The Binomial Distribution

The binomial distribution counts how many times something happens in a fixed number of independent yes/no trials: how many heads in 10 flips, how many of 100 orders fill, how many of 50 bonds default. Whenever you repeat the same simple gamble a set number of times and tally the wins, the count follows a binomial. It is usually the first distribution worth memorizing cold.

The setup

You need three things, and only three:

• a fixed number of trials, $n$;
• each trial independent of the others;
• each trial a success with the same probability $p$ (and failure with probability $1-p$).

A single such yes/no trial is called a Bernoulli trial. The binomial random variable $X$ is the total number of successes across the $n$ trials, so $X$ can be any whole number from $0$ to $n$.

The formula

$$P(X = k) = \binom{n}{k}\, p^k\, (1-p)^{n-k} .$$

Read it in three pieces. The term $p^k$ is the chance of $k$ particular successes; $(1-p)^{n-k}$ is the chance the other $n-k$ trials all fail; and $\binom{n}{k}$, "$n$ choose $k$," counts how many different ways those $k$ successes could be positioned among the $n$ trials (see Counting: Permutations and Combinations). We multiply the first two because the trials are independent, and multiply by the count because any of those arrangements would do.

The chart below is the same mass function, but live. Drag $n$ and $p$ and watch two things move: the mound recenters on its mean $np$ (the dashed line), and it widens or narrows. It's symmetric at $p = 0.5$; push $p$ toward $0$ or $1$ and it skews and bunches against the wall.

Mean and variance

Because $X$ is just a sum of $n$ independent Bernoulli trials, and expectation adds (see Expected Value), the mean and variance drop out with no messy sum:

$$E[X] = np, \qquad \operatorname{Var}(X) = np(1-p) .$$

Each trial contributes $p$ to the expected count and $p(1-p)$ to the variance; add up $n$ of them. For $n = 10, p = 0.5$ that gives a mean of $5$ and a variance of $2.5$ (standard deviation about $1.58$), matching the chart's balance point and spread.

Worked example

Flip a fair coin 10 times. What is the probability of exactly 6 heads? Here $n = 10$, $p = 0.5$, $k = 6$:

$$P(X = 6) = \binom{10}{6}(0.5)^6 (0.5)^4 = \binom{10}{6}(0.5)^{10} .$$

Now $\binom{10}{6} = \frac{10!}{6!\,4!} = 210$ and $(0.5)^{10} = \frac{1}{1024}$, so

$$P(X = 6) = \frac{210}{1024} \approx 0.205 .$$

About a 20.5% chance, the same height as the $k = 6$ bar above. Meanwhile the expected number of heads is simply $np = 10 \times 0.5 = 5$, the tallest bar. Notice that the single most likely outcome (5 heads, near 24.6%) is still far from guaranteed: the count varies.

Where it misleads

• Trials must be independent and identical. If one order filling makes the next more likely, or $p$ drifts over time, the binomial is wrong. Correlated defaults, in particular, fatten the tail far beyond what it predicts.
• It counts, it doesn't order. $X$ is the number of successes, not which trials succeeded. If the sequence matters, you are back to counting arrangements.
• Don't confuse it with its limits. When $n$ is large and $p$ tiny, the count is better handled by the The Poisson Distribution; when $n$ is large and $p$ moderate, it looks Gaussian, which is the The Central Limit Theorem at work. A common rule of thumb: the normal approximation is safe only when both $np$ and $n(1-p)$ exceed about 10.
• Sampling without replacement isn't binomial. Drawing from a finite pool without putting items back changes $p$ each draw; that is the hypergeometric distribution, not the binomial.

The binomial sits at the center of a small family of related laws collected in Common Distributions. Learn to derive its mean and variance from a sum of Bernoullis, and you have the template for reasoning about almost every count a quant meets.