Sure, a bullseye is a bullseye, but what's the average?

Suppose each point on the dartboard is equally likely to be hit by a dart. On average, what score would you expect a single dart to earn?

Let $E = \mathbb{E} [ S ]$ be the average score of a single dart. Based on the setup of the Apollonian gasket, we see that there are infinitely many circles which are only contained within the unit circle. Let's say that we enumerate the radii as $r_n, n \in \mathbb{N},$ say with $r_0 = 1$, $r_1 = r_2 = \frac{1}{2},$ etc. One important point is that the smaller circles entirely contained within the unit eventually mutually exclusively cover the entirety of the unit circle, so however we enumerate them we must have $$\sum_{n=1}^\infty r_n^2 = 1.$$ Additionally, we see that the Apollonian gasket is symmetric about the $x$-axis and about the $y$-axis, so as we are filling things up we can focus on say just enumerating all of the cicles in say the first quadrant and then using symmetrical arguments to fill in the rest.

Additionally, since each circle with radius, say $r_n,$ is then filled with a scaled copy of the Apollonian gasket we see that $$\mathbb{E} [ S \mid r_n ] = \pi + r_n^2 E,$$ since in addition to the scaled copy of the expectation the only other circle that adds to the score is the unit circle. From a law of total expectation, since $$\mathbb{P} [ r_k ] = \frac{\pi r_n^2}{\pi} = r_n^2,$$ we have $$E = \sum_{n=1}^\infty r_n^2 \left( \pi + r_n^2 E \right) = \pi + E \sum_{n=1}^\infty r_n^4,$$ or equivalently $$E = \frac{\pi}{1 - \sum_{n=1}^\infty r_n^4}.$$

Actually, with all of that setup out of the way, let's speak in terms of curvatures $k_n = \frac{1}{r_n}, n \in \mathbb{N},$ where $k_0 = -1$ (since it is encompassing all of the other circles of the gasket). From Descartes' theorem -- or from Soddy's The Kiss Precise, whichever format speaks to your soul best -- we see that if you have three mutually tangent circles, with curvatures $k_\ell$, $k_m,$ $k_n$ then you can find two circles which are mututally tangent to the other three with curvatures $k_\pm,$ satisfying the equation $$(k_\ell + k_m + k_n + k_\pm)^2 = 2 \left( k_\ell^2 + k_m^2 + k_n^2 + k_\pm^2 \right).$$ In particular, we have $$k_\pm = k_\ell + k_m + k_n \pm 2 \sqrt{ k_\ell k_m + k_\ell k_n + k_m k_n }.$$ In particular, if we start with smaller curvatures (bigger radii) and work our way smaller and smaller, we can focus on just the solution $k_+$ since $k_-$ will have already been enumerated. In this way we can enumerate the first $N=84$ largest circles, once symmetry is taken into account, which correspond to the $20$ smallest curvatures. Note that the next largest circle has a curvature of $m_N = \min_{n \geq N+1} k_n = 62.$

Now we see that $$\theta_N = \sum_{n=1}^N r_n^2 \approx 0.955398049382331\dots$$ and $$\phi_N = \sum_{n=1}^N r_n^4 \approx 0.153282259694304\dots.$$ We additionally have $$\sum_{n=N+1}^\infty r_n^4 \leq \left(\max_{n \geq N+1} r_n^2 \right) \left( \sum_{n=N+1}^\infty r_n^2 \right) = \frac{1 - \theta_N}{m_N^2} \approx 0.0000116030048433062\dots.$$ Therefore, we have $$ 1.181031 \pi \approx \frac{\pi}{1-\phi_N} \leq E \leq \frac{\pi}{1 - \phi_N - \frac{1-\theta_N}{m_N^2}} \approx 1.181047 \pi.$$ Therefore, despite the fact that I could do a lot more enumeration and even get some good infinite sequences going based on the recursion formulae above, we see that the expected value of the score of one dart is approximately $$ E \approx 1.1810 \pi \approx 3.7103\dots.$$