Coin-flip questions are the classic introduction to probability. Each toss has two outcomes, the tosses do not influence each other and the chance of heads stays the same, so the number of heads in n tosses follows a binomial distribution. Probability is then the number of favourable sequences divided by the number of possible sequences – 2ⁿ for a fair coin.
For example, there are 2³ = 8 possible sequences of three flips and three of them contain exactly two heads (HHT, HTH, THH), so the probability is 3/8 = 37.5%. For larger numbers counting by hand becomes impractical, and the binomial coefficient C(n, k) does the counting for you.
Enter the number of flips, the condition and the number of heads. The calculator returns the probability in every common format, the exact fraction for a fair coin, the expected number of heads and its standard deviation, and a chart of the whole distribution with the outcomes you asked about highlighted. Change the chance of heads to model a weighted coin or any repeated yes/no event.
How to use the Coin Flip Probability Calculator
- Number of flips
- I want to get…
- Number of heads (k)
- Chance of heads on one flip
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
The tool evaluates the binomial probability mass function for every count of heads and adds up the ones that meet your condition. For a fair coin it also counts the favourable sequences exactly with whole-number arithmetic, so the fraction is exact even for long runs of flips.
Worked example
What is the chance of at least 8 heads in 10 fair flips? Count the sequences: C(10, 8) = 45, C(10, 9) = 10 and C(10, 10) = 1, a total of 56 out of 2¹⁰ = 1,024. So P(X ≥ 8) = 56/1,024 = 7/128 ≈ 5.47%, or about 1 in 18.3. The expected number of heads is 10 × 0.5 = 5 with a standard deviation of √(10 × 0.25) ≈ 1.58.
Quick reference for 10 fair flips
| Heads | Sequences | Probability |
|---|---|---|
| 0 or 10 | 1 | 0.098% |
| 1 or 9 | 10 | 0.977% |
| 2 or 8 | 45 | 4.39% |
| 3 or 7 | 120 | 11.72% |
| 4 or 6 | 210 | 20.51% |
| 5 | 252 | 24.61% |
Formula
P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ · fair coin: P(X = k) = C(n, k) ÷ 2ⁿ · P(X ≥ k) = Σ P(X = j) for j = k … n
Worked example
Inputs — Number of flips: 10; I want to get…: Exactly k heads; Number of heads (k): 8; Chance of heads on one flip: 50 %.
Results — Probability = 4.3945 %; Probability as a decimal = 0.043945; Odds (1 in …) = 1 in 22.76; Exact fraction (fair coin) = 45 / 1,024; Expected number of heads = 5; Standard deviation = 1.5811.
Key terms
- Binomial coefficient C(n, k)
- The number of ways to choose which k of the n flips are heads.
- Binomial distribution
- The distribution of the number of successes in n independent trials with the same success chance.
- Complement
- The event that something does not happen; P(not A) = 1 − P(A).
Frequently asked questions
What is the probability of getting 8 heads in 10 flips?
Exactly 8: C(10, 8) ÷ 1,024 = 45/1,024 ≈ 4.39%. At least 8: 56/1,024 = 7/128 ≈ 5.47%.
What is the chance of 2 heads in 3 flips?
3/8 = 37.5%, from the sequences HHT, HTH and THH.
What is the chance of at least one head in 4 flips?
1 − (1/2)⁴ = 15/16 = 93.75%.
What is the chance of 10 heads in a row?
(1/2)¹⁰ = 1/1,024 ≈ 0.098%.
How do I calculate "at least" probabilities?
Add the exact probabilities from k up to n, or use 1 − P(at most k − 1). The calculator does both.
What if the coin is not fair?
Enter the chance of heads, for example 60%; the binomial formula uses pᵏ(1 − p)ⁿ⁻ᵏ instead of 1/2ⁿ.
What is the most likely number of heads?
n/2 for a fair coin (or the two middle values when n is odd); the chart shows the peak.