Coin Toss Probability Formula with Examples

Key Points

• The coin toss probability formula gives a specific sequence of $n$ tosses the probability $\left(\frac{1}{2}\right)^n$.

• For a particular number of heads rather than a particular order, the count of arrangements multiplies in: $P(k \text{ heads}) = \binom{n}{k}\left(\frac{1}{2}\right)^n$.

• Tosses are independent, so previous results never change the next toss's probability, which is what makes the gambler's fallacy a fallacy.

• Questions asking for "at least one" are usually quickest through the complement, $1 - P(\text{none})$.

• With $n$ tosses there are $2^n$ equally likely sequences, which is the denominator behind every one of these calculations.

What Is the Coin Toss Probability Formula?

For a fair coin, each toss has probability $\frac{1}{2}$ of heads, and tosses are independent. Any one specific sequence of $n$ tosses therefore has probability:

$$P(\text{a given sequence}) = \left(\frac{1}{2}\right)^n$$

When the order does not matter and only the number of heads does, the number of arrangements enters:

$$P(k \text{ heads in } n \text{ tosses}) = \binom{n}{k}\left(\frac{1}{2}\right)^n$$

Why Order Changes the Answer

There is only one way to toss three heads in a row, so $P(HHH) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

But there are three ways to get exactly two heads in three tosses: HHT, HTH and THH. Each has probability $\frac{1}{8}$, so together they give $\frac{3}{8}$.

That factor of 3 is $\binom{3}{2}$, the number of ways of choosing which tosses are heads. Specifying an order pins down a single sequence; specifying only a count collects several, and the formula has to count them.

Variable Key

Examples of the Coin Toss Probability Formula

Example 1

Find the probability of three heads in three tosses of a fair coin.

$$P(HHH) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$$

Final answer: $\frac{1}{8}$

Example 2

A fair coin has landed heads five times in a row. Find the probability the next toss is tails.

Five heads in a row feels overdue for a correction, so a probability above $\frac{1}{2}$ seems reasonable.

The coin has no memory. Each toss is independent, so the five previous results are irrelevant to the sixth, and nothing about the coin has changed.

$$P(\text{tails on the next toss}) = \frac{1}{2}$$

Final answer: $\frac{1}{2}$. Expecting a correction is the gambler's fallacy, and it survives precisely because long runs feel so unlikely.

Example 3

Find the probability of exactly two heads in three tosses.

Three arrangements give two heads, each with probability $\frac{1}{8}$:

$$P = \binom{3}{2}\left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8}$$

Final answer: $\frac{3}{8}$

Example 4

Find the probability of at least one head in four tosses.

Work with the complement, since "no heads" is a single sequence:

$$P(\text{no heads}) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$ $$P(\text{at least one head}) = 1 - \frac{1}{16} = \frac{15}{16}$$

Final answer: $\frac{15}{16}$

Listing the fifteen favourable sequences would give the same answer far more slowly, which is why the complement is the standard move for "at least one".

Example 5

Find the probability of exactly three heads in five tosses.

$$\binom{5}{3} = 10, \qquad P = 10 \times \left(\frac{1}{2}\right)^5 = \frac{10}{32} = \frac{5}{16}$$

Final answer: $\frac{5}{16}$

Four Coin Questions Compared

The first two rows are the pair that matters most. A question naming an order wants the first; a question naming only a total wants the second, and the difference is a factor of $\binom{n}{k}$.

Where Do Students Trip Up on Coin Toss Probability?

Mistake 1: Expecting past tosses to influence the next

Where it slips in: A long run of one result feels so unlikely that a correction seems due.

Don't do this: Raise the probability of tails after five heads.

The correct way: Each toss stays at $\frac{1}{2}$. The coin has no memory, and the unlikeliness of five heads was already settled before those tosses happened.

Students who ask what physically changed about the coin get past this quickly, because the answer is always nothing. The run feels significant, but significance is not a property the coin can carry into the next toss.

Mistake 2: Forgetting to count arrangements

Where it slips in: The powers-of-a-half part is memorable, and the combination factor is easy to omit.

Don't do this: $P(2 \text{ heads in } 3) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$

The correct way: Multiply by $\binom{3}{2} = 3$ to get $\frac{3}{8}$. Several sequences satisfy a count, and all of them must be included.

Mistake 3: Listing outcomes for an "at least" question

Where it slips in: The phrase suggests adding up all the qualifying cases.

Don't do this: Add fifteen separate sequence probabilities for at least one head in four tosses.

The correct way: Subtract the complement, $1 - \frac{1}{16}$. The complement of "at least one" is "none", which is always a single case.

Bottom Line

• The coin toss probability formula gives a specific sequence of $n$ tosses probability $\left(\frac{1}{2}\right)^n$.

• For a count of heads rather than an order, multiply by $\binom{n}{k}$ to include every arrangement.

• Tosses are independent, so no run of results alters the next toss.

• "At least one" questions are fastest through the complement, since "none" is always one case.

• To work through probability with a teacher, explore a 7th grade math tutor or math classes online.

Quick Self-Check, Try These

Find the probability of four tails in four tosses, then of exactly two heads in four tosses. Finally find the probability of at least one tail in three tosses, using the complement.

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Read More

• Conditional Probability Formula

• Empirical Probability Formula

• What Is Probability

• Expected Value Formula

• Pascal's Triangle