## Wednesday, June 22, 2011

### Ngon Dartboard Probability Limit

Continuing from our last post on simulations where we compared our solution analytically and from simulations to answer the following question:
Given an integer n>2, what is the probability that a randomly-thrown dart will land closer to the center of the dartboard than to an edge of a regular polygon with n sides? What is the probability as n approaches infinity?

This post will be focused on the limit as n approaches infinity.

We evaluated our formula:
\begin{align*} P \left ( d_{c}\lt d_{e} \right) &= \frac{ tan^3 \frac{\pi}{2n} + 3 tan \frac{\pi}{2n} }{ 6 tan \frac{\pi}{n} } \\ \end{align*}
...for several different values of n.

Let's also try to evaluate using the built-in definite integral function in the TI calculator. Doing this way, we should be able to bypass all steps to get to the end by just setting up the question.
\begin{align*} P \left ( d_{c}\lt d_{e} \right) &= \frac{A_{shaded} }{A_{triangle} } \\ &= \frac { \frac{1}{2} \int_{0}^{\frac{\pi}{n}} \left ( \frac{ 1 }{1 + cos \theta} \right ) ^{2} d\theta }{ \frac{1}{2} tan \frac{\pi}{n} } \\ &= \frac { \int_{0}^{\frac{\pi}{n}} \left ( \frac{ 1 }{1 + cos \theta} \right ) ^{2} d\theta }{ tan \frac{\pi}{n} } \end{align*}

To do it quickly on a TI calculator we'll be using the TABLE feature. We'll enter the equations into Y1 and Y2 and we'll go TABLE SETUP (TBLSET) and change Indpnt: to Ask and enter the values of n:
Y1=(tan(π/(2X))^3+3tan(π/(2X)))/(6tan(π/X))
For the definite integral we have:
Y2=fnInt(1/(1+cos(A))2,A,0,π/X)/tan(π/X)
Using a TI-83 Plus we have:
n Probability
Y1 Y2
3 .185185185185.185185185185
4 .218951416497.218951416497
5 .231475730333.231475730333
6 .237604307034.237604307034
7 .241091327922.241091327922
8 .243275188464.243275188464
9 .244737577398.244737577398
10 .245766620771.245766620771
100 .249958864808.249958864808
1,000 .249999588765.249999588765
10,000 .249999995888.249999995888
100,000 .249999999959.249999999959
1,000,000 .25.25
109 .25.25
1025 .25.25
1050 .25.25
1075 .25.25
1098 .25.25
1099 .250
That's not a typo in the last cell of the table, but it's an issue for some other post. It would seem that the limit as n approaches infinity of our probability is .25.

It matches nicely the graph of prob vs n created by Andy Rundquist:
The graph of the function is asymptotic to 0.25 as n increases. Besides looking at the graph, we can also evaluate:
\begin{align*} L &= \lim_{x \rightarrow \infty } \frac{ tan^3 \frac{\pi}{2x} + 3 tan \frac{\pi}{2x} }{ 6 tan \frac{\pi}{x} } \\ &= \frac{ 0 + 0 }{ 0 } \\ &= \frac{ 0 }{ 0 } \\ \end{align*}
But we encounter an indeterminate form, so we'll apply l'Hôpital's rule:
\begin{align*} \lim_{x \rightarrow \infty } \frac{ f(x) }{ g(x) } &= \lim_{x \rightarrow \infty } \frac{ f'(x) }{ g'(x) } \\ \end{align*}
Applying it, we obtain the following:
\begin{align*} L &= \lim_{x \rightarrow \infty } \frac{ tan^3 \frac{\pi}{2x} + 3 tan \frac{\pi}{2x} }{ 6 tan \frac{\pi}{x} } \\ &= \lim_{x \rightarrow \infty } \frac{ \frac{d}{dx} \left [ tan^3 \frac{\pi}{2x} + 3 tan \frac{\pi}{2x} \right ] }{ \frac{d}{dx} \left [ 6 tan \frac{\pi}{x} \right ] } \\ &= \lim_{x \rightarrow \infty } \frac{ 3 tan^2 \frac{\pi}{2x} sec^2 \frac{\pi}{2x} \left ( - \frac{\pi}{2x^2} \right ) + 3 sec^2 \frac{\pi}{2x}\left ( - \frac{\pi}{2x^2} \right ) }{ 6 sec^2 \frac{\pi}{x} \left ( - \frac{\pi}{x^2} \right )} \\ &= \lim_{x \rightarrow \infty } \frac{ \left ( - \frac{\pi}{x^2} \right ) \left( \frac{3}{2} tan^2 \frac{\pi}{2x} sec^2 \frac{\pi}{2x} + \frac{3}{2} sec^2 \frac{\pi}{2x} \right ) }{\left ( - \frac{\pi}{x^2} \right ) 6 sec^2 \frac{\pi}{x} } \\ &= \lim_{x \rightarrow \infty } \frac{ \frac{3}{2} tan^2 \frac{\pi}{2x} sec^2 \frac{\pi}{2x} + \frac{3}{2} sec^2 \frac{\pi}{2x} }{ 6 sec^2 \frac{\pi}{x} } \\ &= \frac{ \frac{3}{2} tan^2 (0) sec^2 (0) + \frac{3}{2} sec^2 (0) }{ 6 sec^2 (0) } \\ &= \frac{ \frac{3}{2} (0)^2 (1)^2 + \frac{3}{2} (1)^2 }{ 6 (1)^2 } \\ &= \frac{ 0 + \frac{3}{2} }{ 6 } \\ &= \frac{ 1 }{ 4 } = .25 \\ \end{align*}
Looks like we get the same answer.

Intuitively we know the answer should be one-fourth as n (the number of sides) increases. Because as n grows without bound, the polygon gets closer and closer to a circle. If you want a copy of the applet, click on File then Save As. Be sure to move some of the controls and check and uncheck the checkboxes.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Given that the distance from the focus (center of polygon) to the sides of the polygon (directrix) is twice that of the focus to the parabola (edge of shaded region). We have R=2r. So using area of a circle, we have:
\begin{align*} P \left ( d_{c}\lt d_{e} \right) &= \frac{A_{shaded} }{A_{polygon} } \\ &= \frac { \pi r^2 }{ \pi R^2 } \\ &= \frac { \pi r^2 }{ \pi (2r)^2} \\ &= \frac { \pi r^2 }{ 4 \pi r^2}\\ &= \frac { 1 }{ 4} \\ \end{align*}
Turns out, the case that is easiest to solve is when n approaches infinity where the polygon gets closer to a circle.

Square Dartboard Probability
Ngon Dartboard Probability
Ngon Dartboard Probability Simulation
Ngon Dartboard Probability Limit