Mind-Boggling Bullseye

Consider the area enclosed by the largest tan circle. If you scaled up this area to fill the dartboard, then this new area would look exactly like the old dartboard with the colors reversed.

Now consider that since this area (enclosed by the largest tan circle) is $\frac{1}{2}$ the radius of the entire dartboard, it is $\frac{1}{4}$ the area of the entire dartboard. Thus, we can conclude that the total tan area is $\frac{1}{4}$ as large as the total red area. And so, the probability that the dart strikes tan is $\frac{1}{4}$ the probability that the dart strikes red.

Let $p$ be the probability that the dart strikes red. Then,

$\begin{aligned} p+\frac{1}{4}p &= 1 \\ p &= \frac{4}{5} \end{aligned}$

Thus, the probability that the dart strikes red is $\frac{4}{5}=\boxed{0.8}.$

The radius of the dartboard is $r = 1+\frac 12 + \frac 14 + ... = \dfrac 1{1-\frac 12} = 2$ . Therefore, the area of the dartboard is $A = \pi r^2 = 4\pi$ .

The area of the red region is given by:

$\begin{aligned} A_{\color{#D61F06}red} & = \pi \left(2^2 - 1^2 + \frac 1{2^2} - \frac 1{4^2} + \frac 1{8^2} - \frac 1{16^2} + ... \right) \\ & = \pi \left(\left( 2^2 + \frac 1{2^2} + \frac 1{8^2} + ... \right) - \left(1^2 + \frac 1{4^2} + \frac 1{16^2} +... \right)\right) \\ & = \pi \left(4\left( 1+ \frac 1{16} + \frac 1{16^2} + ... \right) - \left( 1+ \frac 1{16} + \frac 1{16^2} + ... \right) \right) \\ & = \frac {3\pi}{1-\frac 1{16}} = \frac {16\pi}5 \end{aligned}$

Since the probability of a dart strike is uniformly distributed, the probability of striking a colored region is proportional to its area. Therefore the probability of striking the red region is:

$\begin{aligned} P_{\color{#D61F06}red} & = \frac {A_{\color{#D61F06}red}}{A} = \frac {16\pi}5 \times \frac 1{4\pi} = \boxed{0.8} \end{aligned}$