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))Using a TI-83 Plus we have:^{2},A,0,π/X)/tan(π/X)

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 |

10^{9} | .25 | .25 |

10^{25} | .25 | .25 |

10^{50} | .25 | .25 |

10^{75} | .25 | .25 |

10^{98} | .25 | .25 |

10^{99} | .25 | 0 |

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.

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

awesome! But don't stop! Let's go to 3D and do some more "If you teleport randomly into a room, what are the chances you'll be closer to the center than to a wall?"

ReplyDeleteI could be wrong, but this kind of question does not extend as nicely into 3D, at least not what we did here where we extended to an Ngon. In 2D, we can have regular polygons with an arbitrarily large number of sides and have symmetry that extends nicely into any number of sides. We can also have the number of sides approach infinity and solve for the case where we have a circle. In 3D, we may work out specific solution to any of the 5 platonic solids, but there isn't a solid where we can increase the number of regular sides to the point where you get a sphere.

ReplyDeleteI'll find time in the future to work on the platonic solid cases but, for now, I think I'll take a little break from this question (partly, I'm a little rusty with some of this math, it took me a long time before I found a trig identity that simplified in a way that made integration possible).