Statistics From Zero · Lesson 20 of 22
The Binomial Distribution: How Likely Is 6 Out of 10?
In one sentence: When you repeat the same yes-or-no attempt a fixed number of times, the binomial distribution tells you how likely each possible number of successes is, so you can tell a good day from a real improvement.
The idea
Lots of questions in life have the same shape: I try something n times, each try either works or doesn't, so how many will work? Free throws, multiple-choice guesses, seeds that sprout, customers who click "buy". The answer isn't one number but a list of chances: so likely to get none, so likely to get one, and so on. That list is the binomial distribution.
It's the first named distribution most courses teach after the bell curve, and it's the one that sits quietly behind the coin test in Lesson 8.
The hoop above the garage
Uncle Gary has a basketball hoop above the garage. Over the summer, Aunt Linda has kept a tally on the kitchen calendar: Gary makes about 3 shots in 10. A 30% shooter.
On Sunday he takes 10 shots and makes 6. He announces that he has "found his shooting form" and asks whether the local team holds open trials.
Linda asks a statistical question instead: if Gary were still a 30% shooter, how often would he make 6 out of 10 just by luck?
The four conditions
The binomial distribution fits when all four of these are true. A common way to remember them is BINS:
- Binary: each try has two outcomes, success or failure (in or out).
- Independent: one try doesn't change the next (Lesson 4).
- Number fixed: you decide n in advance (10 shots, not "until I get bored").
- Same chance: the probability of success, called p, is the same on every try (30%).
The chance of exactly 3 out of 10, by hand
There are two parts to the calculation.
- The chance of one particular order. Say in, in, in, then seven misses. That's 0.3 × 0.3 × 0.3 × 0.7 × 0.7 × 0.7 × 0.7 × 0.7 × 0.7 × 0.7, which is 0.3³ × 0.7⁷ = about 0.0022.
- How many orders give 3 in. The three makes could be shots 1, 2 and 3, or 1, 5 and 9, and so on. The number of ways to choose 3 shots out of 10 is written "10 choose 3", and it's 120.
Multiply them: 120 × 0.0022 = about 0.267. So Gary makes exactly 3 out of 10 about 27% of the time. That's the whole formula, in words: number of orders × chance of one order.
The whole list
Do the same for every possible number of makes and you get the binomial distribution for 10 shots at 30%:
| Makes out of 10 | 0 | 1 | 2 | 3 | 4 | 5 | 6 or more |
|---|---|---|---|---|---|---|---|
| Chance | 2.8% | 12.1% | 23.3% | 26.7% | 20.0% | 10.3% | 4.7% |
So a 30% shooter makes 6 or more about 4.7% of the time: roughly one Sunday in twenty. Gary's 6 is unusual for a 30% shooter, but not wildly so, and it sits right on the 5% line from Lesson 8. One good Sunday is a reason to keep counting, not a reason to call the local team. Linda has added a new column to the calendar.
Average and spread without the list
Two shortcuts save a lot of work:
- Average number of successes = n × p. 10 × 0.3 = 3. Exactly what the calendar says.
- Standard deviation = √(n × p × (1 - p)). √(10 × 0.3 × 0.7) = √2.1 = about 1.45. So a typical Sunday is 3 makes, give or take one or two.
With more tries, the list starts to look like a bell curve (Lesson 17 explains why: a count of successes is a sum of many small yes-or-no tries). A common rule of thumb: when both n × p and n × (1 - p) are at least 10, the bell curve is a good shortcut. For 100 shots at 30%, the exact chance of 40 or more is 2.1%, and the bell curve gives 1.9%. Close enough for most purposes.
Where it fails
- The chance changes. If Gary shoots better after warming up, or worse after his third beer, p isn't the same on every try and the list is wrong. The calendar should probably record the time of day.
- The tries aren't independent. If one miss makes him rush the next, the shots are linked. Real data often has more streaks than the binomial expects.
- The number of tries isn't fixed. "I'll keep shooting until I make 6" is a different question with a different answer. Stopping when the result looks good is a classic way to fool yourself.
Try it yourself
Cousin Kevin has a multiple-choice quiz with 10 questions, each with 4 options. He didn't study and guesses every answer. The pass mark is 5.
1. How many does he get right on average? 2. Roughly how likely is he to pass? 3. Is it more likely he gets zero right or that he passes?
Show the answer
1. n × p = 10 × 0.25 = 2.5.
2. Adding up the binomial chances for 5, 6, 7, 8, 9 and 10 correct gives about 7.8%, roughly 1 in 13.
3. Zero right has a chance of 0.75¹⁰ = about 5.6%, so passing is slightly more likely, but both are unlikely. Guessing is a strategy in the same sense that a raincoat is a roof. And if a whole class of 30 guessed, about two of them would pass, which is how a few lucky guessers end up convinced they understood the material.
The words your lecturer uses
| Lecturer says | It means |
|---|---|
| Bernoulli trial | One try with two outcomes, success or failure |
| Binomial distribution B(n, p) | The chances of each possible number of successes in n independent tries, each with success chance p |
| Probability mass function (PMF) | The list of chances for each exact count (the table above) |
| Binomial coefficient, "n choose k" | The number of ways to pick which k of the n tries are successes |
| Cumulative probability, P(X ≥ k) | The chance of k or more successes, found by adding up the list |
| Mean np, variance np(1 - p) | The shortcuts for the average and the spread |
| Normal approximation | Using a bell curve instead of the exact list when n is large |
Three things to remember
- Fixed number of independent yes-or-no tries with the same chance: that's binomial.
- Exactly k successes = number of orders × chance of one order. Average n × p.
- An unusual result is a reason to keep counting. Change the chance or the stopping rule and the list no longer applies.
Part of Statistics From Zero, a free beginner course by WrongStat. Not a substitute for your lecturer, but hopefully a good translator.