Suppose we flip a coin until we see a head. What is the expected value of the number of flips we will take?
I am pretty new to expected value, so I tried to evaluate it by multiplying the probability of each scenario with the number of flips it took to get there (like taking the arithmetic mean). This didn't work though, because of the infinite possibilities. I'm really confused, and if someone were to provide an answer, I would really appreciate it if they could go into detail.
Best Answer
Let $X$ be a discrete random variable with possible outcomes: $x_1, x_2, x_3,\dots, x_i,\dots$ with associated probabilities $p_1,p_2,p_3,\dots,p_i,\dots$
The expected value of $f(X)$ is given as:
In your specific example, $X$ could be one of the values: $1,2,3,\dots ,i,\dots$ with corresponding probabilities $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots,\frac{1}{2^i},\dots$ (seen easily from the beginnings of a tree diagram)
So, the expected value of $X$ is:
$\sum\limits_{i=1}^\infty i(\frac{1}{2})^i$
This is a well-known infinite sum of the form $\sum\limits_{i=1}^\infty i p (1-p)^{i-1}$, in this case with $p=1-p=\frac{1}{2}$. You will likely be expected to simply memorize the result, and it is included in most formula lists. $\sum\limits_{i=1}^\infty i p (1-p)^{i-1}=\frac{1}{p}~~~~~~~(\dagger)$
Using this result without proof, we get our expected number of flips is $\frac{1}{0.5}=2$
The proof of $(\dagger)$:
$$\sum\limits_{i=1}^\infty i p (1-p)^{i-1} = p\sum\limits_{i=1}^\infty i (1-p)^{i-1}\\ = p\left(\sum\limits_{i=1}^\infty (1-p)^{i-1} + \sum\limits_{i=2}^\infty (1-p)^{i-1} + \sum\limits_{i=3}^\infty (1-p)^{i-1} + \dots\right)\\ = p\left[(1/p)+(1-p)/p+(1-p)^2/p+\dots\right]\\ = 1 + (1-p)+(1-p)^2+\dots\\ =\frac{1}{p}$$