I have a probability question here. Suppose I throw a coin and I stop when I have a pair of "up" and "down". I want to calculate the expectation of the number of "up" before I got a pair. Here is my strategy:
Suppose we have $k$ heads ups, before we get a pair. And we will have $$ \mathbb{E}(k) = \sum_{k=1}^{\infty} k(\frac{1}{2})^k(1/2) = 2/2 = 1 $$
So the expectation value of k will be $1$, which means we only need to have one heads up when we get a pair of up and down. Does that seems right?
If $1$ up is required, the sequence needed is either $UD$ or $DU$ or $DDU$ or $DDDU$ and so on.
If $2$ "up"s is needed, the sequence needed must be $UUD$.
If $3$ "up"s is required, the sequence needed is $UUUD$.
\begin{align}E[K]=1\cdot \left(\frac14 +\frac14\sum_{k=0}^\infty \frac1{2^k}\right)+ \sum_{k=2}^\infty \frac{k}{2^{k+1}}=\frac14 \cdot \frac{1}{1-\frac12}+\sum_{k=1}^\infty \frac{k}{2^{k+1}}=1.5 \end{align}