How this instrument works
Flipping a fair coin n times and asking for the probability of exactly k heads is one of the oldest questions in probability, and it's answered by the binomial probability formula at its simplest setting: a fixed success probability of exactly 0.5. The number of ways to land k heads among n flips is the binomial coefficient C(n,k), counting every distinct ordering of heads and tails that has exactly k heads; each of those orderings is equally likely, at (1/2)ⁿ, so multiplying the count by that shared probability gives P(X=k) = C(n,k)/2ⁿ.
This calculator deliberately fixes the coin at p = 0.5 rather than exposing a general success-probability input, unlike this site's separate binomial-distribution calculator. That's a real, distinct scope choice, not an accident: a 'coin flip' in ordinary language always implies a fair coin, so someone searching for coin-flip probability is asking a narrower, more specific question than someone searching for a general binomial probability with an arbitrary p. The arithmetic underneath is identical to the binomial-distribution calculator evaluated at p=0.5 — this page just removes the extra input that isn't relevant to the fair-coin question.
Worth remembering: this is the probability of exactly k heads, not 'at least k' or 'at most k.' For 10 flips, the probability of exactly 5 heads (0.24609375) is actually the single most likely individual outcome, yet it's still under 25%, because the remaining probability is spread thinly across all the other possible head counts from 0 through 10. Cumulative questions — the chance of 5 or more heads, say — need the individual probabilities summed across that whole range, which this page doesn't do automatically.
- Enter the total number of coin flips into Number of flips (n).
- Enter the exact number of heads you want the probability for into Number of heads (k) — k cannot exceed n.
- Read P(exactly k heads) — fair coin for the exact probability, using a truly fair coin (p = 0.5) fixed by this calculator.
- Try a few different k values at the same n to see how the probability peaks near n/2 and tapers off toward the extremes of all-heads or all-tails.
Worked example — exactly 5 heads in 10 fair coin flips
Entering n = 10 into Number of flips and k = 5 into Number of heads — the classic question of getting exactly 5 heads in 10 fair coin flips — the instrument first finds C(10,5) = 252, the number of distinct orderings of 5 heads and 5 tails among 10 flips.
P(exactly k heads) — fair coin then reads 0.24609375, computed as 252/2¹⁰ = 252/1024. So even though 5 heads out of 10 is the single most probable individual outcome for a fair coin, it still happens well under a quarter of the time — the remaining roughly 75% of the probability is spread across getting anywhere from 0 to 4 or 6 to 10 heads instead.
Questions
How is this different from the site's binomial-distribution calculator?
The math is identical at p = 0.5 — this page is a special case of the general binomial formula. The difference is scope: binomial-distribution exposes a general success-probability input p for any weighted coin or any pass/fail scenario, while this calculator fixes p at exactly 0.5 and removes that input entirely, since a plain 'coin flip' in everyday language always means a fair coin. If you need an unfair or weighted coin, or any other fixed per-trial probability, the binomial-distribution calculator is the right tool instead.
Why isn't the probability of the most likely outcome closer to 100%?
Because there are many possible outcomes to spread the total probability across, not just two. With 10 flips there are 11 possible head counts (0 through 10), and even the single most likely one — 5 heads — only accounts for about a quarter of the total probability; the rest is distributed, unevenly, across the other 10 possible counts. As n grows larger, this spreading effect gets more pronounced: the single most likely outcome's probability actually shrinks even as it stays the most likely one.
How do I find the probability of at least k heads, not exactly k?
Sum this calculator's exact-probability result across every value from k up to n. For instance, P(at least 5 heads in 10 flips) adds together the individual P(X=5) through P(X=10) probabilities, which comes out well above the 0.246 figure for exactly 5 heads alone, since it includes every outcome from 5 heads all the way up to 10. This calculator reports one k value at a time; running it repeatedly and adding the results is the way to build a cumulative total.
Does the order of heads and tails matter for this probability?
No — P(X=k) already accounts for every possible ordering of k heads among n flips through the C(n,k) term, so it answers 'what's the chance of ending up with exactly k heads total, in any order' rather than 'what's the chance of one specific sequence.' A single specific sequence of n flips (heads, tails, heads, heads, ...) always has probability exactly (1/2)ⁿ on its own, regardless of how many heads it contains.
Can k be 0 or equal to n?
Yes — both are valid, representing all-tails (k=0) or all-heads (k=n). Both have the same probability, 1/2ⁿ, since there's only exactly one way to arrange either outcome (C(n,0) = C(n,n) = 1); these are the two least likely individual outcomes for any n greater than a couple of flips, since every other k value has more than one possible ordering contributing to it.