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I know how to calculate the probability of throwing at least one die of a given face with a set of dice, but can someone tell me how to calculate more than one (e.g., at least two)?

For example, I know that the probability of throwing at least one 4 with two 6-sided dice is 27/216, or 1 - (3/6 x 3/6 x 3/6). How do I calculate throwing at least two 4s with four 6-sided dice?

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    @ihatetoregister: that is correct. Now if we want the chance of at least one $4$ in four throws, I can do $1-(\frac 56)^4$. Your approach would be $\frac 16 +\frac 56 \frac 16 +\frac 56 \frac 56 \frac 16 +\frac 56 \frac 56 \frac 56 \frac 16$. They get the same place. Sometimes one is easier, sometimes the other. Sometimes it is a matter of taste.2013-04-03

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You are asking for the distribution of the number $X_n$ of successes in $n$ independent trials, where each trial is a success with probability $p$. Almost by definition, this distribution is binomial with parameters $(n,p)$, that is, for every $0\leqslant k\leqslant n$, $ \mathrm P(X_n=k)={n\choose k}\cdot p^k\cdot(1-p)^{n-k}. $ The probability of throwing at least two 4s with four 6-sided dice is $\mathrm P(X_4\geqslant2)$ with $p=\frac16$. Using the identity $\mathrm P(X_4\geqslant2)=1-\mathrm P(X_4=0)-\mathrm P(X_4=1)$, one gets $ \mathrm P(X_4\geqslant2)=1-1\cdot\left(\frac16\right)^0\cdot\left(\frac56\right)^4-4\cdot\left(\frac16\right)^1\cdot\left(\frac56\right)^3=\frac{19}{144}. $

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The probability of no 4 with four 6-sided dice$(p_1)=(\frac{5}{6})^4$

The probability of exactly one 4 with four 6-sided dice$(p_2)$ $=4\frac{1}{6}(\frac{5}{6})^3$ as here the combinations are $4XXX$ or $X4XX$ or $XX4X$ or $XXX4$ where $X$ is some other face$≠4$

So, the probability of at least two 4s with four 6-sided dice$=1-p_1-p_2$ $=1-((\frac{5}{6})^4+4\frac{1}{6}(\frac{5}{6})^3)$ $=1-(\frac{5}{6})^3(\frac{5}{6}+\frac{4}{6})=1-\frac{125}{144}=\frac{19}{144}$

The probability of throwing at least 4 by one 6-sided dice $=\frac{3}{6}=\frac{1}{2}$

The possible combinations are $XXYY$, $XYXY$, $XYYX$, $YXXY$, $YXYX$, $YYXX$ where $1≤Y≤3,4≤X≤6$

So, the required probability of throwing exactly two occurrences of at least 4 is $^4C_2\frac{1}{2}\frac{1}{2}(1-\frac{1}{2})(1-\frac{1}{2})=\frac{3}{8}$ using Binomial Distribution.

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    Could you please explain your approach? By the way, I've modified the answer.2012-09-09