Coin Flip Probability Calculator

Compute the exact probability of getting a given number of heads in a series of coin flips — exactly k, at least k, at most k, all heads, or no heads — for a fair coin or any biased coin you specify.

Before You Calculate

  • A fair coin has probability of heads p = 0.5; change it to model a biased coin (say p = 0.6) — everything else works the same.
  • Flips are independent: the coin has no memory, so a run of tails does not make heads “due”.
  • Counting heads in n independent flips is exactly a binomial experiment, so all results use exact binomial probabilities, not approximations.

The Math Behind Coin Flips

Each flip is a Bernoulli trial, and the number of heads in n flips follows the binomial distribution:

P(X = k) = C(n, k) × pᵇ × (1 − p)ⁿ⁻ᵇ

Fair coin (p = 0.5): P(X = k) = C(n, k) / 2ⁿ

Where:

  • n = number of flips, k = number of heads
  • C(n, k) = number of orderings with exactly k heads
  • p = probability of heads on one flip

For a fair coin every specific sequence of n flips is equally likely (probability 1/2ⁿ), so counting is all that matters: the chance of exactly k heads is just the number of sequences containing k heads, C(n, k), divided by the 2ⁿ total sequences. The cumulative answers sum these exact terms — no normal approximation is involved.

Why 5 Heads in 10 Flips Is Not 50-50

The single most common surprise: the chance of getting exactly half heads in 10 flips of a fair coin is only C(10, 5)/2¹⁰ = 252/1024 ≈ 24.6%. The 50% intuition belongs to a different quantity — the expected proportion of heads. Individual counts spread around that center: 4, 5, or 6 heads together cover about 65.6% of outcomes, while extreme results stay rare (all 10 heads happens once per 1024 runs, about 0.098%).

As n grows, the count of heads concentrates relatively — the proportion of heads converges to p (the law of large numbers) — even though the probability of hitting the exact center shrinks. That is also the regime where the binomial starts to look like the bell curve of the normal distribution.

Worked Example: 10 Flips of a Fair Coin

Enter n = 10, k = 5, p = 0.5. The calculator reports:

  1. Exactly 5 heads: C(10, 5) = 252 orderings, each with probability 1/1024, so P = 252/1024 = 0.246094.
  2. At most 5 heads: summing k = 0…5 gives 0.623047.
  3. At least 5 heads: by symmetry of the fair coin, also 0.623047.
  4. All heads: (1/2)¹⁰ = 1/1024 ≈ 0.000977; no heads is the same by symmetry.
  5. At least one head: 1 − 1/1024 ≈ 0.999023.

The distribution summary reads mean np = 5 heads and standard deviation √(10 × 0.5 × 0.5) ≈ 1.5811 — so results between about 3 and 7 heads are entirely ordinary, and a 9- or 10-head run is the only outcome that should genuinely raise an eyebrow about fairness.

Frequently Asked Questions

What is the probability of getting heads 5 times in 10 flips?

For a fair coin, P(exactly 5 heads in 10 flips) = C(10, 5) / 2^10 = 252/1024, which is about 0.2461 or 24.6%. Although 5 is the most likely single count, it is far from certain - three quarters of the time the count lands somewhere else.

What are the odds of flipping heads 3 times in a row?

Independent flips multiply: (1/2)^3 = 1/8 = 12.5%. Ten in a row is (1/2)^10 = 1/1024, about 0.098%. Note this is the probability of all heads in exactly that many flips; in a longer session, runs of 3 or more heads somewhere in the sequence are much more common than intuition suggests.

After several tails in a row, is heads more likely on the next flip?

No. Each flip is independent, so the probability of heads stays exactly p on every flip regardless of history. The feeling that a result is 'due' is the gambler's fallacy. What is true is that long runs of one side become evidence worth testing when they persist - which is a statement about the coin's fairness, not about the next flip being self-correcting.

How does the calculator handle a biased coin?

Set the probability of heads to any value between 0 and 1. With p = 0.6 and 10 flips, the chance of exactly 5 heads becomes C(10, 5) x 0.6^5 x 0.4^5 = 0.2007, and the expected count shifts to np = 6. Everything on the page - cumulative probabilities, all-heads, at-least-one - uses the same exact binomial formula with your p.

What is the difference between 'exactly k heads' and 'at least k heads'?

'Exactly k' is the probability of one specific count. 'At least k' accumulates k, k+1, ..., up to n, so it is always at least as large. For 10 fair flips: exactly 8 heads has probability 0.0439, while at least 8 heads (8, 9, or 10) has probability 0.0547. Choose the version that matches how your question is worded.

How many flips give a 90% chance of at least one head?

Using the complement, P(at least one head) = 1 - (1/2)^n reaches 90% at n = 4, since 1 - 1/16 = 93.75% while n = 3 gives only 87.5%. For a biased coin replace 1/2 with 1 - p; the probability-of-at-least-one calculator on this site solves this inverse question for any p and target.