Will the goldfish hit the waves?

Let $r$ be the radius of the outermost wave radius, $R$ be the radius of the whole water surface, $t$ be the time under $T$ that is the total time needed for outermost wave to reach the full circle, as in the question.

The probability of the goldfish hitting the waves at any time $t$ is $P(t) = \dfrac{\pi r^2}{\pi R^2} = \dfrac{r^2}{R^2}$ .

To find the probability for the goldfish hitting the waves over the time lapse $T$ , we can calculate as:

$\dfrac{1}{T} \int_0^T P(t) \ dt$

Now since the waves spread in circles at a constant rate, that means at $\dfrac{\mathrm{d}r}{\mathrm{d}t}= k$ for some constant $k$ . In other words, $r(t) = kt$ for some linear progression since we know $r = 0$ at $t = 0$ . Also, we know that $r(t = T) = R = kT$ because the whole radius is reached at $t = T$ . We can conclude $k = \dfrac{R}{T}$ . Hence, $r = \dfrac{Rt}{T}$ , or $\dfrac{r}{R}=\dfrac{t}{T}$ .

To visualize it, we can set up the stacks of radii as time passes to $T$ as similar triangles, as shown below:

Thus, the desired probability equals: $\dfrac{1}{T} \int_0^T P(t) \ dt = \dfrac{1}{T} \int_0^T \dfrac{r^2}{R^2}\ dt$

Substituting $\dfrac{r}{R}=\dfrac{t}{T}$ , we will get:

$\begin{aligned} \dfrac{1}{T} \int_0^T \dfrac{r^2}{R^2}\ dt &= \dfrac{1}{T} \int_0^T \dfrac{t^2}{T^2}\ dt \\ &= \dfrac{1}{T^3} \int_0^T t^2\ dt \\ &= \dfrac{1}{T^3}\left[\left.\dfrac{t^3}{3} \right|_0^T\right] \\ &= \dfrac{1}{T^3} \big[ \dfrac{T^3}{3}- 0\big] \\ &= \dfrac{1}{3} \end{aligned}$

Thus, the probability equals $\boxed{\dfrac{1}{3}}$ . Alternatively, we can also visualize the circle stacks of "time" at each instant and integrate them all, as followed:

Clearly, the integrated areas will become volumes of a cone and a cylinder of the same radius $R$ and "height" $T$ , and by the conic volume formula, the ratio is simply $\dfrac{1}{3}$ .