Probability of throwing multiple dice of at least a given face with a set of dice 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?
 A: 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
$$
\mathrm P(X_4\geqslant2)=1-1\cdot\left(\frac16\right)^0\cdot\left(\frac56\right)^4-4\cdot\left(\frac16\right)^1\cdot\left(\frac56\right)^3=\frac{19}{144}.
$$
A: 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.
