Expected value of a weighted coin

Question

I'm given a problem with a biased coin where $P(heads) = 0.25$ and $P(tails) = 0.75$. I've seen in other posts that the $E(X) = 1/(1-p)$ where X is the number of tosses until a tails is shown and $p = P(heads) = 0.25$.

When I've tried it on my own, I haven't gotten the same result. $$\sum_{x=1}^{\infty} p^{x-1}(1-p) = (1-p)\sum_{x=1}^{\infty} p^{x-1} = (1-p)\sum_{x=0}^{\infty} p^{x} = (1-p)(1/(1-p)) = 1.$$

If someone can explain where I'm going wrong in my math, that'd be awesome.

Answer 1

The problem is that you're summing over the probability distribution of the number of flips. Naturally, this will give you an answer of one. To calculate the expectation, you need to include an $x$ term:

$$\mathbb{E}(X) = \sum_{x=1}^{\infty} x p^{x-1} (1-p).$$

Your calculations for summing the probabilities were correct, so you should be able to take it from here.

Answer 2

\begin{eqnarray*} \sum_{x=1}^{\infty} \color{red}{x} p^{x-1}(1-p) &=& (1-p)\sum_{x=1}^{\infty}x p^{x-1} \\ &=& (1-p) \left(\frac{1}{(1-p)^2 } \right) = \frac{1}{1-p} . \end{eqnarray*}