Finding the expected number of success The scenario goes like this: a cell phone user has 1/3 of a chance to send a text message, and a 2/3 of a chance to receive a text message.
Question: what is the expected number of text messages received by the cell phone user before they send a text?
Is the answer 1/p = 1/(2/3) = 3/2 = 1.5 messages? (the expected value for a geometric random variable).
 A: This indeed concerns geometric distribution but in your guess you make 2 mistakes.
Here "sending a message" must be interpreted as "success" so that $p=\frac13$.
We do not deal with the expected number of trials needed to arrive at a success but with the number of failures that preceed the first success. In that case the expectation is not $\frac1p$ but equals:$$\frac1p-1=\frac{1-p}p$$
So the correct answer is:$$\frac{1-\frac13}{\frac13}=2$$
A more direct way is solving the equation:$$\mu=\frac23(1+\mu)+\frac13\cdot0$$Here $\mu$ denotes the expectation.

Informal confirmation: in a situation like this it will not be surprising to meet sequences like:$$\cdots RRSRRSRRSRRSRRS\cdots$$
($R$ for receive and $S$ for send)
A: You correctly noticed that the distribution is geometric, i.e. the distribution for "the number of failures before observing a success". The distribution takes two forms (described in Wikipedia), where the form that would be more useful for you is where $k$ stands for the number of "failures", so the probability mass function is $(1-p)^k p$ and where the expected value is $\tfrac{1-p}{p}$.
The tricky part is finding out what the "success" happening with probability $p$ is and the "failure" happening with probability $1-p$ mean in the problem.
