There is an excellent answer to the more general question, "Expected Waiting time if the are many buses, which each stop every $m_k$ minutes", can be found here.
As mentioned in the comments on this question and the linked one, the answer depends very much on the model used to describe the passage times of the buses. Here are two possible solutions, each based on slightly different assumptions, for two bus scenario, but which give quite different results.
Case 1. Fully random times.
Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent.
Expected Time is $$\mathbb E(T) = m, \quad \textrm{where } \; \frac{1}{m} = \frac{1}{m_1}+\frac{1}{m_2}$$.
That is, $\frac{1}{m} = (1/8+1/12) =\frac{5}{24}$. So $m$ =4 minutes 48 seconds.
Case 2. Fully periodic passage times with random uniform initializations
Here, buses of type $k$ pass at times in $S_k+mkN$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent.
$$
\mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}\right),
$$
thus, with $m_1 =8, m_2 = 12$, we get
$$m=8-\frac{64}{2}\left(\frac{1}{8}+\frac{1}{12}\right)+\frac{512}{3}\left(\frac{1}{96}\right) = 28/9$$
So $m=$ 3 minutes 7 seconds.
Thus, we can see from the quite different results that it depends a lot on what model/assumptions you use.
Best Answer
Okay here is my answer. Say that the bus comes every $12$ minutes. Where you arrive on a twelve minute interval can be modeled by a uniform distribution. However the bus at the start of the twelve minute interval (bus 1) can be early/late and the bus at the end of the interval (bus 2) also early or late. Model these with a normal distributions (iid) centered at $0$ with standard deviation $\sigma$. (These are only approximately normal.) Now if bus 1 is late and you're early enough on the interval, you get bus 1. If not, and if you beat bus 2, you will get on bus 2. However, if you arrive at the end of the interval and bus 2 is early, then you will have to wait through another $12$-ish minutes to get on bus 3. You find that you will wait on average longer than $6$ minutes. This is the bus paradox.
(I used to ride the bus a lot. It gives you a lot of time to think.)