Will Riddler Nation Win Gold In Archery?
A FiveThirtyEight Riddler puzzle.
By Vamshi Jandhyala in mathematics Riddler
July 30, 2021
Riddler Express
R?
Solution
Let $p_w$ be the probability of Riddler Nation winning. Riddler Nation either wins the first time with a probability of $p_{wf}$ or Riddler Nation draws with a probability of $p_d$ and then wins with a probability $p_w$. Therefore we have,
$$
\begin{aligned}
p_w &= p_{wf} + p_d \cdot p_w \\
\implies p_w &= \frac{p_{wf}}{1 - p_d} \end{aligned} $$
Calculating probabilities
There are a couple of ways by which $p_{wf}$ and $p_d$ can be calculated.
One approach is to use brute-force enumeration of all the $3^3$ $3$-tuples of the points earned in each of the $3$ shots and count the number of tuples whose sum is greater or equal to $24$. The Python code for counting the tuples is given below:
from itertools import product sum([1 for p in product(*[[10,9,5]]*3) if sum(p) > 24])
The other approach is to use $\textbf{generating functions}$ i.e the coefficient of $x^{24}$ in $(x^{10} + x^9 + x^5)^3$ gives the count of the number of cases where the sum of the scores of the three shots is $24$.
Computational Solution
From the simulation below, we see that the $\textbf{Riddler Nation}$ is favoured to win and the probability of win is $\approx \textbf{0.524}$.
from random import random def shot_pts_riddler(): p = random() if p < 0.33333333: return 10 elif p > 0.33333333 and p < 0.66666666: return 9 else: return 5 def shot_pts_conundrum(): return 8 def prob_win_riddler(runs=1000000): total_wins = 0 for _ in range(runs): pts_riddler, pts_conundrum = 0, 0 while pts_riddler == pts_conundrum: pts_riddler = sum([shot_pts_riddler() for _ in range(3)]) pts_conundrum = sum([shot_pts_conundrum() for _ in range(3)]) if pts_riddler > pts_conundrum: total_wins += 1 return total_wins/runs print(prob_win_riddler())
Riddler Classic?
Solution
The lengths of the links in the chain for a geometric progression $1, f, f^2,\dots$.
Using the properties of $\textbf{complex numbers}$, the position $p_i$ for $i \geq 2$ of the end point of each link in the chain can be represented as a complex function of $\theta$ (the angle between each chain link measured anticlockwise). We have,
$$
\begin{aligned}
p_1 &= 1 \\
p_2(\theta) &= 1 + fe^{i\theta} \\
\vdots\\
p_n(\theta) &= 1 + fe^{i \theta} + \dots + f^{n-1}e^{(n-1)i\theta} = \sum_{k=0}^{n-1} f^k e^{i k \theta}\\
p_{\infty}(\theta) &= \frac{1}{1 - fe^{i\theta}} = \frac{1}{1-f\cos\theta - if\sin\theta} \end{aligned} $$
If $x$ and $y$ are the coordinates of the end point of the last link in the chain, we have
$$
\begin{aligned}
x = \frac{1 - f\cos\theta}{1+ f^2 - 2f\cos\theta} \\
y = \frac{f\sin\theta}{1+ f^2 - 2f\cos\theta} \end{aligned} $$
Eliminating $\theta$ from the above using brute force, we get
$$ \left(x - \frac{1}{1-f^2}\right)^2 + y^2 = \left(\frac{f}{1-f^2} \right)^2 $$
which is the equation of a circle centered at $\left(\frac{1}{1-f^2},0\right)$ and radius $\frac{f}{1 - f^2}$.
Using Möbius transformation
The mapping $\frac{1}{1 - fe^{i\theta}}$ is a Möbius Transformation of $e^{i\theta}$ a complex number on the unit circle centered at the origin.
Here are a couple of fundamental results related to Möbius Transformations that we will use:
- Möbius transformations map circles to circles
- If two points symmetric with respect to a circle, then their images under a Möbius transformation are symmetric to the image circle. This is called the “Symmetry Principle”.
Two end points of a diameter $(-1,0)$ and $(0,1)$ of the unit circle get mapped to the two end points $\frac{1}{1+f}$ and $\frac{1}{1-f}$ of a diameter of the mapped circle by the symmetry principle. Therefore the centre of the mapped circle is at
$$ \frac{1}{2}\left(\frac{1}{1-f} + \frac{1}{1+f}\right) = \frac{1}{1-f^2} $$
The radius of the circle is given by
$$ \frac{1}{2} \bigg \lvert \left(\frac{1}{1-f} - \frac{1}{1+f}\right)\bigg \rvert = \frac{f}{1-f^2} $$