# Five penalties, how many go in? The Binomial distribution

Source: https://www.footballdatascience.co.uk/learn/binomial-distribution
Published: 2026-09-27

> One penalty is a Bernoulli trial. Five penalties, each with the same chance, is the Binomial distribution, and it shows why missing one in five is normal, not a slump.

## The football question

Our penalty taker scores 80% of the time. He steps up five times. **How many go in?**

One penalty on its own is a [Bernoulli trial](/learn/bernoulli-distribution): goal or miss. Five of them, each with the same chance, is the **Binomial distribution**.

## The concept

The Binomial distribution counts successes across a **fixed number of attempts**, each with the **same chance** of success, each independent of the others. It needs two numbers:

- **n**, the number of attempts: 5 penalties.
- **p**, the chance of success each time: 0.80.

## A football example

What is the chance he scores **exactly 4 of the 5**?

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

$$\begin{aligned} P(X = 4) &= \binom{5}{4} \times 0.8^{4} \times 0.2^{1} \\ &= 5 \times 0.4096 \times 0.2 \\ &= 0.4096 \end{aligned}$$

About a **41% chance of exactly four**.

<div class="plain" markdown="1">
In plain football

- \(0.8^{4}\) is four penalties in a row going in: 0.41.
- \(0.2^{1}\) is the one that doesn't: saved, wide or over.
- \(\binom{5}{4} = 5\) is the number of ways that can happen. The miss could be the first kick, the second, the third, the fourth or the fifth.
- Multiply them together: \(5 \times 0.41 \times 0.2 \approx 0.41\).
</div>

Here is every possible outcome for our 80% taker:

<div class="bars" markdown="1">

| Penalties scored (of 5) | Probability |
|---|---|
| 0 | 0.03% |
| 1 | 0.6% |
| 2 | 5.1% |
| 3 | 20.5% |
| **4** | **41.0%** |
| 5 | 32.8% |

</div>

### A perfect five isn't the most likely result

Even for a reliable taker, scoring all five happens only about **one time in three**. Four out of five is more likely than five out of five. So when a good penalty taker misses one, that isn't a crisis of confidence. It's the most normal outcome there is.

## The same idea, all over the pitch

Anything with a fixed number of attempts and a steady success rate is Binomial:

- How many penalties out of 5 are scored?
- How many passes out of 20 are completed?
- How many shots on target out of 8 test the keeper?

Take a midfielder who completes 85% of his passes and attempts 20 in a match. On average he completes 17. But he will complete **14 or fewer** in about **one match in fifteen** (6.7%) with no change in form at all. Before anyone writes "off the pace today", it's worth asking whether the numbers are simply doing what numbers do.

<details markdown="1">
<summary><span>Show the maths<small>Where the formula comes from, and its mean and spread. Optional.</small></span></summary>

Each specific sequence with *k* goals and *n − k* misses, such as goal, goal, miss, goal, goal, has probability \(p^{k}(1-p)^{n-k}\), because the kicks are independent. The number of such sequences is

$$\binom{n}{k} = \frac{n!}{k!\,(n - k)!}$$

so the total probability is the two multiplied together.

A Binomial count is the sum of *n* Bernoulli trials, so its mean and variance are *n* times a single trial's:

$$E[X] = np = 4$$

$$\text{Var}(X) = np(1 - p) = 0.8$$

Our taker averages 4 from 5, give or take about 0.9.

</details>

## Why it matters

The Binomial distribution is how we tell a real change from ordinary variation. A striker's conversion rate, a keeper's save percentage, a team's penalty record: each is a count of successes from attempts. Knowing how much a count naturally wobbles stops us reading too much into one match, or one shoot-out.

## Limitations

- **p isn't really fixed.** A penalty taker's chance changes with the keeper, the pressure and the pitch. Binomial assumes the same p every time.
- **Attempts aren't always independent.** Miss one and the next might be harder, or easier. Binomial assumes each kick knows nothing about the last.
- **n has to be fixed in advance.** In open play we don't know how many shots a team will take. That's the question the next distribution answers.

## Try it yourself

Pick a player and one of their stats with a fixed number of attempts: penalties taken, passes attempted, shots on target. Use their season success rate as p and ask how often a "bad" match would happen by chance alone.

Then put two sets of penalty takers against each other in the [Penalty Shootout Simulator](/models/penalty-shootout): the first five kicks each are exactly the Binomial distribution above.

## Further reading

- [Seeing Theory: probability distributions](https://seeing-theory.brown.edu/probability-distributions/index.html), Brown University. Interactive: drag p and n and watch the Binomial distribution change.
- [Binomial distribution: properties, proofs, exercises](https://www.statlect.com/probability-distributions/binomial-distribution), StatLect. The derivations in full, with worked exercises.
- [Binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution), Wikipedia. The formal definition and properties.