Two in a Row

After the first flip, think about what happens on each subsequent flip relative to the previous one.

Every flip after the first independently matches the previous flip with probability $1/2$, so the number of extra flips needed is geometric.

1. 1. Reduce to a geometric wait

   The first flip is free. Every flip after that matches the immediately preceding flip with probability $1/2$, independent of history, and the process stops the first time a match occurs.

2. 2. Compute the expectation

   Let $N-1$ be the number of flips after the first, which is geometric with success probability $1/2$, so $E[N-1]=2$.

   $$E[N]=1+E[N-1]=1+2=3$$

Answer: 3