Will Riddler Nation Win Gold In Archery?

Riddler Express

Each Riddler archer’s arrow scores $10$, $9$ or $5$, each with probability $\tfrac13$; each Conundrum archer scores exactly $8$. Three arrows per team per round, higher total wins, ties replay another three-arrow round until broken. Which team is favoured, and with what probability?

Solution

Conundrum scores $3 \times 8 = 24$ every round, dead certain. Riddler’s three arrows give a random total, so in one round Riddler wins, ties, or loses against the fixed $24$. Let $p_w$ be Riddler’s overall chance of winning. A round either settles it in Riddler’s favour (probability $p_{wf}$) or ties (probability $p_d$), and after a tie the situation resets, so $$p_w = p_{wf} + p_d\, p_w \;\Longrightarrow\; p_w = \frac{p_{wf}}{1 - p_d}.$$ Among the $3^3 = 27$ equally likely arrow triples, the totals beat $24$ in $11$ cases, equal $24$ in $6$ (the rearrangements of $10+9+5$), and fall short in $10$. So $p_{wf} = \tfrac{11}{27}$, $p_d = \tfrac{6}{27}$, and $$p_w = \frac{11/27}{1 - 6/27} = \frac{11}{21} \approx \boxed{0.524}.$$ Riddler Nation is favoured: its risky $10/9/5$ archers clear $24$ slightly more often than they miss it.

The computation

Enumerate all $27$ arrow triples, count wins and ties against $24$, and apply the replay formula exactly.

from fractions import Fraction as F
from itertools import product
totals = [sum(t) for t in product([10, 9, 5], repeat=3)]
win = sum(s > 24 for s in totals); draw = sum(s == 24 for s in totals)
pwf, pd = F(win, 27), F(draw, 27)
print(pwf / (1 - pd))                          # 11/21 = 0.5238...

Riddler Classic

A chain has links of lengths $1, f, f^2, \dots$ ($f < 1$), laid in a straight line with the longest link pinned. The chain bends so that every consecutive pair of links makes the same angle $\theta$. As $\theta$ ranges over all values, what shape does the tail (the far, infinitesimal end) trace?

Solution

Treat the plane as the complex numbers. The first link points along the real axis; each next link is shorter by a factor $f$ and turned by a further angle $\theta$, so the $k$-th link is $f^k e^{ik\theta}$. The tail of the infinite chain sits at the sum of all links: $$p_\infty(\theta) = \sum_{k=0}^{\infty} f^k e^{ik\theta} = \frac{1}{1 - f e^{i\theta}}.$$ As $\theta$ runs around the circle, $e^{i\theta}$ traces the unit circle, and $w \mapsto \tfrac{1}{1 - f w}$ is a Möbius transformation, which sends circles to circles. So the tail traces a circle. Its two extreme points come from $\theta = 0$ and $\theta = \pi$, namely $\tfrac{1}{1-f}$ and $\tfrac{1}{1+f}$, which are ends of a diameter; their midpoint and half-distance give $$\boxed{\text{a circle centred at } \Big(\tfrac{1}{1 - f^2},\,0\Big) \text{ with radius } \tfrac{f}{1 - f^2}.}$$

The computation

Sweep $\theta$, plot the tail position $1/(1 - f e^{i\theta})$, and check every point lies on the predicted circle (residual near zero).

import numpy as np
f = 0.5
th = np.linspace(0, 2 * np.pi, 100_000, endpoint=False)
z = 1 / (1 - f * np.exp(1j * th))
c, r = 1 / (1 - f**2), f / (1 - f**2)
residual = np.abs((z.real - c)**2 + z.imag**2 - r**2).max()
print(c, r, residual)                          # 1.333, 0.667, ~1e-15