Can You Feed The Hot Hand?

The Riddler for October 23, 2020. The Express exposes the famous “hot hand” selection bias in a short sequence of shots, and the Classic balances a one-on-one basketball game with rebounds and steals.

Riddler Express

Sue Bird makes every shot with probability exactly $\tfrac12$, independent of history. In each trial she takes three shots. Stewart keeps only the trials containing at least one shot that was immediately preceded by a made shot, picks one such trial at random, then picks at random one of its shots that was immediately preceded by a made shot. What is the probability that the chosen shot was made?

The Riddler, FiveThirtyEight, October 23, 2020(original post)

Solution

It is not $\tfrac12$. The eight equally likely sequences (X = made, O = missed) are OOO, OOX, OXO, XOO, OXX, XOX, XXO, XXX. The two without any shot following a made shot (OOO, OOX) are discarded, leaving six, each chosen with probability $\tfrac16$. Within a trial, look at the shots immediately preceded by a made shot and record the fraction of them that were made: $$\begin{aligned} \text{OXO} &: \{O\} \to 0, &\quad \text{XOO} &: \{O\} \to 0, &\quad \text{OXX} &: \{X\} \to 1,\\ \text{XOX} &: \{O\} \to 0, &\quad \text{XXO} &: \{X, O\} \to \tfrac12, &\quad \text{XXX} &: \{X, X\} \to 1. \end{aligned}$$ Averaging these per-trial fractions (Stewart picks a trial first, then a shot within it), $$\frac{1}{6}\left(0 + 0 + 1 + 0 + \tfrac12 + 1\right) = \frac{5/2}{6} = \boxed{\frac{5}{12}} \approx 0.417.$$ The selection rule is biased toward misses: conditioning on “the previous shot was made” and then sampling shots makes a made shot look less than $50\%$ likely. This is the Miller–Sanjurjo bias behind the hot-hand fallacy.

The computation

Encode the procedure: enumerate the eight sequences, and for each keep the shots preceded by a make, recording the made-fraction; average over the qualifying trials.

import itertools
from fractions import Fraction
fractions = []
for seq in itertools.product('XO', repeat=3):
    after_make = [i for i in range(1, 3) if seq[i - 1] == 'X']
    if not after_make:
        continue                                  # no qualifying shot: discard trial
    made = sum(1 for i in after_make if seq[i] == 'X')
    fractions.append(Fraction(made, len(after_make)))
print(sum(fractions) / len(fractions))            # 5/12

Averaging the per-trial made-fractions gives $5/12$, below one half.

Riddler Classic

LeBron and Davis play sudden-death one-on-one; a coin decides first possession, and the first basket wins. Each makes a shot with probability $\tfrac12$. Davis rebounds every miss (his or LeBron’s) and resets to the three-point line. Before each of Davis’s shots, LeBron steals with probability $p$, regaining possession. For what $p$ is the game even?

The Riddler, FiveThirtyEight, October 23, 2020(original post)

Solution

Let $J$ be LeBron’s win probability when he has the ball and $D$ his win probability when Davis has it. When LeBron has possession he either scores ($\tfrac12$) or misses ($\tfrac12$); a miss is rebounded by Davis, so $$J = \tfrac12 + \tfrac12 D.$$ When Davis has possession, LeBron either steals (probability $p$, moving to state $J$) or fails to steal ($1 - p$), after which Davis must miss ($\tfrac12$) for LeBron to get another chance (back to state $D$): $$D = pJ + (1 - p)\tfrac12 D.$$ The coin makes LeBron’s overall win probability $\tfrac12 J + \tfrac12 D$, and fairness requires it to equal $\tfrac12$, i.e. $J + D = 1$. From $J = \tfrac12 + \tfrac12 D$ and $J + D = 1$ we get $D = \tfrac13$, $J = \tfrac23$. Substituting into the middle equation, $$\tfrac13 = \tfrac23 p + (1 - p)\tfrac16 \;\Longrightarrow\; 2 = 4p + (1 - p) \;\Longrightarrow\; p = \boxed{\tfrac13}.$$ LeBron needs to steal a third of the time to offset Davis’s monopoly on rebounds.

The computation

Encode the three equations ($J$, $D$, and the fairness condition $J + D = 1$) and solve for $p$; cross-check with a direct simulation.

import sympy as sp
J, D, p = sp.symbols('J D p')
sol = sp.solve([J - (sp.Rational(1, 2) + D / 2),
                D - (p * J + (1 - p) * D / 2),
                J + D - 1], [J, D, p], dict=True)[0]
print(sol[p], sol[J], sol[D])                # 1/3 2/3 1/3

import random
def lebron_wins(pp, rng):
    davis_ball = rng.random() < 0.5          # who starts
    while True:
        if not davis_ball:                   # LeBron shoots
            if rng.random() < 0.5: return True
            davis_ball = True                # miss -> Davis rebounds
        else:                                # Davis has the ball
            if rng.random() < pp:            # LeBron steals
                davis_ball = False; continue
            if rng.random() < 0.5: return False   # Davis scores
            # Davis misses -> he rebounds, still his ball
rng = random.Random(0); n = 400_000
print(round(sum(lebron_wins(1/3, rng) for _ in range(n)) / n, 3))   # ~0.5

Solving gives $p = \tfrac13$, and the simulation confirms a coin-flip game at that steal rate.