Why HHH Takes Longer to Appear Than HHT

The problem. Flip a fair coin repeatedly. On average, how many flips until you first see HHH? How many until you first see HHT? Both are specific 3-flip sequences, each with probability $\frac{1}{8}$ of occurring in any given window of 3 flips — so it's tempting to guess the waiting times are equal. They aren't: HHT appears after 8 flips on average, while HHH takes 14. The gap comes entirely from what happens when a near-miss occurs.

The key idea: what a failed attempt leaves behind

Suppose you're chasing HHH and you've just flipped H, H, and then the coin comes up T. You needed HHH but got HHT — total failure, and you have to start over from scratch, because a T wipes out any H-streak. Now suppose you're chasing HHT and you've flipped H, H, then the third flip comes up H again (not T). You haven't completed HHT, but you haven't lost anything either — you still have "HH" as your current progress, exactly where you'd want to be if you're two flips into building toward HHT. A failed attempt at HHT still leaves useful partial progress behind; a failed attempt at HHH (specifically, hitting a T after any number of H's) destroys all progress instantly.

Worked example: setting up the states for HHT

Model progress toward HHT as a Markov chain with states: $S_0$ (no useful progress), $S_H$ (just saw an H), $S_{HH}$ (just saw HH), and "done" (saw HHT). From $S_0$: flip H moves to $S_H$ (prob $\frac12$), flip T stays at $S_0$ (prob $\frac12$). From $S_H$: flip H moves to $S_{HH}$, flip T moves back to $S_0$. From $S_{HH}$: flip H stays at $S_{HH}$ (because "HH" is still valid progress toward HHT after another H), flip T completes to "done." Let $E_0, E_H, E_{HH}$ be expected additional flips from each state:

$$E_{HH} = 1 + \tfrac12 E_{HH} + \tfrac12(0) \;\Rightarrow\; E_{HH} = 2.$$ $$E_H = 1 + \tfrac12 E_{HH} + \tfrac12 E_0, \qquad E_0 = 1 + \tfrac12 E_H + \tfrac12 E_0.$$

Solving: from the $E_0$ equation, $\tfrac12 E_0 = 1 + \tfrac12 E_H$, so $E_0 = 2 + E_H$. Substituting into $E_H = 1 + \tfrac12(2) + \tfrac12 E_0 = 2 + \tfrac12(2+E_H) = 3 + \tfrac12 E_H$, giving $E_H = 6$, and $E_0 = 8$. Starting from scratch, HHT takes 8 flips on average.

Why HHH is worse: self-overlap costs you the progress

Redo the chain for HHH: from $S_{HH}$ (just saw HH), a flip of H completes HHH, but a flip of T sends you all the way back to $S_0$ — unlike HHT's chain, where the failing flip only cost you the "T" attempt while H kept you at $S_{HH}$. Solving the analogous equations gives $E_0 = 14$ for HHH — a full 6 flips worse than HHT, purely because every near-miss on HHH (a T after any H's) wipes the slate clean, while HHT's near-misses (extra H's) do not.

Think of each sample path above as a run of coin flips: the "waiting time" question asks how long, on average, before a specific short pattern first appears somewhere along the path — a natural first-passage question the explorer's paths make visually intuitive.