How Rare Was Houston's 27-Miss Three-Point Streak? A secondary analysis

1. What Actually Happened (HSR vs HOU, Game 7, 2018)

On May 28, 2018, the Houston Rockets set an NBA record for futility: They missed 27 consecutive three-pointers in Game 7 of the Western Conference Finals against the Golden State Warriors.

Fans and analysts called it one of the most improbable cold streaks in playoff history, and headlines shouted odds like "1 in 72,000."

But how rare was it, really? Let's go beyond the headlines and do the math.

2. Where Did the "1 in 72,000" Headline Come From?

Multiple major outlets, mostly taking their info from FiveThirtyEight, cited the number "1 in 72,000" as the odds of this happening. But what does that really mean?

That number is closely based on the probability that a 36% three-point shooting team (Houston's average that season) would miss 27 threes in a row — if they took only 27 shots:

P = (1 - 0.36)²⁷ = 0.64²⁷ ≈ 0.000015 (about 1 in 66,000)

That's mind-blowingly rare. But Houston attempted 44 threes in that game. With more attempts, you get more chances for a streak like this to happen.

3. The Union Bound: "Just Multiply By The Number of Windows!"

Some people, including sharp Redditors at the time (check this reddit thread) pointed out that in 44 attempts, there are multiple places where a 27-miss streak could occur:

There are 18 windows (from shot 1 to 27, 2 to 28, …, 18 to 44) where you could miss 27 in a row.

If you multiply the probability by the number of windows (known as the union bound):

P ≈ 18 × 0.000015 = 0.00027 (about 1 in 3,600 games)

This is much more common than the headline number—but it massively overestimates the true probability.

Why the Union Bound Fails Here

The thing is , for you to miss exactly 27 consecutive shots (say shots 10-36), you actually only satisfy one window.

The union bound overcounts because longer streaks satisfy multiple windows simultaneously. For example:

If you miss 30 consecutive shots (shots 8-37), you satisfy windows 8, 9, 10, and 11 all at once. The union bound counts this as 4 separate events, when it's really one unlucky streak.

4. The Actual Probability (Python + Dynamic Programming)

The real question is:

What's the probability that a 36% shooter, over 44 threes, will suffer at least one streak of 27 consecutive misses, anywhere in the game?

To answer this exactly, you need to account for overlapping streaks and dependencies. This is where dynamic programming (DP) and simulation come in.

Monte Carlo Simulation Approach

We can simulate thousands of "games," and count how often a 27-miss streak shows up:

import random

def monte_carlo_streak(N=44, streak_len=27, p=0.36, trials=100_000):
    streak_found_count = 0
    for trial in range(trials):
        shots = [1 if random.random() > p else 0 for _ in range(N)]
        max_streak = 0
        current_streak = 0
        for shot in shots:
            if shot == 1:
                current_streak += 1
                max_streak = max(max_streak, current_streak)
            else:
                current_streak = 0
        if max_streak >= streak_len:
            streak_found_count += 1
    probability = streak_found_count / trials
    print(f"Monte Carlo estimated probability: {probability:.8f}")
    print(f"That's about 1 out of every {1/probability:.1f} games.") 
    return probability

    <- Expect ~1 out of 24,000

monte_carlo_streak()

Dynamic Programming Approach

The mathematically exact way is to use DP, which efficiently counts every possible scenario without double-counting overlaps:

import numpy as np

def dp_streak_exact(N=44, streak_len=27, p=0.36):
    q = 1 - p
    dp = np.zeros((N+1, streak_len+1))
    dp[0][0] = 1.0
    for i in range(1, N+1):
        for j in range(streak_len):
            dp[i][0]    += dp[i-1][j] * p
            dp[i][j+1]  += dp[i-1][j] * q
        dp[i][streak_len] += dp[i-1][streak_len]
    probability = dp[N][streak_len]
    print(f"DP exact probability: {probability:.8f}")
    print(f"That's about 1 out of every {1/probability:.1f} games.")
    return probability

     <- This will always be 1 out of 24024
dp_streak_exact()

What's the Answer?

The probability comes out to about:

1 in 24,000

Still historic, but much more likely than the "1 in 72,000" you might see in the headlines back in the day.

5. Why Are the Numbers So Different?

The "1 in 72,000" is for missing 27 straight with only 27 attempts.

The union bound ("1 in 3,600") overcounts by pretending each possible window is independent, when they actually overlap.

The real answer (1 in 24,000) is higher than the pure-streak case, but much less frequent than the union bound estimate.

This difference is all about overlapping windows—if you miss 28 or more straight, you're not just unlucky once, you fulfill the streak for several windows at the same time!!