Bull's-eye!

Relevant wiki: Expected Value

The probability P of hitting the bullseye is the ratio of the area of the whole board and the bullseye. So $P =\frac{12.7^2}{451^2}$ . We can reformulate the question to when is the expected value of hitting the board equal to 1. This leads to $P \times n =1 \Rightarrow n = \frac{1}{P} = \frac{451^2}{12.7^2} \approx 1261$

Let H be the event of hitting the bull's-eye, N be the event of no-hit.

The probability of H (i.e. bull's-eye at trial 1) is $\frac{\frac{\pi}{4}(12.7)^2}{\frac{\pi}{4}(451)^2}\approx7.9297\times 10^{-4}$ . [Note: The dimensions should be of the same units !]

Similarly, the probability of N is $1 - 7.9297\times 10^{-4}\approx0.9992$ .

Now, the probability of the sequence NH (i.e. bull's-eye at trial 2) is $(0.9992)(7.9297\times 10^{-4})$ ,

The probability of the sequence NNH (i.e. bull's-eye at trial 3) is $(0.9992)^2 (7.9297\times 10^{-4})$ ,

The probability of the sequence NNNH (i.e. bull's-eye at trial 4) is $(0.9992)^3 (7.9297\times 10^{-4})$ ,

... and so on.

We can show that this forms a probability distribution, since $\sum_{k=1}^\infty (0.9992)^{k - 1} (7.9297\times 10^{-4}) = 1$ Therefore, the expected value is $E(X) = \sum_{k=1}^\infty k(0.9992)^{k - 1} (7.9297\times 10^{-4}) =\frac{1}{7.9297\times 10^{-4}} \approx \boxed{1261}$