When it applies
The binomial distribution describes the number of successes in a fixed number of repeated trials. It applies when four things hold:
- There is a fixed number of trials, n.
- Each trial has two outcomes, which we call success and failure.
- The chance of success, p, is the same on every trial.
- The trials are independent: one result does not change the next.
Flipping a coin 10 times and counting heads fits. So does rolling a die 6 times and counting sixes. Drawing cards from a deck without putting them back does not fit exactly, since the chance changes after each draw.
The formula
The probability of exactly k successes in n trials is:
P(k) = C(n, k) × p^k × (1 − p)^(n − k)
There are two pieces. The factor p^k × (1 − p)^(n − k) is the chance of one particular sequence with k successes and n − k failures. The factor C(n, k) counts how many different sequences contain exactly k successes. This is where the binomial coefficients from Pascal’s triangle come in.
Example 1: coin flips
Flip a fair coin 5 times. What is the chance of exactly 3 heads?
- C(5, 3) = 10
- Each specific sequence has probability (1/2)^3 × (1/2)^2 = 1/32
So P(3) = 10 × 1/32 = 10/32 = 0.3125, or 31.25%.
For 10 flips and exactly 5 heads: C(10, 5) = 252 and each sequence has probability 1/1,024. So P(5) = 252/1,024 ≈ 0.246. Even though 5 heads is the most likely single count, it happens only about a quarter of the time.
Example 2: free throws
Suppose a player is assumed to make 80% of free throws in the long run, and every attempt is treated as independent. This is an illustrative simplification, not a claim about any real player. What is the chance of making exactly 4 of 5?
- C(5, 4) = 5
- p^4 = 0.8^4 = 0.4096
- (1 − p)^1 = 0.2
P(4) = 5 × 0.4096 × 0.2 = 0.4096, or about 41%.
Making all 5: C(5, 5) = 1, and 0.8^5 = 0.32768, or about 32.8%.
So the chance of making at least 4 of 5 is 0.4096 + 0.32768 = 0.73728, or about 73.7%. Adding the exact cases is how “at least” and “at most” questions are handled.
Example 3: sixes on a die
Roll a fair die 6 times. What is the chance of exactly one six?
- C(6, 1) = 6
- p = 1/6, so the success factor is 1/6
- (5/6)^5 = 3,125/7,776
P(1) = 6 × (1/6) × 3,125/7,776 = 3,125/7,776, about 0.402 or 40.2%.
The chance of no sixes in 6 rolls is (5/6)^6 = 15,625/46,656, about 33.5%. This could also be found as a complement counting step: the chance of at least one six is 1 − 0.335 = 0.665, roughly two in three.
The average and the spread
The expected number of successes is n × p. For 5 free throws at 80%, that is 4. For 10 fair coin flips, it is 5. Of course, an expected value is an average over many repetitions, not a prediction for any single run. Actual counts vary around it, and the typical size of that variation is described by the standard deviation, which for the binomial is the square root of n × p × (1 − p). For 10 fair flips that is the square root of 2.5, about 1.58.
With many trials, the shape of the distribution becomes bell-like, which connects to the normal distribution. And the law of large numbers says that the proportion of successes tends to settle near p over a very large number of trials, which does not mean that any short run will “balance out”.
Checking the conditions
Before applying the formula, ask:
- Is the chance really constant? A player getting tired, or a game that changes odds as it goes, breaks this.
- Are the trials independent? Streaks caused by real factors, such as a change in conditions, may break this.
- Is there truly a success-or-failure outcome? If there are several outcomes, a different model is needed.
If the model is wrong, the arithmetic can be perfect and the answer still misleading. The sample size also matters: estimating p from only a handful of attempts gives a very rough figure.
A caution about games of chance
The binomial distribution can tell you how likely a particular number of wins is over repeated plays of a game with fixed odds, but it cannot make unfavourable odds favourable. With a house edge, the expected result per play is a loss, and more plays tend to increase the total expected loss, even though a short run of wins is possible. Past outcomes do not change the chance of the next one. Gambling can be harmful, and anyone struggling should contact a local support service.