Coin Flip Probability Calculator
Getting exactly 5 heads in 10 fair flips has a 24.6094% chance, while getting at least 5 heads has a 62.3047% chance. This coin flip probability calculator answers three related questions about a chosen number of heads: the chance of exactly that many, at least that many, and at most that many. It also converts the exact probability into an approximate 1-in-N figure. The calculation uses the binomial distribution for independent flips of a fair coin. Every sequence of heads and tails has the same probability, but many different sequences can contain the same total number of heads. For example, HHT and HTH are separate sequences that both contain two heads. That distinction is why the calculation counts arrangements before dividing by the total number of possible sequences. Use the exact result when a question asks for one specific count. Use at least when the chosen count is a lower boundary, such as 60 or more heads in 100 flips. Use at most when it is an upper boundary, such as no more than 3 heads in 10 flips. These statements sound similar, but they cover different sets of outcomes and can produce very different percentages.
Quick answer
Each fair flip is 50/50, and n flips produce 2^n equally likely sequences.
Exactly this many heads
24.6094%
At least
62.3047%
At most
62.3047%
About 1 in
4.1
What this tells you
- •Each fair flip is 50/50, and n flips produce 2^n equally likely sequences.
- •The chance of exactly k heads is the number of orderings, C(n, k), divided by 2^n.
- •At-least and at-most probabilities add up the exact chances across a range of counts.
- •More flips concentrate results near half heads, but the chance of exactly half actually shrinks.
- •Counts equally far from half have equal exact probabilities, so 4 and 6 heads in 10 flips are equally likely.
- •The calculator accepts whole counts only and rejects a heads count greater than the number of flips.
How to Use
- 1Enter the total number of flips.
- 2Enter the number of heads you care about.
- 3Read the exact probability, plus the at-least and at-most cumulative chances.
- 4The 1-in-N figure translates small probabilities into plainer language.
- 5Match the result to the wording of your question. Exactly includes one count, at least includes that count and every larger count, and at most includes that count and every smaller count.
- 6Keep the fair-coin assumption in mind. If the chance of heads differs from 50%, this calculator does not model that bias.
- 7For a quick reasonableness check, compare a count with half the flips. Counts near half should usually be more likely than counts near either extreme.
How It Works
Formula
P(exactly k) = C(n, k) / 2^nC(n, k) counts how many distinct sequences place k heads among n flips. It can be written as n! / (k!(n - k)!). For 5 heads in 10 flips, C(10, 5) = 252. Ten fair flips have 2^10 = 1,024 equally likely sequences, so 252 / 1,024 = 0.24609375, or 24.6094% after the calculator rounds to four decimal places. The at-least result adds P(k), P(k + 1), and every exact probability through P(n). The at-most result adds the exact probabilities from P(0) through P(k). The implementation calculates binomial coefficients in logarithmic form instead of constructing large factorials directly. This avoids ordinary factorial overflow for large flip counts, although the displayed percentages still use finite floating-point precision.
Calculation note: values are processed in the order shown above, using the current input units.
Worked Examples
Exactly 5 heads in 10 flips
252 of the 1024 possible sequences have exactly 5 heads.
10 heads in a row
Only one sequence of 1024 is all heads.
At least 60 heads in 100 flips
A 60/40 split in 100 flips is already rare enough to raise an eyebrow about the coin.
At most 2 heads in 5 flips
The favorable sequences contain 0, 1, or 2 heads. Their counts are C(5,0) + C(5,1) + C(5,2) = 1 + 5 + 10 = 16. Dividing 16 by 2^5 = 32 gives 0.5, or 50%.
Exactly 0 heads in 8 flips
Zero heads means every flip is tails. Only one of the 2^8 = 256 sequences meets that condition, so the exact probability is 1 / 256 = 0.00390625, or 0.3906% after rounding.
Exactly 50 heads in 100 flips
Fifty is the most likely single head count in 100 fair flips, but it is only one point in a distribution spanning 0 through 100. The exact probability is about 7.9589%, while a wider range around 50 captures much more probability.
Exactly k Heads in 10 Flips
The full probability distribution for 10 fair flips.
| Heads | Probability |
|---|---|
| 0 or 10 | 0.10% each |
| 1 or 9 | 0.98% each |
| 2 or 8 | 4.39% each |
| 3 or 7 | 11.72% each |
| 4 or 6 | 20.51% each |
| 5 | 24.61% |
The distribution is symmetric. Replacing every head with a tail pairs k heads with 10 - k heads, so each pair has the same probability.
Why exactly half is not the same as about half
A fair coin has an expected head count of n / 2, but an expected value is not a promise about one trial. Ten flips can produce any whole count from 0 through 10. Exactly 5 heads is the most likely individual count, yet its probability is only 24.6094%. The remaining 75.3906% is spread across all the other possible counts.
As the number of flips grows, the proportion of heads tends to stay closer to one half, but the number of possible head counts also grows. That is why exactly 50 heads in 100 flips has a smaller probability than exactly 5 heads in 10 flips. A range such as 45 through 55 heads is a better way to express about half than one exact count.
Cumulative results answer range questions directly. At least k includes k itself, while more than k begins at k + 1. At most k includes k itself, while fewer than k ends at k - 1. Read those boundary words carefully before selecting the result.
Common mistakes
- Expecting exactly half heads to be likely. It is only the single most likely count, 24.61% for 10 flips and shrinking as flips grow.
- Believing a streak changes the next flip. After 9 heads, the 10th is still 50/50. The 1-in-1024 rarity belonged to the whole sequence in advance.
- Confusing exactly with at least. Exactly 5 in 10 is 24.61%, at least 5 is 62.3%, very different questions.
- Adding 50% per flip. The chance of at least one head in two flips is 75%, not 100%.
- Treating the expected count as a guaranteed result. Five is the expected number of heads in 10 flips, but most 10-flip trials do not produce exactly 5.
- Reading at least k as more than k. At least 5 includes 5, while more than 5 begins at 6.
- Using the 1-in-N figure as a schedule. A 1-in-256 event can happen twice close together or fail to happen in far more than 256 trials.
- Applying the fair-coin result to a weighted or mechanically biased coin. A different probability of heads requires the general binomial formula with that probability included.
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