The September 11, 2026 Lotto Max draw includes 44–45. Having that specific pair and having any consecutive pair are different events: the latter includes many more combinations. Neither category raises the probability of an exact selection. This article describes the result; it does not explain its cause or predict another draw.
Analysis revised 2026-09-14: the previous example, whose draw was not authenticated, was replaced with the 2026-09-11 draw. The original publication date is retained.
The verified draw and its rules
Source: saved text from the official Loto-Québec page, consulted September 14, 2026, for the 2026-09-11 draw. Main numbers: 1, 38, 41, 44, 45, 49, 51; bonus number: 47, excluded from calculations. This is one draw, not a frequency study. The saved text is not an archive of the original HTTP response. Official result.
This draw uses 7 distinct main numbers from 52. The Loto-Québec rules describe a play as 4 selections; a selection and a complete play are different units. Bonus and auxiliary draws are excluded here, as are probabilities of winning any prize across all prize categories.
The official number-availability note records number 50 from May 14, 2019, and numbers 51 and 52 from April 14, 2026. For comparison, C(50,7)=99884400 and C(52,7)=133784560, where C(N,k) counts ways to choose k distinct numbers from N without regard to order. One exact main-number selection therefore has theoretical probability 1/99884400 in the former model and 1/133784560 in the model used here. The former is a mathematical comparison, not verification of all historical product conditions or whole-play odds.
Read the whole set
A successive gap is the difference between neighbouring numbers in sorted order. The complete list shows that a large gap and a consecutive pair can occur in the same draw. The ranges are descriptive choices, not privileged zones.
| Measure | Result |
|---|---|
| Parity | 2 even / 5 odd |
| Ranges → counts | 1–10 / 11–30 / 31–52 → 1, 0, 6 |
| Successive gaps | 37, 3, 3, 1, 4, 2 |
| Consecutive pairs | 1 (44–45) |
| Sum of numbers | 269 |
| Span = sum of gaps | 50 = 50 |
This describes one draw. An empty range does not have to be filled in the next draw. The gaps sum to the largest number minus the smallest: 51 − 1 = 50.
A specific pair versus any consecutive pair
The pair 44–45 was chosen for explanation after observing this result; it is not a recommended pair. “At least one consecutive pair” means that at least one pair of sorted neighbours differs by 1. Under a uniform main-number draw, the events compare as follows; these are theoretical probabilities, not historical frequencies.
Each fraction gives the number of combinations in that category in the numerator, divided by the total number of possible combinations in the denominator.
| Event | Exact theoretical probability |
|---|---|
| Both 44 and 45 present | 2118760/133784560 |
| At least one consecutive pair | 80259880/133784560 |
| One exact 7-number selection | 1/133784560 |
For both 44 and 45 to be present, the remaining numbers can be chosen in C(50,5) ways, giving C(50,5)/C(52,7).
To count combinations with no consecutive numbers, put the chosen numbers in increasing order, then subtract zero from the first, one from the second, and so on. This removes the required gaps between neighbours: the last number shifts down six places, reducing the upper limit from 52 to 46.
This gives a one-to-one correspondence with choosing seven distinct values from the compressed range; adding the same offsets recovers each original choice. The no-neighbour count is therefore C(46,7).
For any consecutive pair, subtract the combinations with no consecutive neighbours from all possible combinations: 1-C(46,7)/C(52,7).
The first two numerators count whole categories of combinations; the last row, 1/C(52,7), concerns one exact selection matching all main numbers, not an entire ticket.
Categories do not favour individual numbers
C(...) expressions written next to each other are multiplied.
The observed 2 even and 5 odd numbers belong to a category with theoretical probability C(26,2)C(26,5)/C(52,7)=21378500/133784560.
The range counts 1,0,6 belong to a separate category with probability C(10,1)C(20,0)C(22,6)/C(52,7)=746130/133784560.
These arbitrary ranges were applied after observing the draw, and the probabilities are neither a joint probability nor a significance test.
How many numbers from a group are drawn on average in this model? The expected count is that theoretical average over repetitions of the same model.
The separate set 20–40 contains 21 numbers and has expected count 147/52 = 21×7/52 under this model; it does not use the table’s 11–30 and 31–52 boundaries.
This expectation is not an observed count or an advantage for those numbers: every specified number has the same inclusion probability, 7/52.
What this draw cannot tell us
The sample is one draw, described through categories chosen after the result. It cannot establish historical frequencies or a forecasting method. Uniformity and independence are assumptions of the model, not properties proved by this snapshot. See Choosing Lotto Numbers: What Statistics Can and Cannot Change.