Five penalties, how many go in? The Binomial distribution

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.

27 September 2026 Beginner Part 2 of Statistics Through Football

Contents

The football question
The concept
A football example
A perfect five isn't the most likely result
The same idea, all over the pitch
Why it matters
Limitations
Try it yourself
Further reading

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: 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.

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: $5 \times 0.41 \times 0.2 \approx 0.41$.

Here is every possible outcome for our 80% taker:

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

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.

Show the mathsWhere the formula comes from, and its mean and spread. Optional.

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.

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: the first five kicks each are exactly the Binomial distribution above.