How coin flip probability is calculated
Each flip is an independent 50/50 event, so the number of heads in n flips follows the binomial distribution. The chance of any single sequence with exactly k heads is 0.5ⁿ, and there are C(n,k) such sequences.
- n — number of flips
- k — number of heads
- C(n, k) — number of ways to choose which k of the n flips are heads
More detail
Exactly k vs. at least k
"Exactly k" (the PMF) only counts the single outcome of k heads. "At least k" adds up every outcome from k heads through n heads, so it's always the larger — or equal — number. For n = 10, exactly 5 heads is about 24.6%, but at least 5 heads is about 62.3% because it includes 6, 7, 8, 9, and 10 heads too.
Why the probabilities peak near n/2
Because C(n, k) is largest when k is close to n/2, the exact-probability curve peaks around half heads and shrinks toward the extremes. Getting all heads (k = n) has probability 0.5ⁿ — for 10 flips that's about 0.1% (shown as 0.10%), vanishingly small even though each single flip is 50/50.
Quick sanity check. In theory P(X ≥ 0) is 100%, and P(X ≥ k) plus the probability of fewer than k heads sums to 100%. Enter n = 10, k = 0 and the at-least figure reads 100.00%; so does n = 100,000, k = 10,000, because getting at least 10,000 heads in 100,000 flips is all but certain. The exact-k figure beside it reads 0.00% at that size — landing on one specific count out of 100,001 possibilities really is that unlikely, so the two numbers disagreeing is the arithmetic working, not breaking.
Frequently asked questions
What's the probability of getting exactly 5 heads in 10 flips?
C(10,5) × 0.5¹⁰ = 252 / 1024 ≈ 24.6%. Enter n = 10, k = 5 above to confirm — this is the single most likely outcome, but it's still under a quarter of the time because 10 heads can land in many other ways too.
What's the difference between "exactly k" and "at least k" heads?
"Exactly k" is one specific count of heads: P(X=k) = C(n,k) × 0.5ⁿ. "At least k" sums that formula for every count from k up to n, so it's always equal to or greater than the exact probability — for n=4, k=2 exact is 37.5% but at least is 68.75%.
How likely is it to get all heads in 4 flips?
P(X=4) with n=4, k=4 is C(4,4) × 0.5⁴ = 1/16 = 6.25%. In general, all-heads probability is 0.5ⁿ, which halves with every additional flip — by 10 flips it's under 0.1%.
Does this assume a fair coin?
Yes — every flip is treated as an independent 50/50 event. A biased coin (say 55% heads) would need a different success probability in place of 0.5, which this calculator does not model.