Yes. Numbers from the last Lotto 6/49 Classic draw can appear again. Under a model of independent draws in which every six-number combination from 1 to 49 is equally likely, the chance that at least one of the previous six numbers appears in the next draw is about 56.40%. That is a different question from whether one specified number will return. The calculation below separates the two.
Analysis revised 2026-09-18: this analysis adds results from June to September 2026. The original publication date, March 12, 2026, is retained; those later results were not part of the original article.
One number appearing in two real draws
The official Classic results for June 3, 2026 and June 6, 2026 share the number 30. Only the six main numbers are compared; the bonus number and Gold Ball draw are excluded.
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One shared main number: 30
| Date and official source | Main numbers |
|---|---|
| 2026-06-03 | 3, 5, 6, 28, 30, 37 |
| 2026-06-06 | 9, 27, 30, 33, 45, 47 |
To check the repetition:
- Read the six main numbers in each row.
- Find the numbers present in both rows. Here, the intersection is {30}.
- Count them: this pair has one repeated number.
This example was selected after inspecting the data to make the comparison easy to follow. It was not selected in advance or presented as representative of all draws. The emphasis on 30 simply identifies the match; it gives that number no special status.
Here, “repeat” means appearing in two successive draws. Consecutive integers within one draw, such as 5 and 6 in the June 3 result, describe a different pattern.
Three questions with different probabilities
The official game description sets out the Classic selection of six numbers from 1 to 49. For the following calculations, assume that draws are independent and that all possible six-number combinations are equally likely. Fix the preceding draw's six main numbers, then compare the next draw with that set.
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Three events under the independent uniform 6-of-49 model
| Event | Exact probability | Readable value |
|---|---|---|
| At least one of the previous six repeats | 1 − C(43, 6)/C(49, 6) | About 56.40% |
| One specified previous number repeats | 6/49 | About 12.24% |
| The same six-number set repeats, ignoring order | 1/13,983,816 | 1 in 13,983,816 |
C(n, r) means the number of ways to choose r different items from n without counting their order. Thus C(49, 6) = 13,983,816 possible combinations. The percentages are rounded; the fractions and formulas are exact.
The first event includes many different combinations: some share one number, some share two, and others share more. A specified number such as 30 has only a 6/49 chance of appearing in the next draw, about 12.24%. It does not inherit the 56.40% probability of the broader event.
A category containing many combinations can be common without giving any particular combination better odds. Under this model, including or excluding numbers from the previous draw does not improve a particular selection's chance of matching all six main numbers.
What the 29 comparisons show
The historical calculation uses a fixed window of 30 Lotto 6/49 Classic draws, from June 3 through September 12, 2026, inclusive. It includes every Wednesday and Saturday in that interval. The window was inherited from an existing archive for continuity, before the repeat counts were calculated. It is not a latest-results feed.
Each draw is compared with the immediately preceding draw within that window. The first draw has no earlier draw inside the window, so 30 draws produce 29 comparisons. Neither the bonus numbers nor Gold Ball results enter the calculation. The full dated appendix provides both sets and the official sources for every pair.
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Observed overlaps in 29 pairs, June 3–September 12, 2026
| Shared main numbers | Number of pairs |
|---|---|
| 0 | 7 |
| 1 | 15 |
| 2 | 7 |
| 3 | 0 |
| 4 | 0 |
| 5 | 0 |
| 6 | 0 |
There is at least one shared number in 22 of the 29 pairs: 22/29 ≈ 75.86%. Adding the shared-number counts across all pairs gives 29, so the observed mean is 29/29 = 1 shared number per pair. That total counts occurrences across comparisons, not 29 distinct number values.
No pair in this window shares three or more main numbers. A zero count in this sample does not make such an outcome impossible.
Why this sample does not predict the next draw
The observed share, 75.86%, is higher than the model probability, 56.40%. They answer different questions: one describes these 29 historical comparisons; the other follows from the stated draw model. The historical percentage does not replace the model probability for the next draw.
This is a small, fixed historical window, not a claim of representative random sampling. No significance test or confidence interval is calculated here. The difference alone establishes neither a change in draw probabilities nor a conclusion about the fairness of the lottery.
