We flip a biased coin (P (head) = p) continously until we observe k heads, and then we stop. Let X be the number of flips. Find E[X].
My intuition tells me this should be a very simple problem, but I'm somehow struggling with it. I've tried to derive the distribution for X, then summing that over k to infinity, but the sum is not easy to evaluate.
Thanks for any help.
Best Answer
The trick is to write $X = X_1 + X_2 + \dots + X_k$, where $X_i$ is "the expected number of flips to get the $i$th head after getting the $(i-1)$th head".
Now it is not too hard to see that each $X_i$ is identically distributed and that the probability that $X_i = j$ (for $j \geq 1$) is equal to $p(1-p)^{j-1}$. (In other words, each $X_i$ is a geometric random variable). A simple calculation shows that $\mathbb{E}[X_i] = \frac{1}{p}$, so by linearity of expectation $\mathbb{E}[X] = \frac{k}{p}$.