A coin toss is the archetype of a Bernoulli trial—a discrete experiment with exactly two mutually exclusive outcomes: success (heads) and failure (tails). While a single flip of a fair coin has an intuitive $50\%$ ($1/2$) chance, calculating compound odds across multiple tosses requires the Binomial Distribution. Use this Coin Flip Probability Calculator to solve exact probabilities, cumulative thresholds (at least $k$ or at most $k$ heads), and consecutive streak odds alongside our Probability Calculator and Dice Probability Calculator.
Binomial Coin Flip Probability Formulas
1. Mathematical Foundations of Coin Toss Probability
When flipping a coin $n$ times, each flip is completely independent of preceding results. The sample space size expands exponentially with each toss:
- $1\text{ flip} \implies 2^1 = 2\text{ outcomes } (\text{H}, \text{T})$
- $2\text{ flips} \implies 2^2 = 4\text{ outcomes } (\text{HH}, \text{HT}, \text{TH}, \text{TT})$
- $3\text{ flips} \implies 2^3 = 8\text{ outcomes } (\text{HHH}, \text{HHT}, \text{HTH}, \text{HTT}, \text{THH}, \text{THT}, \text{TTH}, \text{TTT})$
- $n\text{ flips} \implies 2^n\text{ total outcomes}$
Because sequence order often does not matter when calculating total heads, we use combinations $\binom{n}{k}$ to count the number of unique sequences that contain exactly $k$ heads out of $n$ trials.
2. Binomial Distribution Table for 10 Coin Flips ($n = 10$, $p = 0.5$)
Total possible binary sequences: $2^{10} = 1,024$. The following matrix details exact, cumulative “at least”, and cumulative “at most” probabilities for every possible outcome:
| Heads ($k$) | Combinations $\binom{10}{k}$ | Exact Probability $P(X = k)$ | At Least $k$ Heads $P(X \ge k)$ | At Most $k$ Heads $P(X \le k)$ |
|---|---|---|---|---|
| 0 Heads | 1 / 1,024 | 0.098% | 100.000% | 0.098% |
| 1 Head | 10 / 1,024 | 0.977% | 99.902% | 1.074% |
| 2 Heads | 45 / 1,024 | 4.395% | 98.926% | 5.469% |
| 3 Heads | 120 / 1,024 | 11.719% | 94.531% | 17.188% |
| 4 Heads | 210 / 1,024 | 20.508% | 82.813% | 37.695% |
| 5 Heads (Mode) | 252 / 1,024 | 24.609% | 62.305% | 62.305% |
| 6 Heads | 210 / 1,024 | 20.508% | 37.695% | 82.813% |
| 7 Heads | 120 / 1,024 | 11.719% | 17.188% | 94.531% |
| 8 Heads | 45 / 1,024 | 4.395% | 5.469% | 98.926% |
| 9 Heads | 10 / 1,024 | 0.977% | 1.074% | 99.902% |
| 10 Heads | 1 / 1,024 | 0.098% | 0.098% | 100.000% |
3. Consecutive Streak Odds: The Math of Coin Runs
The probability of getting a consecutive run of $k$ heads on $k$ flips is $(1/2)^k$:
| Consecutive Streak Length | Exact Fraction | Percentage | Odds Against |
|---|---|---|---|
| 3 in a row | 1 / 8 | 12.50% | 1 in 8 |
| 5 in a row | 1 / 32 | 3.125% | 1 in 32 |
| 7 in a row | 1 / 128 | 0.781% | 1 in 128 |
| 10 in a row | 1 / 1,024 | 0.0977% | 1 in 1,024 |
| 15 in a row | 1 / 32,768 | 0.00305% | 1 in 32,768 |
| 20 in a row | 1 / 1,048,576 | 0.000095% | 1 in 1,048,576 |
4. Step-by-Step Worked Calculation Examples
Example 1: Exactly 7 Heads in 10 Flips
Problem: Calculate the exact probability of tossing exactly 7 heads in 10 flips of a fair coin ($n = 10$, $k = 7$, $p = 0.5$).
Example 2: At Least 2 Heads in 5 Flips (Complement Rule)
Problem: Find the probability of flipping at least 2 heads in 5 coin tosses ($P(X \ge 2)$).
Frequently Asked Questions
What is the binomial formula for coin flip probability?
The binomial formula is $P(X = k) = \binom{n}{k} \times p^k \times (1 – p)^{n – k}$, where $\binom{n}{k} = \frac{n!}{k!(n – k)!}$, $n$ is the total flips, $k$ is the number of heads, and $p$ is the probability of heads on each toss ($0.50$ for a fair coin).
What is the Gambler’s Fallacy in coin flips?
The Gambler’s Fallacy is the mistaken belief that previous random outcomes influence future flips. A fair coin has no memory: even if tails has appeared 10 times consecutively, the probability of heads on the 11th toss remains precisely $50\%$.
Why is getting 5 heads in 10 flips not a 50% probability?
While 5 heads is the single most probable outcome in 10 flips, its exact probability is $24.61\%$ ($252 / 1,024$). The remaining $75.39\%$ probability is distributed among all other possible counts ($0, 1, 2, 3, 4, 6, 7, 8, 9, 10$).
What is the difference between ‘exact’ and ‘at least’ probability?
Exact probability $P(X = k)$ computes the chance of getting strictly $k$ heads. “At least” probability $P(X \ge k)$ computes the cumulative likelihood of getting $k$ or more heads by summing $P(X = k) + P(X = k+1) + \dots + P(X = n)$.
How do you calculate odds for a weighted or biased coin?
For a weighted coin where $p \neq 0.50$, substitute the specific bias value into the binomial equation. For example, if a coin lands on heads $60\%$ of the time ($p = 0.60$), the probability of getting 3 heads in 3 flips is $(0.60)^3 = 0.216$ ($21.6\%$).