Coin flips are the binomial distribution in its simplest clothes
Flip a coin n times and count the heads. Every sequence of flips is equally likely with a fair coin, so the only question is counting: how many of the 2n sequences contain exactly k heads? That count is the binomial coefficient C(n, k), and the probability is C(n, k) ÷ 2n. Tilt the coin (a bent coin, a weighted game, a click-through rate relabelled) and each sequence gets priced by pk(1 − p)n−k instead: the full binomial formula, with the fair coin as the special case p = 0.5.
Ten flips of a fair coin. Exactly 6 heads: C(10, 6) = 210 sequences out of 1,024, so 0.2051. At least 6 heads: 0.3770. At most 6: 0.8281. The single most likely count is 5 heads at 0.2461, yet even it happens less than a quarter of the time: the distribution spreads its mass, which is why "exactly average" outcomes are rarer than intuition expects.
Streaks and the gambler's fallacy
The same maths prices streaks. Five heads in a row has probability (0.5)5 = 0.031 in any specified run of five flips, but in 100 flips there are 96 overlapping windows of five, and at least one five-head run appears with probability near 0.81. Streaks are not evidence of a hot coin; they are what randomness looks like when you watch it long enough. What does not change: after any streak, the next flip is still 50/50. The coin has no memory, and the conditional probability P(heads | five heads just happened) stays 0.5.
Computing "at least one head in 10 flips" by adding ten 0.5s. Probabilities of overlapping events do not add. Use the complement: 1 − P(no heads) = 1 − (0.5)10 = 0.9990.
Frequently asked questions
What is the probability of getting all heads in 10 flips?
(0.5)^10 = 1/1024, about 0.001. The same as any other fully specified sequence, all tails or alternating included.
How many heads should I expect in 100 flips?
On average 50, with a standard deviation of 5. Counts between 40 and 60 cover roughly 95% of experiments; outside that range, start suspecting the coin.
Does a long run of heads make tails due?
No. That is the gambler's fallacy. Independent flips have no memory, so P(tails) stays 0.5 whatever came before.