Coin Flip Probability Calculator
Calculate the exact probability of getting a specific number of heads in a fixed number of fair coin flips. It applies the binomial formula, so it is ideal for statistics homework, game design, and making sense of random streaks.
Last updated: May 2026
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About this calculator
This calculator uses the binomial probability formula for a fair coin. For n independent flips, the chance of getting exactly k heads is P(X = k) = C(n, k) · p^k · (1 − p)^(n − k), where p = 0.5 for a fair coin, so the expression simplifies to P = C(n, k) · 0.5^n. Here n is the number of flips, k is the number of heads you want, and C(n, k) = n! / (k! · (n − k)!) is the binomial coefficient — the number of distinct orderings in which those k heads can appear among the n flips. The result is a probability between 0 and 1 (multiply by 100 for a percentage). A key property is symmetry: P(k) = P(n − k), because heads and tails are equally likely. Edge cases matter: if k > n the event is impossible (probability 0), and if k = 0 or k = n the probability collapses to 0.5^n, the chance of one specific all-tails or all-heads sequence. The model assumes the flips are independent and the coin is unbiased; it gives the probability of EXACTLY k heads, not 'k or more'.
How to use
Example 1 — 5 heads in 10 flips. Enter Number of Flips = 10 and Number of Heads = 5. Compute C(10, 5) = 252, and 0.5^10 = 1/1024 ≈ 0.000977. Multiply: 252 × 0.000977 = 0.2461, or 24.61%. Verify: this is the single most likely outcome, yet it still happens less than a quarter of the time — a good reminder that 'half heads' is typical but not guaranteed. Example 2 — 2 heads in 4 flips. Enter Number of Flips = 4 and Number of Heads = 2. Compute C(4, 2) = 6 and 0.5^4 = 1/16 = 0.0625. Multiply: 6 × 0.0625 = 0.375, or 37.5%. Verify by listing the 16 equally likely sequences: exactly six of them (HHTT, HTHT, HTTH, THHT, THTH, TTHH) contain two heads, and 6/16 = 0.375. The two methods agree.
Frequently asked questions
What is the probability of getting exactly half heads in an even number of flips?
For 2m flips the chance of exactly m heads is C(2m, m) · 0.5^(2m), and counterintuitively it shrinks as the number of flips grows. With 2 flips it is 50%, with 10 flips about 24.6%, and with 100 flips only about 8%. The reason is that there are more and more possible outcomes to spread the probability across, even though 'half heads' stays the single most likely count. People often assume 'half' should become more certain with more flips, but that intuition applies to the proportion of heads, not to hitting the exact count. If you care about being near half rather than exactly half, you should sum a range of outcomes instead.
What is the difference between exactly k heads and at least k heads?
This calculator returns the probability of EXACTLY k heads in n flips. 'At least k heads' is a cumulative probability that adds up P(k) + P(k+1) + ... + P(n), and it is almost always larger than the single-point value. For example, exactly 5 heads in 10 flips is 24.6%, but at least 5 heads in 10 flips is about 62.3%. Confusing the two is the single most common mistake when people apply these numbers to real decisions. If a homework problem or game rule says 'or more', 'at most', or 'between', you must add the relevant individual probabilities rather than reading off one value here.
Does this work for a biased or unfair coin?
No — this calculator assumes a fair coin with p = 0.5 for heads on every flip. A biased coin uses the full binomial formula P(X = k) = C(n, k) · p^k · (1 − p)^(n − k) with p set to the true head probability, which this tool does not let you change. If you used it for, say, a coin that lands heads 70% of the time, the answers would be wrong because the 0.5^n term no longer applies. For weighted coins, loaded dice, or any event whose two outcomes are not equally likely, use a general binomial probability calculator instead. The fair-coin assumption is what makes the convenient 0.5^n simplification valid.
Why are the flips assumed to be independent, and when does that fail?
Independence means the result of one flip does not influence any other, which is what lets us multiply probabilities and use the binomial formula. Real coin flips are an excellent approximation of independence, so the model is very accurate for them. However, the formula is wrong for events that look coin-like but are actually dependent — for instance, drawing cards without replacement, where each draw changes what remains. The 'gambler's fallacy' (believing a long run of heads makes tails 'due') is a misreading of independence: past flips never change the next flip's 50/50 odds. If your scenario has memory or sampling without replacement, switch to a hypergeometric or conditional-probability model.
When should I NOT use a coin-flip probability calculator?
Avoid it whenever the two outcomes are not equally likely, the trials are not independent, or you actually need a cumulative or 'streak' probability rather than an exact count. It is also the wrong tool for questions about the proportion of heads converging over time — that is the law of large numbers, better explored with a normal-distribution approximation for large n. Do not use it to model real-world 'fair or not' decisions where the underlying probability is unknown; estimate p first. Finally, for very large n the exact binomial values become tiny and a normal approximation is usually more informative. Pick the model that matches whether you need exactly-k, at-least-k, or near-k.