Neighbouring comparisons share a draw. That fact alone is not a proof that their overlap counts must be statistically dependent under the uniform, independent-draw model. This article uses the pairs descriptively and makes no inferential claim based on their dependence or independence.
Loto-Québec's explanation of lottery randomness states that past appearances do not alter a number's chance in a subsequent draw. Under the model used here, a specified number has the same 6/49 chance whether it appeared last time or did not. The observed repetition is no reason to call it “due,” “hot,” or less likely to return.
What to check in a claim about repeated numbers
Before interpreting a percentage, check:
- What repeats: one specified number, any of the previous six, or the entire set?
- Which game and date window the claim covers, and whether bonus numbers are included.
- Whether the denominator counts draws or pairs of draws.
- Whether the figure describes recorded results or a probability derived from a model.
A useful repeat table lets you verify its dates, sets and denominator. It cannot turn past results into a number-selection advantage. Treat lottery play as an optional entertainment expense, not an income strategy, and keep any spending within an amount you can afford to lose.
Sources and method
The calculations use archived official Loto-Québec result pages. The archive records acquisition on September 15, 2026 (UTC). Local verification on September 17, 2026 (UTC) re-read the saved pages and checked their hashes, dates and numbers; it did not repeat the original HTTP retrieval or authenticate the historical transport anew. Each appendix date links to the exact official URL recorded for that draw.
The official game-description extraction was recorded on September 15, 2026 (UTC). The lottery-independence explanation was retrieved through a web tool on September 17, 2026 (UTC). These are tool extractions, not raw HTTP captures. This article revision does not claim a fresh retrieval of those sources.
The hero is a conceptual illustration of two groups of unnumbered balls, not a picture of actual drawn numbers. The reading-time estimate covers the main explanation, not a row-by-row review of the appendices.
View all 29 comparisons and official sources
Each row compares successive draws within the fixed window. Both dates link to the recorded official results. Sets are sorted for reading; order does not affect the intersection. Only main numbers count. Bonus and Gold Ball results are excluded.
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All 29 Lotto 6/49 Classic draw pairs
| Previous date and source | Previous main numbers | Next date and source | Next main numbers | Shared numbers | Count |
|---|---|---|---|---|---|
| 2026-06-03 | 3, 5, 6, 28, 30, 37 | 2026-06-06 | 9, 27, 30, 33, 45, 47 | 30 | 1 |
| 2026-06-06 | 9, 27, 30, 33, 45, 47 | 2026-06-10 | 4, 9, 33, 39, 40, 48 | 9, 33 | 2 |
| 2026-06-10 | 4, 9, 33, 39, 40, 48 | 2026-06-13 | 1, 5, 9, 17, 36, 46 | 9 | 1 |
| 2026-06-13 | 1, 5, 9, 17, 36, 46 | 2026-06-17 | 2, 14, 23, 26, 42, 48 | None | 0 |
