Why HH takes longer to appear than HT

Question

Flip a fair coin until a target two-flip pattern first appears.

Show that the expected wait for "HH" is $6$ flips, but the expected wait for "HT" is only $4$ flips, even though any specific pair of flips is HH or HT with equal probability $\tfrac14$ .

Accepted Answer

Set up small Markov chains tracking progress. Write e\varnothing for the expected wait from scratch and eH for the wait once you have a lone H that could start the pattern. Waiting for HT. From scratch: flip H (prob 	frac12) 	o state H; flip T (prob 	frac12) 	o still nothing useful: e\varnothing = 1 + 	frac12 eH + 	frac12 e\varnothing. From state H: flip T 	o done; flip H 	o still state H (a fresh H, still a valid start): eH = 1 + 	frac12(0) + 	frac12 eH \implies eH = 2. Then e\varnothing = 1 + 	frac12(2) + 	frac12 e\varnothing \Rightarrow 	frac12 e\varnothing = 2 \Rightarrow e\v…

Why HH takes longer to appear than HT

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