Answer: HH vs HT: Waiting Times for Coin Patterns

Key Idea / Intuition

At first glance, HH and HT seem symmetric — each is a sequence of two fair-coin outcomes. But they behave very differently because of what happens when a "near-miss" occurs. If you're waiting for HH and you see HT, you've wasted both flips. But if you're waiting for HT and you see HH, the second H can still serve as the start of a future HT. This asymmetry in "overlap structure" makes HH harder to achieve, giving it a longer expected waiting time.

Formal Proof / Solution

Expected time to HH

Let $E$ = expected number of flips to see HH from a fresh start.
Let $E_H$ = expected additional flips to see HH, given the last flip was H.

Setting up equations:

From a fresh start, flip once:

With prob $\tfrac{1}{2}$ , get T → back to fresh start. Cost: 1 flip.
With prob $\tfrac{1}{2}$ , get H → now in state $E_H$ . Cost: 1 flip.

$E = 1 + \tfrac{1}{2} E + \tfrac{1}{2} E_H$

From state $E_H$ (last flip was H), flip again:

With prob $\tfrac{1}{2}$ , get H → done! Cost: 1 flip.
With prob $\tfrac{1}{2}$ , get T → back to fresh start. Cost: 1 flip.

$E_H = 1 + \tfrac{1}{2}(0) + \tfrac{1}{2} E = 1 + \tfrac{1}{2} E$

Substituting into the first equation:

$E = 1 + \tfrac{1}{2}E + \tfrac{1}{2}\!\left(1 + \tfrac{1}{2}E\right) = 2 + \tfrac{3}{4}E$

$\tfrac{1}{4}E = 2 \implies \boxed{E_{HH} = 6}$

Expected time to HT

Let $F$ = expected flips to HT from fresh start.
Let $F_H$ = expected additional flips, given last flip was H.

From a fresh start:

With prob $\tfrac{1}{2}$ , get T → back to fresh start. Cost: 1 flip.
With prob $\tfrac{1}{2}$ , get H → state $F_H$ . Cost: 1 flip.

$F = 1 + \tfrac{1}{2}F + \tfrac{1}{2}F_H$

From state $F_H$ (last flip was H), flip again:

With prob $\tfrac{1}{2}$ , get T → done! Cost: 1 flip.
With prob $\tfrac{1}{2}$ , get H → stay in $F_H$ (the new H can still start HT). Cost: 1 flip.

$F_H = 1 + \tfrac{1}{2}(0) + \tfrac{1}{2}F_H$

$\tfrac{1}{2}F_H = 1 \implies F_H = 2$

Substituting back:

$F = 1 + \tfrac{1}{2}F + \tfrac{1}{2}(2) = 2 + \tfrac{1}{2}F$

$\tfrac{1}{2}F = 2 \implies \boxed{E_{HT} = 4}$

Why the difference? (Intuition recap)

| Pattern | Near-miss behavior | Expected time | |---------|-------------------|---------------| | HH | Getting HT resets you completely | 6 | | HT | Getting HH keeps you in a "H seen" state | 4 |

For HH: a failure (getting T after H) wastes everything.
For HT: a "false start" (getting HH) is not fully wasted — the second H still counts as a potential head toward HT.

This is an instance of the general theory of pattern waiting times, where the overlap structure of the pattern determines how quickly it appears. HH has a self-overlap (its first H is also a valid start of HH), which paradoxically hurts it, because each near-miss costs more.