3 trams are coming every 10, 15 and 15 minutes. On average, how long do I have to wait for any tram to come? 3 trams are coming to the stop every 10, 15 and 15 minutes.
 On average, how long do I have to wait for any tram to come?

It's a practical problem, not some kind of a riddle for which I have a surprising magic trick or an answer. I really don't know. I was waiting for a tram when this question come to my mind. So, if you ask me for example "how the trams are driving?" my answer will be I don't know, I have the same (or lesser) tram knowledge as you. Assume some accurate (probably probabilistic;) model and present the answer, for example "5 minutes" + showing how you obtain this result. Perfect answer will generalize the problem, answering how long do we have to wait when the trams come every $x_1, x_2, x_3...$ minutes. But even the basic problem is not as easy as it is looking, so feel warned.
There are (at least) three reasonable probability models for this problem: (a) the Poisson process model (see heropup's answer), (b) the assumption that the trams arrive on time according to a scheme known to the user (for  examples see PhiNotPi's and user1008646's answers), and (c) the assumption that the trams arrive on time with unknown but equidistributed phases. In the following I shall treat model (c).

Model (c) is equivalent to the following: A random point $P=(X,Y,Z)$ is chosen in the block $$B:=\{(x,y,z)\&gt;|\&gt;0\leq x\leq15,\ 0\leq y\leq 15,\ 0\leq z\leq 10\}\ .$$
The waiting time $T$ is then given by $T=\min\{X,Y,Z\}$. The points $P$ with waiting time between $t$ and $t+dt$ are lying in the union of three rectangular panels of thickness $dt$ and having a distance $t$ from the planes $x=0$, $y=0$, and $z=0$ respectively. The area of the two vertical panels is $(10-t)(15-t)$, and the area of the horizontal one is $(15-t)(15-t)$. It follows that the probability distribution function $f_T$ of the waiting time is given by
$$f_T(t)={1\over2250}(525-80t+3t^2)\qquad(0\leq t\leq10)\ .$$
From this we obtain the expected waiting time as
$$E(T)=\int_0^{10} t\&gt;f_T(t)\ dt={85\over27}\ .$$
Suppose that none of the trains come simultaneously. Wait significant amount of time, say $N &gt;&gt; 1$ hours. Then the interval of $N$ hours will be punctured by $6N$ arrivals of the 10-min train, $4N$ arrivals of the 1st 15-min train, and $4N$ arrivals of the 2nd 15-min train. This breaks the $N$-hour interval into $14N$ waits between arrivals. Therefore the average wait is $60/14$ minutes.

EDIT: as pointed out in the comment, $60/14$ is the average interval length between the trains; however, the average wait should be $1/2$ of the interval length. Therefore the correct answer for the wait is $60/28$.

EDIT 2: as pointed out by the commenter, the answer is, generally speaking, incorrect. On the, eh, 3rd thought I realized the cause of my error: the average wait is not proportional to the average size of the interval; it's proportional to the 2nd moment of the interval.

Let me elaborate on the above a bit: suppose that the interval lengths are $X_i,1\le i
\le n$. Let $\Sigma_{i=1}^n X_i=1$ for simplicity. The probability of landing in the interval $X_i$ is $X_i$. The average wait on the assumption that you landed in $X_i$, is $X_i/2$. Therefore the average wait among all intervals is $\Sigma_{i=1}^n X_i\cdot X_i/2=1/2\cdot M_2$, where $M_2$ is the 2nd moment of the distribution $X_i$. The prior computations applied to the 1st moment $M_1$, also known as average.

So the correct answer should make further assumption on the distribution of the intervals.