A specific run is 0.5 to the power of the flips
One flip of a fair coin is 50/50. Because flips are independent, the chance of a named sequence multiplies: two heads is 0.5 × 0.5 = 1 in 4, and each extra flip halves it again. The same math applies to any exact pattern — HTHTH is just as likely as HHHHH.
Odds of a run of heads
| Flips in a row | Probability | Odds |
|---|---|---|
| 2 | 25% | 1 in 4 |
| 3 | 12.5% | 1 in 8 |
| 5 | 3.13% | 1 in 32 |
| 7 | 0.78% | 1 in 128 |
| 10 | 0.098% | 1 in 1024 |
These are the odds set before the first flip. Once you have already thrown four heads, the fifth is still a plain 50/50.
Counting heads across many flips: the binomial rule
For "how many heads out of n," a run is just one path to the answer. The binomial formula counts every path to exactly k heads, so the middle counts get the most ways and become the most likely.
- Fair coin, p = 0.5. Expected heads is n × 0.5, with spread of about half the square root of n. In 100 flips you expect 50, and 95% of the time land between 40 and 60.
- Biased coin, p ≠ 0.5. Same formula, just swap in the real p. A coin with p = 0.6 over 100 flips centers on 60 heads, not 50.
- Probability vs odds. A 25% chance is odds of 1 to 3 (one win to three losses), not 1 in 25. Odds compare wins to losses; probability is wins out of the total.
Common questions
What are the odds of flipping heads several times in a row?
For a fair coin the chance of a specific run is 0.5 raised to the number of flips. Two in a row is 1 in 4, five in a row is 1 in 32, and ten in a row is 1 in 1024. Each flip is independent, so a run does not make the next flip any more or less likely.
What is the probability of exactly 5 heads in 10 flips?
About 24.6%. It is the single most likely count because 5 is the expected number, but it is still under one in four, since results fan out around the average.
Is the coin "due" for tails after a streak of heads?
No. That is the gambler's fallacy. A fair coin has no memory, so after any streak the next flip is still 50/50. Streaks look surprising but are exactly as likely as any other specific sequence of the same length.