| 2026-06-17 | 2, 14, 23, 26, 42, 48 | 2026-06-20 | 9, 15, 26, 27, 35, 42 | 26, 42 | 2 |
| 2026-06-20 | 9, 15, 26, 27, 35, 42 | 2026-06-24 | 9, 20, 28, 29, 36, 42 | 9, 42 | 2 |
| 2026-06-24 | 9, 20, 28, 29, 36, 42 | 2026-06-27 | 14, 22, 27, 29, 37, 46 | 29 | 1 |
| 2026-06-27 | 14, 22, 27, 29, 37, 46 | 2026-07-01 | 3, 10, 14, 20, 30, 45 | 14 | 1 |
| 2026-07-01 | 3, 10, 14, 20, 30, 45 | 2026-07-04 | 4, 5, 27, 29, 38, 47 | None | 0 |
| 2026-07-04 | 4, 5, 27, 29, 38, 47 | 2026-07-08 | 15, 19, 25, 37, 41, 42 | None | 0 |
| 2026-07-08 | 15, 19, 25, 37, 41, 42 | 2026-07-11 | 5, 10, 11, 33, 34, 41 | 41 | 1 |
| 2026-07-11 | 5, 10, 11, 33, 34, 41 | 2026-07-15 | 12, 15, 17, 21, 28, 38 | None | 0 |
| 2026-07-15 | 12, 15, 17, 21, 28, 38 | 2026-07-18 | 15, 27, 29, 33, 46, 49 | 15 | 1 |
| 2026-07-18 | 15, 27, 29, 33, 46, 49 | 2026-07-22 | 9, 25, 29, 30, 33, 44 | 29, 33 | 2 |
| 2026-07-22 | 9, 25, 29, 30, 33, 44 | 2026-07-25 | 1, 8, 9, 24, 36, 37 | 9 | 1 |
| 2026-07-25 | 1, 8, 9, 24, 36, 37 | 2026-07-29 | 6, 8, 13, 15, 20, 37 | 8, 37 | 2 |
| 2026-07-29 | 6, 8, 13, 15, 20, 37 | 2026-08-01 | 8, 22, 37, 41, 43, 44 | 8, 37 | 2 |
| 2026-08-01 | 8, 22, 37, 41, 43, 44 | 2026-08-05 | 12, 30, 31, 35, 41, 48 | 41 | 1 |
| 2026-08-05 | 12, 30, 31, 35, 41, 48 | 2026-08-08 | 3, 7, 18, 24, 45, 48 | 48 | 1 |
| 2026-08-08 | 3, 7, 18, 24, 45, 48 | 2026-08-12 | 6, 13, 28, 34, 45, 48 | 45, 48 | 2 |
| 2026-08-12 | 6, 13, 28, 34, 45, 48 | 2026-08-15 | 1, 9, 17, 34, 36, 43 | 34 | 1 |
| 2026-08-15 | 1, 9, 17, 34, 36, 43 | 2026-08-19 | 6, 7, 10, 32, 33, 36 | 36 | 1 |
| 2026-08-19 | 6, 7, 10, 32, 33, 36 | 2026-08-22 | 11, 13, 21, 31, 34, 45 | None | 0 |
| 2026-08-22 | 11, 13, 21, 31, 34, 45 | 2026-08-26 | 2, 14, 30, 33, 41, 45 | 45 | 1 |
| 2026-08-26 | 2, 14, 30, 33, 41, 45 | 2026-08-29 | 16, 20, 23, 24, 25, 33 | 33 | 1 |
| 2026-08-29 | 16, 20, 23, 24, 25, 33 | 2026-09-02 | 6, 8, 9, 14, 17, 25 | 25 | 1 |
| 2026-09-02 | 6, 8, 9, 14, 17, 25 | 2026-09-05 | 5, 10, 17, 34, 42, 49 | 17 | 1 |
| 2026-09-05 | 5, 10, 17, 34, 42, 49 | 2026-09-09 | 14, 18, 27, 35, 41, 47 | None | 0 |
| 2026-09-09 | 14, 18, 27, 35, 41, 47 | 2026-09-12 | 1, 3, 17, 20, 26, 32 | None | 0 |
View the probability calculation
Let K be the number of main numbers shared by two successive draws. Fix the preceding six numbers and assume independent, uniform draws of six different numbers from 1 to 49. There are C(49, 6) = 13,983,816 equally likely sets for the next draw.
To obtain exactly k matches, choose k of the preceding six numbers and 6 − k of the other 43. For k = 0, 1, …, 6:
P(K = k) = C(6, k) × C(43, 6 − k) / C(49, 6)
This is the hypergeometric distribution. Its seven numerators add up to C(49, 6), so the probabilities add up exactly to 1.
For no repeats, all six numbers must come from the other 43: P(K = 0) = C(43, 6)/C(49, 6) = 6,096,454/13,983,816. Subtracting from 1 gives P(K ≥ 1) = 7,887,362/13,983,816 ≈ 56.40%.
For a specified number to appear, choose the other five numbers from the remaining 48: C(48, 5)/C(49, 6) = 6/49 ≈ 12.24%. For all six to return, only one unordered set qualifies, giving 1/13,983,816. These are exact model calculations, not estimates from the historical sample.