from $S$ independent uniform distributions over $R$ items, the answer is $\sum_{n=X}^S\binom{S}{n}p^n(1-p)^{S-n}$, where $p=1/R$. – joriki Jun 16 '11 at 4:38
@joriki: sorry for the semantic errors, I'm not too familiar with math jargon. But how does rolling 4 dice not generate a permutation? – Matt Munson Jun 16 '11 at 4:51
See A permutation is an arrangement of different elements without repetition; rolling $4$ dice doesn't generate all of the $6$ possible values and may lead to repetitions. An example of a permutation of the values $1$ to $6$ would be $2,5,4,3,1,6$; an example of the result of rolling $4$ dice would be $2,3,6,3$. – joriki Jun 16 '11 at 4:57
@joriki: Ah, ok. My definition of the word permutations was clearly way off. – Matt Munson Jun 16 '11 at 4:59
up vote 1 down vote accepted

For your example, the probability is $$ \sum_{n=2}^4\binom{4}{n}\left(\frac{1}{6}\right)^n\left(\frac{5}{6}\right)^{4-n} $$ as this is the sum of the probabilities that you roll exactly $2$ sixes, exactly $3$ sixes, or exactly $4$ sixes in four rolls using the binomial distribution. In general, the formula will given by $$ \sum_{n=X}^S\binom{S}{n}p^nq^{S-n} $$ assuming there is a probability $p=\frac{1}{R}$ of the object having value $V$, and probability $q=1-p$ of the object not having value $V$.

share|cite|improve this answer
@yunone: I have to admit I don't quite understand the notation, and I didn't expect the answer to be this complicated. But I will figure it out. Thanks a bunch :) – Matt Munson Jun 16 '11 at 4:57
@Matt: To understand the notation, see and For why this is the correct answer, see Note that despite the similarity in appearance, in $\binom{4}{n}\left(\frac{1}{6}\right)^n\left(\frac{5}{6}\right)^{4-n}$ the first factor is a binomial coefficient whereas the other two are just fractions in parentheses. – joriki Jun 16 '11 at 5:01
@Matt, sure thing. What notation do you not understand? It's all straightforward multiplication and addition, so it's not as bad as it may seem at first. – yunone Jun 16 '11 at 5:04
@ joriki: So if I understand correctly then, the notation is equivalent to $$ \left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{4-2}+ \left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{4-3}+ \left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{4-4} $$ , which simplifies to 25/1296 + 5/1296 + 1/1296 = 31/1296. But that can't be right. I don't know what I'm supposed to do with the binomial coeffeicient. – Matt Munson Jun 16 '11 at 5:43
@Matt, you evaluate the binomial coefficient in each term as well. $\binom{n}{k}$ is shorthand for $n!/(k!(n-k)!)$. So for example, $\binom{4}{2}=4!/(2!(4-2)!)=4!/(2!2!)=24/(2\cdot 2)=6$. I take it you're familiar with the factorial? If not, $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n$, the product of all positive integers up to $n$. – yunone Jun 16 '11 at 5:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.