Coin Flip Probability Calculator
Set the number of flips and how many heads you want, then read the binomial probability. Adjust the heads chance for a biased coin and switch between exactly, at least, or at most.
How to calculate coin flip probability
- Enter the number of flips and the target number of heads.
- Set the heads probability (0.5 for a fair coin) if needed.
- Choose exactly, at least, or at most, and read the probability.
Examples
Five heads in ten fair flips
10 flips, exactly 5 heads, fair coin
About 24.61 percent
Eight or more heads
10 flips, at least 8 heads, fair coin
About 5.47 percent
Frequently asked questions
How is coin flip probability calculated?
It uses the binomial formula. The chance of exactly k heads in n flips is C(n, k) times p to the power k times (1 minus p) to the power (n minus k), where p is the chance of heads on one flip and C(n, k) counts the ways to choose which flips are heads.
What is the chance of exactly 5 heads in 10 fair flips?
About 24.61 percent. There are C(10, 5) = 252 ways to get 5 heads out of 10 flips, and each specific sequence has probability 0.5 to the power 10, which is 1 in 1024. So the chance is 252 divided by 1024.
What does at least or at most mean here?
At least adds up the chances of the target heads count and every higher count, so at least 8 of 10 covers 8, 9, and 10 heads. At most adds the target and every lower count. Exactly counts only the single target number of heads.
Can I model a biased or unfair coin?
Yes. Set the heads probability to any value between 0 and 1. A coin that lands heads 70 percent of the time uses 0.7. The formula weights each sequence by that probability, so the results reflect the bias automatically.
Why is getting exactly half heads not the most likely on every count?
For a fair coin the single most likely outcome is the count nearest half the flips, but its probability is still well under 50 percent because it competes with many nearby counts. As you flip more coins the distribution spreads, so any one exact count becomes less likely even as the average stays at half.
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