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?
Best Answer
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 $$ \mathrm P(X_4\geqslant2)=1-1\cdot\left(\frac16\right)^0\cdot\left(\frac56\right)^4-4\cdot\left(\frac16\right)^1\cdot\left(\frac56\right)^3=\frac{19}{144}. $$