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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.

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    @joriki: Ah, ok. My definition of the word permutations was clearly way off.2011-06-16

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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 $V$.

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    Thanks allot guys. I learned much more than I e$x$pected here. Cheers.2011-06-16