Coin Flip Probability Calculator

What is a coin flip probability calculator?

A coin flip probability calculator works out how likely a particular number of heads is when you toss a coin several times. Each toss is an independent trial with two possible outcomes — heads or tails — so a sequence of flips follows the binomial distribution. The calculator answers questions such as “What is the chance of getting exactly 5 heads in 10 flips?” or “What is the chance of at least 2 heads in 3 flips?”.

It works for both fair and biased coins. You set the probability of heads $p$ to any value between 0 and 1, so the same tool also covers loaded coins, weighted dice with two outcomes, and any other yes/no experiment repeated a fixed number of times.

How does the calculator work?

You provide three inputs and choose what to calculate:

• Number of flips ($n$) — how many times the coin is tossed (an integer $\ge 1$).
• Number of heads ($k$) — the count of heads you are interested in (an integer with $0 \le k \le n$).
• Probability of heads ($p$) — the chance of heads on a single toss, between 0 and 1 (0.5 for a fair coin).

The Calculate option selects one of three questions:

• Exactly k heads — the probability of getting precisely $k$ heads.
• At most k heads — the cumulative probability of getting $k$ or fewer heads.
• At least k heads — the cumulative probability of getting $k$ or more heads.

The result is shown as a probability between 0 and 1 (rounded to six decimal places) and also as a percentage.

Formulas

The probability of exactly $k$ heads in $n$ flips is the binomial probability mass function:

where the binomial coefficient is

The cumulative cases sum the individual terms:

Worked examples

1. Exactly 5 heads in 10 fair flips. With $n = 10$, $k = 5$, $p = 0.5$: $\binom{10}{5} = 252$, so $P = 252 \times 0.5^{5} \times 0.5^{5} = 252 / 1024 \approx 0.246094$ (about 24.61%).

2. Exactly 1 head in 2 fair flips. With $n = 2$, $k = 1$, $p = 0.5$: $\binom{2}{1} = 2$, so $P = 2 \times 0.5 \times 0.5 = 0.5$ (50%).

3. At least 2 heads in 3 fair flips. With $n = 3$, $k = 2$, $p = 0.5$: $P(X \ge 2) = P(2) + P(3) = 0.375 + 0.125 = 0.5$ (50%).

Practical notes

• Setting $k = 0$ with “at least” always returns 1, and $k = n$ with “at most” always returns 1, because every outcome satisfies the condition.
• For a biased coin, change $p$. For example, $n = 5$, $k = 2$, $p = 0.3$ gives $\binom{5}{2} \times 0.3^{2} \times 0.7^{3} = 0.3087$.
• The binomial model assumes the flips are independent and that $p$ stays the same on every toss.

To explore related ideas, see the Bayes’ theorem calculator for updating probabilities with evidence, or the average calculator for summarising data.