Even stopping time for a first repeated face

A fair coin is tossed repeatedly, stopping as soon as two successive tosses agree (HH or TT). Find the chance that an even number of tosses was needed.

$2/3$. Stopping at toss $n$ forces the first $n-1$ tosses to alternate and the $n$th to repeat, so $\mathbb{P}(N=n)=2^{-(n-1)}$ for $n\ge2$. Summing over even $n$: $\tfrac12+\tfrac18+\cdots=\tfrac23$.