Multinomial Distribution — Definition, Formula & Examples

The multinomial distribution gives the probability of observing a specific combination of counts across multiple categories when you repeat an experiment a fixed number of times. It generalizes the binomial distribution from two possible outcomes to any number of outcomes.

If a trial has $k$ mutually exclusive outcomes with probabilities $p_1, p_2, \ldots, p_k$ where $\sum_{i=1}^{k} p_i = 1$ , and the trial is repeated $n$ independent times, then the joint probability that outcome $i$ occurs exactly $x_i$ times (with $\sum_{i=1}^{k} x_i = n$ ) follows the multinomial distribution.

Key Formula

P(X_1 = x_1, \ldots, X_k = x_k) = \frac{n!}{x_1!\, x_2!\, \cdots\, x_k!}\; p_1^{x_1}\, p_2^{x_2} \cdots p_k^{x_k}

Where:

$n$ = Total number of independent trials
$k$ = Number of possible outcome categories
$x_i$ = Number of times outcome i occurs, where x₁ + x₂ + ⋯ + xₖ = n
$p_i$ = Probability of outcome i on any single trial, where p₁ + p₂ + ⋯ + pₖ = 1

How It Works

You use the multinomial distribution whenever an experiment has more than two categories and each trial is independent with the same probabilities. First, identify the number of trials

n

, the number of categories

k

, and each category's probability

p_i

. Then specify the count

x_i

you want for each category, making sure the counts sum to

n

. Plug these values into the multinomial probability formula. The multinomial coefficient in front accounts for all the different orderings that produce the same set of counts.

Worked Example

Problem: A fair die is rolled 6 times. What is the probability of getting exactly 2 ones, 2 twos, and 2 threes (and zero of everything else)?

Identify parameters: Here n = 6 trials, k = 6 faces. The counts are x₁ = 2, x₂ = 2, x₃ = 2, x₄ = 0, x₅ = 0, x₆ = 0. Each face has probability 1/6.

Compute the multinomial coefficient: The coefficient counts the number of orderings that give these counts.

\frac{6!}{2!\,2!\,2!\,0!\,0!\,0!} = \frac{720}{2 \cdot 2 \cdot 2} = 90

Compute the probability product: Raise each category's probability to the power of its count.

\left(\frac{1}{6}\right)^2 \left(\frac{1}{6}\right)^2 \left(\frac{1}{6}\right)^2 = \left(\frac{1}{6}\right)^6 = \frac{1}{46656}

Multiply: Combine the coefficient and the probability product.

P = 90 \times \frac{1}{46656} = \frac{90}{46656} \approx 0.00193

Answer: The probability is approximately 0.00193, or about 0.19%.

Why It Matters

The multinomial distribution underpins the chi-squared goodness-of-fit test, which compares observed category counts to expected ones. It arises naturally in genetics, survey analysis, natural language processing, and any setting where outcomes fall into more than two groups.

Common Mistakes

Mistake: Using the binomial formula when there are more than two outcome categories.

Correction: The binomial distribution handles only two outcomes (success/failure). When there are three or more categories, you need the multinomial distribution, which accounts for how counts are split across all categories simultaneously.