Expected Longest Streak of Heads

Question. Flip a fair coin 100 times. What's the expected length of the longest run of consecutive heads somewhere in that sequence? Intuition badly underestimates this one — most people guess something small, like 4 or 5, imagining that "long streaks are rare." The actual answer is around 7, and the reasoning behind it — approximate scaling with $\log_2 n$ — is a genuinely useful mental tool, not just a memorized fact.

Why the answer is bigger than intuition suggests

The mistake in the naive intuition is thinking about one specific run of, say, 6 heads in a row, and correctly noting that has probability $(1/2)^6 = 1/64$ — rare. But the question isn't about one specific position; it's about the longest run anywhere among 100 flips, and there are roughly 95 different starting positions where a run of length 6 could begin. With that many chances, even a $1/64$-per-position event becomes likely to occur somewhere.

The scaling argument

A run of length $k$ starting at a given position has probability $(1/2)^k$. Across roughly $n$ starting positions in a sequence of $n$ flips, the expected number of length-$k$ runs is about $n \times (1/2)^k$. The longest run tends to sit right around the length $k^*$ where this expected count transitions from "many" to "essentially zero" — i.e., where $n \times (1/2)^{k^*} \approx 1$:

$$n \cdot 2^{-k^*} \approx 1 \quad \Rightarrow \quad k^* \approx \log_2 n .$$

In plain English: the longest streak you should expect to see in $n$ flips is roughly $\log_2 n$, because that's the run length at which "how many places could this length of streak start" and "how rare is any one streak of that length" roughly balance.

Worked example: checking $n=100$

$$\log_2 100 = \frac{\ln 100}{\ln 2} \approx \frac{4.605}{0.693} \approx 6.64 .$$

The more careful classical result adds a small correction (roughly $\log_2 n - \log_2\log_2 n$ type refinements plus a constant, and simulation confirms an expected longest streak close to 6.6–7 for $n=100$), landing at around 7. A quick sanity check: for $n=1{,}000$ flips, $\log_2 1000 \approx 9.97$, so the expected longest streak only grows to about 10 — going from 100 to 1,000 flips (10× more data) adds only about 3 to the expected longest streak, because the relationship is logarithmic, not linear.

Why this matters beyond the puzzle

This is the exact reasoning that should make you skeptical of "wow, that's an amazing streak" claims without context. A trader who's had 7 winning days in a row out of the last 100 trading days isn't necessarily unusually skilled that stretch — a run of that length is close to what you'd expect to see somewhere in 100 independent fair-coin-like outcomes, purely from chance. The right question is never "is this streak long," it's "is this streak longer than $\log_2 n$ would predict by chance alone," which is a much higher bar than most gut checks apply.