Average time waiting for bus There are two buses: first arrives to the bus top every 8 minutes, and second arrives every 12 minutes. What is the average waiting time on a bus stop if we take whatever came first?
It is expected that buses arrive at regular intervals, so first arrives every 8 minutes and second arrives every 12 minutes, but it is unknown when did they started.
I was thinking in the following way: there is 1/3 probability that second bus arrives at 8-12 minutes, so it is after the first one. In this case probability is 1/3 * 4(average waiting time for the first bus). Then we have 2/3 probability that second bus came at 0-7 minutes, and basically we have two buses that arrive every 8 minutes. In this case I estimated average waiting time as 2 minutes(4 min average waiting time and 2 minutes because there are two 8 minutes buses now) and in this case answer is 2 * 2/3 + 4 * 1/3 = 8/3 = 2 + 2/3.
But I'm not sure that in case of two buses there is actually 2 minutes waiting time and not some other number, and also I think that this is a "standard" problem that should have standard way of solving. Please guide me to the correct answer.
 A: 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.
