Coin Toss Probability Calculator

The Coin Toss Probability Calculator takes two integers: Flips (n) and Heads (k). The defaults are 10 flips and 5 heads — the fair-coin problem that sits at the centre of a Binomial(n, 1/2) distribution. Leave those values, calculate, and SorteCalc returns P(exactly 5 heads in 10 flips) = 24.6094% together with Expected Heads = 5.0. That pair is the entire output: a point probability for the count you asked for, and the mean of the distribution, which does not depend on k.

Work the default by hand so the screen is not a black box. A fair coin has success probability p = 1/2 on every independent toss. The number of binary sequences with exactly five heads in ten trials is the binomial coefficient C(10,5) = 10! / (5! 5!) = 252. Each particular sequence of ten tosses has probability (1/2)^10 = 1/1024. Multiply: 252 / 1024 = 0.24609375, printed as 24.6094%. Expected heads is always n/2 for a fair coin, so 10/2 = 5.0. The Heads field is unused for that second line; it only feeds the exact-count probability.

Change Heads to 0 with Flips still 10. Only the all-tails sequence survives, so the probability collapses to 1/1024 ≈ 0.0977%. Heads = 10 is the same number by symmetry. Heads = 3 yields C(10,3)/1024 = 120/1024 ≈ 11.7188%. The engine clamps k with min(heads, flips), so typing Heads = 12 and Flips = 10 silently computes P(exactly 10 heads). Flips is bounded 1–40 because C(n,k) and 2^n grow fast; forty tosses already give a 2^40 sample space.

The mean 5.0 is not a promise that your next session of ten flips will land on five heads. Variance of Binomial(n, 1/2) is np(1−p) = 2.5, so the standard deviation is √2.5 ≈ 1.58 heads. Roughly two-thirds of ten-flip sessions fall between about 3.4 and 6.6 heads. Streaks of tails do not make heads “due.” The calculator has no memory term because a fair coin has none.

Use the page as a calibration check before you trust more exotic binomial models — biased coins, unequal p, or optional stopping. If 24.6094% at the centre of a ten-flip session feels wrong, the intuition is the problem, not Pascal’s formula. Casino chatter about a coin that “owes” heads is the gambler’s fallacy in its purest form.

A coin toss is the oldest classroom model of a Bernoulli trial: two outcomes, equal probability, independence between tosses. Gamblers used coins to settle disputes long before they used them to teach combinatorics. What changed in 1654 was not the hardware. Blaise Pascal and Pierre de Fermat, writing about the problème des partis — how to split a stake when a game of chance is interrupted — put the binomial coefficient and the equally likely sequence on a firm footing. Christiaan Huygens’s 1657 De ratiociniis in ludo aleae then gave expectation a numerical definition. Jacob Bernoulli’s Ars Conjectandi (published 1713) proved the weak law of large numbers for this exact process: the sample proportion of heads converges in probability to 1/2.

SorteCalc’s tool is deliberately narrow. It computes P(X = k) for X ~ Binomial(n, 1/2) and the mean n/2. It does not simulate a physical coin, estimate bias from data, or implement a sequential test. Real coins are slightly biased; Persi Diaconis and colleagues have measured spinning and flipping coins that land on the start face more often than 50%. A casino “coin flip” promotion is usually a marketing wrapper around a random-number generator, not a minted disc. None of that enters the formula here. The educational value is the opposite of realism: if you cannot compute the fair case, you cannot interpret a biased one.

The calculator exists because people chronically misread sequences. Ten flips with five heads feels “normal”; ten flips with one head feels “broken.” Both are fully legal outcomes of the same measure. The probability mass function is symmetric and peaked at 5, but the tails are fat enough that 2 heads in 10 (C(10,2)/1024 = 45/1024 ≈ 4.39%) happens often in a night of play. Sports showdowns and tournament tie-breaks that use a coin are using this distribution, not luck as a moral quality.

Limits are sharp. Independence fails if a toss is a spin with a conserved angular bias, or if a dealer palming a coin introduces dependence. The binomial model also fails if you stop when you hit five heads — that is a negative-binomial waiting time, a different random variable. Optional stopping is how many “hot streak” stories get manufactured. This page will not save you from that design error; it will only tell you the probability of a fixed (n, k) pair under a fair independent model.

Treat the output as a definition, not a forecast. After one session of ten flips you will see some integer between 0 and 10. After thousands of sessions the average of those integers will sit near 5. That is the law of large numbers, not a system. Pascal and Fermat were splitting a pot, not picking a side.

A lecture on how to calculate the chance of exactly five heads in ten tosses of a fair coin begins with the binomial law, not with a superstition about a coin that “owes” a face. Fix n = 10 independent tosses and k = 5 heads. Each toss is a Bernoulli trial with success chance one half. The number of binary sequences with exactly five heads is the binomial coefficient C(10,5) = 10! / (5! 5!) = 252. Each particular sequence of ten faces has probability (1/2)^10 = 1/1024. Multiply the count by that atom: 252/1024 = 0.24609375, printed as 24.6094 percent. That point mass sits at the centre of the symmetric law on {0, 1, …, 10}.

Expectation is a separate identity. For n tosses with success chance one half, the mean count of heads is n/2. With n = 10 that mean is exactly 5. The heads field never enters this second line; it only feeds the exact-count probability. Variance of the same law is n/4 = 2.5, so the standard deviation is the square root of 2.5, about 1.5811 heads. Roughly two-thirds of ten-toss sessions fall between about 3.4 and 6.6 heads. A normal sketch already sits nearby at this n, but the exact coefficient is cheap and is what we compute.

History sits on the 1654 letters of Blaise Pascal and Pierre de Fermat on the interrupted-game problem, the problème des partis. They treated equally likely sequences as the measure, which is why C(n,k)/2^n is a probability and not merely a combinatorial count. Christiaan Huygens gave expectation a numerical definition in 1657. Jacob Bernoulli’s Ars Conjectandi, published in 1713, proved the weak law of large numbers for this exact process: the sample proportion of heads converges in probability to one half. Abraham de Moivre later wrapped a normal curve around Pascal’s triangle; that approximation is the ancestor of every large-n coin lecture, not a substitute for 252/1024 at n = 10.

Symmetry is immediate. The chance of k heads equals the chance of n minus k heads, so zero heads and ten heads share 1/1024 ≈ 0.0977 percent. Three heads yield C(10,3)/1024 = 120/1024 ≈ 11.7188 percent. Cumulative probabilities are a different object: the chance of at most five heads would sum the atoms k = 0 through 5 and, because of the atom at the mean, equals one half plus half of 24.6094 percent, namely 62.3046875 percent. This page reports only the atom.

Assumptions must be named before the percentage is believed. Tosses are independent, the success chance is constantly one half, and n is chosen before the data are seen. Violate independence with a spinning bias, or stop when you first hit five heads, and you have left this law for a waiting-time law. Optional stopping is how “hot streak” stories are manufactured. A slightly biased physical coin, as Persi Diaconis and colleagues have measured, is a different parameter. After thousands of independent ten-toss sessions the average of the integers will sit near 5; that is Bernoulli’s law, not a system, and not a promise about the next ten flips.