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:

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.