In the September 16, 2026 Powerball drawing, nobody won the $271 million jackpot — again. That outcome is statistically unremarkable: with roughly 90 million tickets sold and 1-in-292-million odds per ticket, the probability of no winner on any given draw is approximately 73%. Here is exactly how that number is calculated.
The Powerball jackpot grows after every no-top-prize-winner Wednesday — and after every no-winner Monday and Saturday draw too — through a mechanism that is fully derivable from first principles.
What are the base odds of winning the Powerball jackpot?
Powerball requires matching five white balls (drawn without replacement from 69) plus the red Powerball (drawn from 26). The number of distinct equally-likely jackpot combinations is:
C(69,5) × 26 = 11,238,513 × 26 = 292,201,338
Every ticket carries exactly a 1-in-292,201,338 probability of hitting the jackpot, independent of all other tickets. No combination is more likely than any other. No ticket becomes "due." The odds reset identically for every draw.
How many Powerball tickets are sold per drawing?
MUSL (Multi-State Lottery Association) does not publish per-draw ticket counts in real time, but state lottery revenue disclosures allow a reasonable reconstruction at $2 per ticket:
- Jackpots in the $50–$150M range: roughly 30–60 million tickets
- Jackpots in the $150–$400M range: roughly 60–100 million tickets
- Jackpots above $500M: frequently 150 million or more tickets per draw
These are estimates derived from publicly reported sales figures; individual draws vary. The September 16 draw at $271M falls in the mid-range, making ~90 million tickets a reasonable working assumption.
The rollover formula: how is P(no winner) derived?
With N independent tickets each carrying probability p = 1/292,201,338 of matching all six numbers, the probability that no ticket wins the jackpot is:
P(no winner) = (1 − 1/292,201,338)^N
For the values involved — N on the order of tens of millions, p on the order of 10−9 — the standard approximation (1 − 1/n)n ≈ e−1 applies cleanly:
P(no winner) ≈ e^(−N / 292,201,338)
Each ticket is treated as an independent Bernoulli trial. The approximation error from truncating the Taylor series for ln(1 − p) is negligible: at p = 1/292,201,338, the second-order correction contributes less than 10−10 to the exponent — far below any rounding in the sales estimate.
Plugging in the numbers: the ~73% rollover probability
Substituting N = 90,000,000:
P(no winner) ≈ e^(−90,000,000 / 292,201,338)
= e^(−0.3080)
≈ 0.7348
Approximately 73.5% probability of no jackpot winner on a draw with ~90 million tickets. The complementary probability — that at least one ticket matches all six numbers — is 1 − 0.7348 = 26.5%.
The September 16 no-winner result is not a surprise. It is the modal outcome. The jackpot rolls over on roughly three out of every four mid-range draws.
How does jackpot size change rollover probability?
As the jackpot grows, more casual players enter, ticket sales rise, and the rollover probability falls. Plugging in higher sales estimates:
- N = 150 million tickets: P(no winner) ≈ e−0.5134 ≈ 59.8%
- N = 200 million tickets: P(no winner) ≈ e−0.6845 ≈ 50.4%
As N approaches 292 million — the total combination count — P(no winner) approaches e−1 ≈ 36.8%, which remains a high rollover probability even in that theoretical limit. In practice, no U.S. Powerball draw has ever sold 292 million tickets, so the rollover probability never collapses to zero.
This feedback loop is self-correcting but not self-terminating: a larger jackpot attracts more tickets, which compresses rollover odds — but even at 200 million tickets, there is still a 50% chance the jackpot survives the draw.
Geometric distribution: expected number of draws until a jackpot is claimed
If each draw is an independent Bernoulli trial with P(jackpot claimed) = q, the number of draws until the first win follows a geometric distribution:
- Expected draws until win: E[X] = 1/q
- Standard deviation: σ = √(1−q) / q
- P(jackpot survives exactly k draws): (1−q)k × q
At q = 0.265 (the 90M-ticket draw), E[X] = 1/0.265 ≈ 3.8 draws. Most jackpots resolve quickly. But the geometric distribution's right tail is heavy: P(jackpot survives ≥ 10 draws at constant q = 0.265) = 0.73510 ≈ 4.9%; at ≥ 20 draws: 0.73520 ≈ 0.24%.
The critical caveat: q is not constant. Early in a rollover streak, the jackpot is small, ticket sales are low, and q can be as small as 6–13%. Each early draw then carries a very high individual rollover probability, and those independent probabilities multiply. That is where long jackpot chains originate.
Case study: the $1.04 billion Illinois jackpot
The $1.04B jackpot claimed in Illinois illustrates the early-cycle dynamics. The Powerball jackpot resets to $20M after each win. At those low levels, ticket sales are modest and rollover probability is high:
- N = 20 million tickets: P(no winner) ≈ e−0.0685 ≈ 93.4%
- N = 40 million tickets: P(no winner) ≈ e−0.1369 ≈ 87.2%
With early-draw rollover probability near 90%, extended chains are structurally expected rather than anomalous. P(20 consecutive rollovers at 90% each) = 0.9020 ≈ 12.2% — roughly 1-in-8, occurring regularly across all active Powerball jackpots.
A billion-dollar jackpot requires the high-rollover early phase to persist for many draws, followed by a mid-cycle run where rising ticket sales gradually compress the odds but not enough to end the streak. It is a genuine tail event — billion-dollar jackpots occur roughly once every several years — but it requires no deviation from pure randomness. The math explains it entirely.
Frequently asked questions
Does buying more Powerball tickets affect whether the jackpot rolls over?
Buying one more ticket shifts the draw's rollover probability by less than 0.000001 percentage points — immeasurably small. Collectively, all players together drive the aggregate: 90 million tickets create a 73% rollover probability, while 150 million tickets reduce it to 60%. No single ticket purchase affects whether the jackpot is claimed that night.
Is a Powerball jackpot ever statistically "overdue" for a winner?
No. Each Powerball draw is statistically independent — prior rollovers carry zero information about the next result. A jackpot that has rolled over 30 consecutive times has the same rollover probability on draw 31, at the same ticket sales volume, as it did on draw 1. Believing otherwise is the gambler's fallacy.
What does rollover probability mean for expected value (EV)?
Rollover probability describes jackpot growth dynamics, not single-ticket EV. Individual EV depends on jackpot size, ticket price, lump-sum discount, tax rate, and split-prize probability. A larger jackpot raises headline EV but also attracts more tickets, increasing split risk — the two effects partially, and sometimes fully, cancel out.
Explore Powerball draw history, current jackpot size, and rollover streaks at Jackpot Teller — free signup, no purchase required: https://jackpotteller.com/w/data