What's the current 4D number with the highest total prize amount for the year? Certainly, these would be among the numbers that have appeared in first prize and perhaps other top prizes too (non-iBet). There are 4 such numbers: 1516, 2490, 2916 and 9248. Out of these, 1516 stood out. When we checked it against the current 4D records for 2026, we found three first-prize appearances:

  • 7 February 2026 — Draw 5441
  • 25 February 2026 — Draw 5449
  • 27 May 2026 — Draw 5488

That's within a 48-draw interval (i.e. ~16 weeks). Remarkably, within the timeframe of 3 January 2026 to 21 October 2026 (126 draws), 1516 also appeared in no other prize position. How improbable is this?

A phenomenon rarer than a Group 1 win?

If a number and a 48-draw interval were specified in advance, the chance for that number to appear at least three times in first prize within that interval is about 0.000001724%, or 1 in 58 million. That's 4 times rarer than winning a typical Group 1 prize!

Including the other observed details (exactly three first-prize appearances in the full 126-draw sample and no appearances in any of the other 22 prize positions) gives probability of about 0.000001295%, or 1 in 77 million. Almost 6 times now!

But that assumes we chose a number in advance. In one fixed 48-draw interval, the chance that any of the 10,000 numbers appears in first prize at least three times is about 0.01724%, or 1 in 5,802. If we also require the fuller observed pattern (exactly three first-prize appearances in the 126-draw sample, all within that interval, and no appearance in any of the other 22 prize positions), the chance is about 0.01295%, or 1 in 7,723.

Show the derivation of all four fixed-window probabilities

Let m = 10,000 possible four-digit numbers and p = 1/m. The model assumes each prize result is independent and uniformly distributed. The four calculations below keep one 48-draw interval fixed in advance. In the formulas, a stacked pair in round brackets is a binomial coefficient; it counts how many ways to select positions without regard to order. “Occupancy” means how the 48 first-prize results are distributed among the 10,000 numbers. The detailed-pattern calculations also use the full 126-draw sample, or 2,898 prize positions.

1. One preselected number: at least three first prizes

The number of first-prize hits for one number in 48 draws is binomial with n = 48 and p = 1/10,000. “At least three” is one minus the probabilities of zero, one, or two hits:

P(X ≥ 3) = 1 − [(9,999/10,000)48 + 48(1/10,000)(9,999/10,000)47 + (1/10,000)2(9,999/10,000)46]

The coefficient here is the binomial coefficient with 48 above 2: 48 × 47 ÷ 2 = 1,128. It counts which two of the 48 draws would contain the number.

This is about 1.7238 × 10−8, or 0.000001724%—about 1 in 58 million.

2. Any of 10,000 numbers: at least three first prizes

The complement is that every number appears at most twice in the 48 first-prize results. The coefficient below counts allocations in which each number appears zero, one, or two times:

P(max ≥ 3) = 1 − 48!10,00048 [x48](1 + x + x2/2!)10,000

Evaluating this occupancy probability gives 0.000172365, or 0.01724%—about 1 in 5,802.

3. One preselected number: the full observed pattern

There must be exactly three first-prize hits, all in the fixed 48-draw interval, and no hit in any other prize position across all 126 draws. There are this many ways to place the three first-prize hits:

P = (1/10,000)3(9,999/10,000)2,895

This is about 1.2948 × 10−8, or 0.000001295%—about 1 in 77.2 million.

4. Any of 10,000 numbers: the full observed pattern

Use inclusion–exclusion across the possible numbers. For k specified numbers to each meet the pattern, assign three of the 48 first-prize slots to each, then require all remaining prize slots to avoid those k numbers. Sum the alternating terms for k = 1 to 16, since 48 slots can hold at most sixteen disjoint triples:

P(any) = ∑k=116 (−1)k+1 48!(3!)k(48 − 3k)! (1/10,000)3k ((10,000 − k)/10,000)2,898 − 3k

This gives 0.000129476, or 0.01295%—about 1 in 7,723. It accounts for scanning all numbers in one fixed interval, but not for choosing the interval after seeing the data.

That looks pretty complicated. Actually, just multiplying by 10,000 (i.e. point 1 vs 2, point 3 vs 4) gives you a good approximation. Doing so will double coin causes where two different numbers meet the condition in the same sample. The formula accounts for that but realistically these overlaps are so rare, you can be happy with the approximation and get on with life.

Has this happened before?

Based on 3,310 available draw records from 18 June 2005 to 24 October 2026, only two numbers appeared in first prize at least three times within any 100 consecutive draw records:

  • 5756 — 14 June 2023 (Draw 5026), 2 July 2023 (Draw 5034) and 28 January 2024 (Draw 5124), spanning 99 consecutive draw records.
  • 1516 — 7 February 2026 (Draw 5441), 25 February 2026 (Draw 5449) and 27 May 2026 (Draw 5488), spanning 48 consecutive draw records.

No other number in the range of records met this criterion so it is indeed a rare occurrence! It's therefore probably not the brightest idea to buy a "hot" number (e.g. once you find it appearing in 1st prize twice) thinking it will give you an advantage in the upcoming draws. Perhaps we will leave this to a future backtesting article.

Takeaway: 1516 now has an unprecedented record (3 first prizes within 48 draws!) that deserves a hall-of-fame mention, but it does not imply any advantage in a future draw. Out of 10,000 numbers, it is not that unlikely for us to witness such repetitions but if you are hoping that your particular number will repeat in the same way, you are likely to be very disappointed.

See the full 1516 number record, including the monthly chart and filters.

Source and scope: 4D results for draws 5426–5551 (3 January–21 October 2026), as well as a broaded dataset of historical draws covering earlier periods (18 June 2005 to 24 October 2026). The probability estimates above are conditional on the assumption of a fair, independent-draw model.