How many shots until he scores? The Geometric distribution

A striker scores with one shot in five. How many shots until his first goal? The Geometric distribution answers it, and shows why a three-match drought is often just bad luck.

Beginner Part 5 of Statistics Through Football

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Contents

The football question
The concept
A football example
The first shot is always the most likely, and still unlikely
The same idea, all over the pitch
Droughts
Why it matters
Limitations
Try it yourself
Reproduce the analysis
Further reading

The football question

How many shots until the striker scores his first goal?

The Exponential distribution measures the wait for a goal in minutes. Here we count attempts instead.

The concept

The Geometric distribution gives the chance that the first success comes on the k-th attempt. Like the Bernoulli trial it's built from, it needs just one number: p, the chance each attempt succeeds.

Exponential: how long until the next event.
Geometric: how many attempts until the first success.

A football example

Our striker scores with 20% of his shots, so p = 0.20. What's the chance his first goal comes on exactly his 4th shot?

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

$$P(X = 4) = 0.8^{3} \times 0.2 = 0.1024$$

About a 10% chance.

In plain football

$0.8^{3}$: his first three shots all miss. Each miss has a 0.8 chance, so three in a row is $0.8 \times 0.8 \times 0.8 = 0.512$.
$0.2$: then the fourth goes in.
Multiply them: $0.512 \times 0.2 \approx 0.10$.

Twenty percent is a sharp finisher, by the way. In the Scottish Premiership and Championship, about 13% of all shots are goals (and about 30% of shots on target).

Here's his whole first-goal picture:

First goal on shot	Probability	Scored by then
1	20.0%	20.0%
2	16.0%	36.0%
3	12.8%	48.8%
4	10.2%	59.0%
5	8.2%	67.2%
6	6.6%	73.8%

The first shot is always the most likely, and still unlikely

The single most likely shot for his first goal is shot 1. It always is, for any p. Yet even after four shots, there's still a 41% chance he hasn't scored. On average he needs 5 shots.

The same idea, all over the pitch

First goal on the 4th shot.
First successful tackle on the 3rd attempt.
First completed long pass on the 2nd try.

Anything that repeats with the same chance of success until it comes off is Geometric.

Droughts

Say our 20% striker takes three shots a game. The chance he goes three matches without scoring, nine shots and no goal, is $0.8^{9} \approx$ 13%. Roughly one run of three games in eight, with his finishing exactly as good as ever.

The Geometric distribution is memoryless, like the Exponential. After nine misses, the chance his next shot goes in is still 20%. He isn't "due" one, and he hasn't lost it either. Before a manager drops a striker for a drought, it's worth checking whether the numbers are simply doing what numbers do.

Show the mathsMean, cumulative probability and memorylessness. Optional.

The first success comes by attempt k unless the first k all fail:

$$P(X \le k) = 1 - (1 - p)^{k}$$

The mean number of attempts is

$$E[X] = \frac{1}{p} = \frac{1}{0.2} = 5$$

And the past doesn't matter:

$$\begin{aligned} &P(X > s + t \mid X > s) \\ &= \frac{(1-p)^{s+t}}{(1-p)^{s}} \\ &= (1-p)^{t} = P(X > t) \end{aligned}$$

Why it matters

Strikers are judged in streaks: "hasn't scored in five", "can't stop scoring". The Geometric distribution gives a baseline for how long a perfectly consistent player should expect to wait. Only a drought much longer than that baseline is real evidence that something has changed.

Limitations

p changes from shot to shot. A tap-in and a 30-yard effort aren't the same attempt. A striker's 20% is an average over very different chances; expected goals gives each shot its own p.
Shots aren't always independent. Confidence, fatigue and the opposition all shift from one attempt to the next.
It only counts to the first success. For "how many shots until his third goal?", we need the next distribution in the series.

Try it yourself

Pick a striker and find his shots and goals for the season. Use goals divided by shots as p, work out $(1 - p)$ raised to his shots per game times three, and you have the chance of a three-match drought by luck alone. Then compare it with how often he actually goes three games without scoring.

Reproduce the analysis

The made-up parts need nothing but Python. The last part, real shot conversion, reads the results files published by football-data.co.uk: download the Premiership (SC0) and Championship (SC1) files for each season from 2000/01 to 2025/26 with the download script, which names them as the snippet expects; they aren't rehosted on this site. Then:

Show the Python20 lines, ready to copy and run.

# From Football Data Science by Bryan McGuire. Free to use with credit.
# https://www.footballdatascience.co.uk/learn/geometric-distribution
import csv

p = 0.20  # a striker who scores with one shot in five
for k in range(1, 7):
    print(f"first goal on shot {k}: {(1 - p) ** (k - 1) * p:.1%}, scored by then {1 - (1 - p) ** k:.1%}")
print(f"still no goal after four shots: {(1 - p) ** 4:.0%}; average shots needed {1 / p:.0f}")
print(f"three games of three shots without a goal: {(1 - p) ** 9:.0%}")

# real conversion: Premiership (SC0) and Championship (SC1) matches with shot data, 2000/01 to 2025/26
goals = shots = on_target = matches = 0
for league in ("SC0", "SC1"):
    for y in range(2000, 2026):
        with open(f"{league}_{y % 100:02d}{(y + 1) % 100:02d}.csv", encoding="latin-1") as f:
            for r in csv.DictReader(f):
                if r.get("HS", "").strip() and r.get("HST", "").strip():
                    matches += 1
                    goals += int(r["FTHG"]) + int(r["FTAG"])
                    shots += int(r["HS"]) + int(r["AS"])
                    on_target += int(r["HST"]) + int(r["AST"])
print(f"{matches:,} matches: {goals / shots:.0%} of shots scored, {goals / on_target:.0%} of shots on target")