Balls are placed into 3 urns. Expected time until some urn has 100 balls. We have $3$ urns. At each round a ball is placed is placed into one of them, at random, with uniform probability. The game stops when some urn has $100$ balls.
What is expected duration of the game (number of rounds)? 
Results from a simulation:

 A: The expected time can be expressed in terms of the incomplete gamma function as follows (inspired by this paper and comments here):
In general: We want the expected value of the time to wait $T$ until one of the $3$ urns contains $n$ ($=100$) balls. Then
$$E_{n}[T] = \sum_{t=1}^\infty P(T\ge t) = \sum_{t=0}^\infty p_{n}(t) \tag1$$
where $p_{n}(t) $ is the probability that after $t$ rounds ($t$ balls) all $3$ urns have less than $n$ balls. But this is equivalent to
$$ \sum_{x=0}^{n-1}\sum_{y=0}^{n-1}\sum_{z=0}^{n-1} \frac{1}{3^{x+y+z}} \frac{(x+y+z)!}{x! \, y! \, z!} \tag2$$
Further, we use a property of the (upper) incomplete gamma function:
$$\begin{align}
\left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^3 &= \left( e^{-a} \sum_{r=0}^{n-1}\frac{a^r}{r!} \right)^3 \\&=  
e^{-3a} \sum_{x=0}^{n-1}\sum_{y=0}^{n-1}\sum_{z=0}^{n-1}\frac{a^{x+y+z}}{x! \, y! \, z!} \tag3
 \end{align}$$
Integrating and using $\int_0^\infty \exp(-3a) a^p da = p!/3^{p+1}$ we get
$$ \int_0^\infty  \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^3 da= 
\sum_{x=0}^{n-1}\sum_{y=0}^{n-1}\sum_{z=0}^{n-1} \frac{1}{3^{x+y+z+1}} 
 \frac{(x+y+z)!}{x! \, y! \, z!} \tag4$$
and finally
$$E_{n}[T]  = 3 \int_0^\infty  \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^3 da \tag5$$
More in general, if we have $d$ urns:
$$E_{n,d}[T]  = d \int_0^\infty  \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^d da \tag6$$
This can be evaluated numerically, I don't know about asymptotics (asked here).
Empirically , it seems that  $E = 3 n - \beta \sqrt{n} +O(1)$ where $\beta \approx 2.5$

And here's a numerical recursive computation (in Java):
public class MSE3368225 {

    static Double[] cache = new Double[(1<<21)];

    static double ex(int x, int y, int z) {
        if (x == 0 || y == 0 || z == 0)
            return 0;
        if (x > 127 || y > 127 || z > 127) 
            throw new RuntimeException("Out of range");
        int k = (x << 14) | (y << 7) | z; // packs three variables in one integer
        Double d = cache[k];
        if (d == null) {
            d = 1 + (ex(x - 1, y, z) + ex(x, y - 1, z) + ex(x, y, z - 1)) / 3.0;
            cache[k] = d;
        }
        return d;
    }

    public static void main(String[] args) {
        System.out.println(ex(100, 100, 100));
    }
}

This solves the recursion 
$$g(x,y,z)=\begin{cases} 0 & \text {if $x=0$ or $y=0$ or $z=0$}\\
1+ \frac13\left(g(x-1,y,z)+g(x,y-1,z)+g(x,y,z-1)\right) & \text{elsewhere}
\end{cases}
$$
where $g(x,y,z)$ is the expected remaining time, when there remains $(x,y,z)$ balls for each urn.
The result is $E_{100}[T]=274.9186440$

Some values
  n     E
  2  2.888889 
  3  5.049383 
  4  7.348270 
  5  9.734204
 10  22.34468
 20  48.99126
 50  132.3676
100  274.9186

