How hard is it to sink 5 balls off the break in 9-ball? (with the math)

Short answer: it depends a lot on how likely each individual ball is to go in on the break. If you model each of the 9 object balls as having the same independent pocket probability $p$ on the break, the probability of pocketing at least 5 balls is

P(≥5)=∑k=59(9k)pk(1−p)9−k.P(\text{≥5})=\sum_{k=5}^{9} \binom{9}{k} p^k(1-p)^{9-k}.P(≥5)=k=5∑9​(k9​)pk(1−p)9−k.

The expected number of balls pocketed on the break is simply $E[\text{balls}]=9p$.

Worked example (step-by-step for $p=0.20$)

Assume each ball has probability $p=0.20$ of going in on the break (this is a simple model — real breaks have correlations and geometry effects).

Compute each term $P(k)$ for $k=5,\dots ,9$:

• P(5)=(95)(0.2)5(0.8)4=126⋅0.00032⋅0.4096≈0.016515072P(5)=\binom{9}{5}(0.2)^5(0.8)^4 = 126 \cdot 0.00032 \cdot 0.4096 \approx 0.016515072P(5)=(59​)(0.2)5(0.8)4=126⋅0.00032⋅0.4096≈0.016515072

• P(6)=(96)(0.2)6(0.8)3=84⋅0.000064⋅0.512≈0.002752512P(6)=\binom{9}{6}(0.2)^6(0.8)^3 = 84 \cdot 0.000064 \cdot 0.512 \approx 0.002752512P(6)=(69​)(0.2)6(0.8)3=84⋅0.000064⋅0.512≈0.002752512

• P(7)=(97)(0.2)7(0.8)2=36⋅0.0000128⋅0.64≈0.000294912P(7)=\binom{9}{7}(0.2)^7(0.8)^2 = 36 \cdot 0.0000128 \cdot 0.64 \approx 0.000294912P(7)=(79​)(0.2)7(0.8)2=36⋅0.0000128⋅0.64≈0.000294912

• P(8)=(98)(0.2)8(0.8)1=9⋅2.56⋅10−6⋅0.8≈0.000018432P(8)=\binom{9}{8}(0.2)^8(0.8)^1 = 9 \cdot 2.56\cdot10^{-6} \cdot 0.8 \approx 0.000018432P(8)=(89​)(0.2)8(0.8)1=9⋅2.56⋅10−6⋅0.8≈0.000018432

• P(9)=(99)(0.2)9(0.8)0=1⋅5.12⋅10−7≈0.000000512P(9)=\binom{9}{9}(0.2)^9(0.8)^0 = 1 \cdot 5.12\cdot10^{-7} \approx 0.000000512P(9)=(99​)(0.2)9(0.8)0=1⋅5.12⋅10−7≈0.000000512

Sum: $P(\text{≥5})\approx 0.016515072+0.002752512+0.000294912+0.000018432+0.000000512\approx 0.01958144$.

So with $p=0.20$, chance to sink at least 5 balls on the break ≈ 1.96%. Expected balls = $9\times 0.2=1.8$.

Quick reference: $P(\text{≥5})$ for some representative $p$ values

(These come from the same binomial sum ∑k=59(9k)pk(1−p)9−k\sum_{k=5}^{9}\binom{9}{k}p^k(1-p)^{9-k}∑k=59​(k9​)pk(1−p)9−k.)

• $p=0.05$ → $P(\ge 5)\approx 0.0000332$ (0.0033%)

• $p=0.10$ → $P(\ge 5)\approx 0.000891$ (0.089%)

• $p=0.15$ → $P(\ge 5)\approx 0.00563$ (0.56%)

• $p=0.20$ → $P(\ge 5)\approx 0.01958$ (1.96%)

• $p=0.25$ → $P(\ge 5)\approx 0.04893$ (4.89%)

• $p=0.30$ → $P(\ge 5)\approx 0.09881$ (9.88%)

• $p=0.40$ → $P(\ge 5)\approx 0.26657$ (26.66%)

• $p=0.50$ → $P(\ge 5)=0.5$ (50%)

Interpretation: if each ball is only a 10–20% chance to drop individually on a typical break, sinking 5 or more is quite rare (sub-2–5% range). If you’re an exceptional breaker (per-ball pocket odds closer to 0.3–0.4), the chance climbs to the tens of percent.

Important caveats (real pool vs this model)

• The independence assumption (each ball independently pockets with probability $p$) is a big simplification. In reality:

  • Pocketing one ball changes cluster dynamics and the chance others follow (positive correlation).

  • Rack alignment, cue speed, cue-ball position, table pockets, and ball deflections make the process highly non-independent and geometry-driven.

  • Some skilled players intentionally aim patterns that increase the chance of multiple follow-in shots.

• The binomial model is a useful first approximation and helps see how sensitive the outcome is to per-ball pocket probability.

I did it in real life its possible and like always you wasted moments of your life before you got to the reality of the human being that did it and don’t I deserve a lottery check because of it?