Probablility of a dart landing closer to the center than the edge of a square dartboard?

The question is: There is a square dart board with a side length of $2$m, and a dart has equal probability to land anywhere on the board. What is the probability that the dart will land closer to the center than to the edges? So I know that if I center the square at the origin, I have to find $f(x)$ such that every point on $f(x)$ is equidistant between the origin and the line $x=1$ or $y=1$ (I'm only working with the first quadrant since by symmetry the probability should be the same). Another reasonable assumption that I made is that $f(x)$ is reflected over $y=x$, so I only have to work with the first $45$ degrees of the first quadrant. So to find $f(x)$ I did the following: $\sqrt{x^2+[f(x)]^2}=1-x$, so $f(x)=\sqrt{1-2x}$. Now I know that I just find the area under this curve and divide it by $1/8$ the area of the square to find the probability. I just wanted to know if my $f(x)$ was okay? Thanks!

• Can you think of an easier way to do this? What would you say if I asked you what the probability is for the dart to land in the upper half of the square? – John Douma Apr 17 '13 at 2:55
• Nice use of the symmetries! The procedure is right. – André Nicolas Apr 17 '13 at 2:56
• Thanks Andre! And @user 69810 of course the probability would be 1/2 for the upper part of the square but the probability is not 1/2 here. Can you elaborate? – Ovi Apr 17 '13 at 3:09
• @Ovi Yes. I wasn't thinking. When I read the problem I was thinking simple shape inside simple shape until I sat down and started calculating the area. – John Douma Apr 17 '13 at 3:16
• O ok thanks for getting back to me haha I was being bothered that there might just be a very simple solution to this somehow related to 1/2. – Ovi Apr 17 '13 at 19:31

Your thought process looks good, but the integration is actually a bit tricky. Here is your dartboard: We'd like to calculate the area of the center shape, and as you say, we can split this into eight slices of equal area. Here we see the two slices in the first quadrant: Let's compute the area of the slice from $45^{\circ}$ to $90^{\circ}$:

$$\int_{0}^{\sqrt{2}-1}(\frac{1-x^2}{2}-x)dx=\frac{1}{6}(4\sqrt{2}-5)$$

Multiplying this by $8$ gives us the area of the center shape. Then, we divide by $4$, the area of the dartboard. The final probability is then:

$$\frac{1}{3}(4\sqrt{2}-5)\approx.2190$$

Judging by our picture above, this seems right. The center shape is about $1/5$ the area of the dartboard.

• I don't really understand your definite integration. The area in the first quadrant would be the definite integral of $xdx$ from 0 to $sqrt(2)-1$+the definite integral of $(1-x^2)/2dx$ from $sqrt(2)-1$ to $1$. Then I would just multiply this by 8. And even more, I don't see where the $pi/4$ and $pi/2$ come in. – Ovi May 16 '13 at 0:24
• @Ovi: I've computed the area of the slice from $45^{\circ}$ to $90^{\circ}$, because I thought the integration was easier. If you want to do the slice from $0^{\circ}$ to $45^{\circ}$, you have to split it up, as you have. Everything you've said is fine, except your second integral should be $\sqrt{1-2x}$ from $\sqrt{2}-1$ to $1$. – Jared May 16 '13 at 0:40
• Oh ok I see. Very nice! – Ovi May 16 '13 at 12:00
• If you do the integration in polar coordinates, then the solution easily generalizes to any regular polygon. See math.stackexchange.com/questions/1688936/… – Dominic108 Mar 15 '16 at 0:48
• Another answer to the same question that also generalizes to a regular $n$-gon: math.stackexchange.com/a/1706984. – joriki Mar 21 '16 at 21:53