Mind Boggling Bullseye Part 2

Consider the area enclosed by the largest white 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 white circle) is $\frac{3}{4}$ the radius of the entire dartboard, it is $\left(\frac{3}{4}\right)^2=\frac{9}{16}$ the area of the entire dartboard. We can conclude that the total white area is $\frac{9}{16}$ as large as the total black area. And so, the probability that the dart strikes white is $\frac{9}{16}$ the probability that the dart strikes black.

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

$\begin{aligned} p+\frac{9}{16}p &= 1 \\ p &= \frac{16}{25} \end{aligned}$

Thus, the probability that the dart strikes black is $\frac{16}{25}=\boxed{0.64}.$

Let the Radius of the white circle annulus at the ith stage be Ri. The radius of the adjoining black circle/annulus would be 3 Ri/4. The area of the ith stage annulus = pi [Ri^2 - (3*Ri/4)^2]

= pi [Ri^2-9Ri^2/16] =pi 7Ri^2/16 So, the sum of the area of the white annuli = pi*7/16[R1^2 + R2^2 +..........Ri^2 + .........] Since what we need is the relative area, the radius of the outer circle can be assumed to be 1 without loss of generality. Also, the factor of pi would drop off.

The radius of the first circle would be 3/4. The radius of the second white circle would be (3/4) * (9/16) (essentially 3/4 * 3/4 because of the alternating black circle):. The radius of the third white circle would be equal to (9/16) * (9/16) * (3/4) and so on....

So, the areas would be proportional to (7/16) [(3/4)^2 + ((9/16) (3/4)^2 + ([9/16]^2*3/4)^2 +......]

This is an infinite geometric progression with starting term (3/4)^2 and common ratio (9/16)^2. Hence, the sum of the infinite sequence of white annuli = (7/16) * [(3/4)^2 / ([1-(9/16)^2]] = (7/16) * (9/16)/(1 - (81/256)) = (7/16) * [(9/16)/ (175/256)] = 7 / 16 * [144/175] = 63/175 So, the area of the black annuli would be 1-(63/175) = 112/175 = 0.64 which is the req probability

The whole black area is:

$B=\pi R^{2}-\pi\left[\frac{3}{4}R\right]^{2}+\pi\left[\left(\frac{3}{4}\right)^{2}R\right]^{2}-\pi\left[\left(\frac{3}{4}\right)^{3}R\right]^{2}+ ...$

Extracting the common factor $\pi R^{2}$ we would have:

$B=\pi R^{2}\left[1-\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{4}-\left(\frac{3}{4}\right)^{6}...\right]$

The whole dartboard area is:

$C=\pi R^{2}$

The probability to hit on the black area is $p=\frac{B}{C}$

So: $p=1-\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{4}-\left(\frac{3}{4}\right)^{6}...$

Which is a geometric progression with common ratio $r=-\left(\frac{3}{4}\right)^{2}$

So $p = \frac{1}{1-r}=\frac{1}{1+\left(\frac{3}{4}\right)^{2}}=0.64$

Consider the outer black ring and the outer white ring. The ratio of their areas is $16:9$ . Thus, if we condition our probability on the dart landing in the union of these two rings, the probability of hitting black is $16/25 = 0.64$ .

Let $B_i$ be the ith black ring and $W_i$ be the ith white ring, with $U_i$ their union. Then the same argument for $U_1$ repeated for all i will show a conditional probability $P[B_i|U_i] =0.64$ .

Let $p_i$ be the probability of landing in $U_i$ The total probability of landing in the black is $\sum_i P[U_i] * P[B_i|U_i] = \sum_i (p_i 0.64) = 0.64 \sum_i p_i = 0.64$ .

The area within each circle is 9/16 the area within the preceding larger one. To hit black, you are working out the area of large black - large white + 2nd largest black - 2nd largest white + ... (note the alternating signs). This is the series 1 - 9/16 + (9/16)^2 - (9/16)^3 + ..., a geometric series with common ratio 9/16.

Put this into 1/(1-r) and you get the answer 16/25 = 0.64

Strategy for solving this could be to divide a black area by the whole area. Notice that area of black area corresponds to following geometric progression:

$R^{2}\pi-(\frac{3}{4})^{2}R^{2}\pi+(\frac{3}{4})^{4}R^{2}-...=\frac{R^{2}\pi}{1+\frac{9}{16}}=\frac{16R^{2}\pi}{25}$

Now, the probability is $\frac{\tfrac{16R^{2}\pi}{25}}{R^{2}\pi}=\frac{16}{25}=0.64$

Holy cow!!! First time I solved an infinite type of question!!! It's so exciting!!! And I leveled up in Discrete Maths!!! XD