The Riddler’s Inaugural Rock-Paper-Scissors Tournament

Riddler Express

Michelle walks toward her departure gate the wrong way on a moving walkway, at her usual walking pace. After advancing $100$ metres closer to the gate, she drops her boarding pass on the walkway. She does not notice and continues walking for another $90$ seconds. She then turns around and jogs in the direction of the walkway’s movement at twice her walking pace. She catches up with the boarding pass $10$ metres from the start of the walkway. How fast does the walkway move?

The Riddler, FiveThirtyEight, June 22, 2018(original post)

Solution

Let $w$ be the walkway speed (in metres per second), and $m$ Michelle’s walking speed. Her walking ground-speed against the walkway is $m - w$ (positive, since she does make progress); her jogging speed is $2m$. Jogging with the walkway, her ground-speed is $2m + w$.

Set the origin at the point where she drops the boarding pass at time $t = 0$. At that instant she is at position $0$ and continues walking against the walkway. After $90$ seconds, at $t = 90$, she has moved to position $+90(m - w)$ in the direction of the gate. The boarding pass, riding the walkway in the direction opposite the gate, has moved to position $-90 w$. The separation between her and the boarding pass is $$90(m - w) - (-90 w) \;=\; 90 m.$$ She turns and jogs after the boarding pass. The pass moves at $+w$ in her (new) direction of travel (toward the start of the walkway); she moves at $2m + w$ in that same direction; the relative speed is $(2m + w) - w = 2m$. She catches up in time $$\frac{90 m}{2 m} \;=\; 45 \text{ seconds.}$$ So she catches the pass at $t = 90 + 45 = 135$ seconds.

In those $135$ seconds the boarding pass has been carried by the walkway from the drop point to a final position $135 w$ from the drop point in the walkway-flow direction. The drop point itself was $100$ metres in from the start of the walkway. So at the moment of the catch, the boarding pass is at $100 - 135 w$ metres from the start, and the problem tells us this is $10$: $$100 - 135 w \;=\; 10 \;\;\Longrightarrow\;\; w \;=\; \frac{90}{135} \;=\; \frac{2}{3} \text{ m/s.}$$ Converting, $$\boxed{\;w \;=\; \tfrac{2}{3} \text{ m/s} \;\approx\; 1.49 \text{ mph.}\;}$$ (Notice Michelle’s walking speed $m$ cancels: the answer depends only on the geometry and the walkway speed.)

The computation

Encode the problem as a small symbolic system, solve for $w$, and confirm the catch-up time by integrating forward from the drop event.

from sympy import symbols, Eq, solve, Rational

w, m, t_chase = symbols('w m t_chase', positive=True)
# After 90 s of walking past the drop point:
# Michelle's position relative to drop = 90 (m - w)
# Boarding pass position relative to drop = -90 w
# Separation she must close = 90 m
sep = 90 * m
# Chase speed relative to pass = (2m + w) - w = 2m
chase_time = sep / (2 * m)
print("chase time (s) =", chase_time)  # 45

# Boarding pass position relative to walkway start when caught:
# pass moved at speed w for (90 + 45) s, starting from 100 m in
# So position = 100 - (90 + 45) * w
pos_eq = Eq(100 - 135 * w, 10)
print("walkway speed (m/s) =", solve(pos_eq, w))
print("                mph =", float(Rational(2, 3) * 3600 / 1609.344))

The script prints chase time = 45, walkway speed = [2/3], and the mph conversion $\approx 1.49$.

Note on the Riddler Classic

The Classic invited readers to submit strategies for an empirical rock-paper-scissors tournament: each submission gave a first-throw distribution over $\{R, P, S\}$ and a conditional distribution for the second throw given the opponent’s first throw, and submissions were paired off in a round-robin. The winning strategy (Kevin Qiao’s pure scissors-on-throw-one) emerged from the actual distribution of $1\,648$ valid submissions; there is no derivable optimum without that empirical population. This is the same category as the Battle for Riddler Nation rounds, the Caffeine King contest, the lowest-unique-integer round, and the Riddler-Nation War deck-design contest: no closed-form Nash equilibrium that survives contact with a specific population of reader strategies. We omit a worked solution.