Probability
2. Discrete Distributions
Multinomial Distribution

Multinomial Distribution

The Multinomial distribution generalizes the Binomial: each trial can land in one of kk categories (not just success/failure).

Setup

  • You run nn independent trials.
  • Each trial falls into exactly one category 1,2,,k1,2,\dots,k.
  • Category probabilities are p1,,pkp_1,\dots,p_k with i=1kpi=1\sum_{i=1}^k p_i = 1.
  • Let XiX_i be the number of outcomes in category ii.

Then:

(X1,,Xk)Multinomial(n;p1,,pk)(X_1,\dots,X_k) \sim \mathrm{Multinomial}(n; p_1,\dots,p_k)

PMF

For counts x1,,xkx_1,\dots,x_k that sum to nn:

P(X1=x1,,Xk=xk)=n!x1!xk!i=1kpixiP(X_1=x_1,\dots,X_k=x_k)=\frac{n!}{x_1!\cdots x_k!}\prod_{i=1}^k p_i^{x_i}

Intuition

  • The factorial term counts how many sequences produce the same category totals.
  • The product term is the probability of any one specific sequence with those totals.

Relationship to Binomial

If k=2k=2 (category A vs not-A), the Multinomial collapses to a Binomial.

Key facts

  • Means: E[Xi]=npiE[X_i] = np_i
  • Variances: Var(Xi)=npi(1pi)\mathrm{Var}(X_i) = np_i(1-p_i)
  • Covariances: Cov(Xi,Xj)=npipj\mathrm{Cov}(X_i,X_j) = -np_ip_j for iji\neq j

That negative covariance matters: if one count goes up, the others must “make room” because totals sum to nn.

When to use

Use Multinomial when you count outcomes across multiple categories with replacement / independent trials:

  • Dice rolls (1–6), survey responses (A/B/C/D), product segments, etc.

If you sample without replacement from a finite population, you’re in Hypergeometric territory.

Test Your Knowledge

Example: Multinomial counts

A bowl of candy has 50% Red, 30% Green, and 20% Blue candies. You grab 5 candies at random (with replacement). What is the probability you get exactly 2 Red, 2 Green, and 1 Blue candy?

View Step-by-Step Solution

Multinomial formula:

n!x1!x2!x3!(p1)x1(p2)x2(p3)x3\frac{n!}{x_1! x_2! x_3!} (p_1)^{x_1} (p_2)^{x_2} (p_3)^{x_3}

  • n=5n = 5
  • Red: x1=2,p1=0.5x_1 = 2, p_1 = 0.5
  • Green: x2=2,p2=0.3x_2 = 2, p_2 = 0.3
  • Blue: x3=1,p3=0.2x_3 = 1, p_3 = 0.2

5!2!2!1!=1204=30\frac{5!}{2! 2! 1!} = \frac{120}{4} = 30

Probability =30×(0.5)2×(0.3)2×(0.2)1= 30 \times (0.5)^2 \times (0.3)^2 \times (0.2)^1

Probability =30×0.25×0.09×0.2=0.135= 30 \times 0.25 \times 0.09 \times 0.2 = 0.135

There is a 13.5% chance of grabbing this exact combination.