## A Binomial Example

This post discusses the limited version of a counting problem that can be solved by using the binomial distribution. The more general problem is also discussed.

Example 1
Suppose 7 dice are rolled. What is the probability that at least 4 of the dice show the same face?

Example 2
Suppose that 6 job assignments are randomly assigned to 5 workers. What is the probability that at least 4 of the job assignments go to the same worker?

Example 2 is left as exercise.

Discussion of Example 1

First, we discuss how to solve using the binomial distribution. Fix a face (say 1). Finding the probability of that at least 4 of the dice show the face 1 is a binomial problem. Then multiplying this answer by 6 will give the desired answer.

Consider obtaining a 1 as a success. Let $X$ be the number of successes when 7 dice are thrown. Then $X$ is $\text{binom}(7,\frac{1}{6})$. We have the following calculation:

\displaystyle \begin{aligned}(1) \ \ \ \ \ P(X \ge 4)&=1-P(X \le 3) \\&=1-P(X=0)-P(X=1) \\&- \ \ \ P(X=2)-P(X=3) \\&=1-\binom{7}{0} \biggl[\frac{1}{6} \biggr]^0 \biggr[\frac{5}{6} \biggr]^7 - \binom{7}{1} \biggl[\frac{1}{6} \biggr]^1 \biggr[\frac{5}{6} \biggr]^6 \\&- \ \ \ \binom{7}{2} \biggl[\frac{1}{6} \biggr]^2 \biggr[\frac{5}{6} \biggr]^5 - \binom{7}{3} \biggl[\frac{1}{6} \biggr]^3 \biggr[\frac{5}{6} \biggr]^4 \\&=\frac{4936}{279936} \end{aligned}

Multiplying $(1)$ by 6 produces the desired answer.

\displaystyle \begin{aligned}(2) \ \ \ \ \ 6 \times P(X \ge 4)&=6 \times \frac{4936}{279936} \\&=\frac{29616}{279936} \\&=0.105796 \end{aligned}

Note that in rolling 7 dice, only one face can appear 4 or more times. The binomial distribution $\text{binom}(7,\frac{1}{6})$ counts the number of ways one particular face (say 1) can appear 4 or more times. There are six such binomial distributions (one for each face). These six binomial distributions do not overlap. So we can obtain the answer by multiplying the binomial result by 6.

If more dice are rolled, the different binomial distributions will overlap. In fact, in rolling 8 dice, it is possible that 2 faces can appear 4 times (e.g. the face 1 appears 4 times and the face 2 appears 4 times). If we multiple the binomial calculated result by 6, the result would not be correct. To work the problem of rolling 8 or more dice, we need to use the multinomial theorem. See this post a discussion of using multinomial theorem.

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$\displaystyle \frac{1325}{15625}=0.0848$

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