The winning Delaware Play 4 numbers for December 7, 2003 were: night 1-0-5-9.
Sum is the digits added together (5+5+0 = 10). Root sum keeps adding until one digit is left (10 → 1+0 = 1) — many regular players track it in their notebooks. Odd/Even marks each digit O or E (5-5-0 = OOE). High/Low marks 0–4 as L and 5–9 as H (5-5-0 = HHL).
| Date | Draw | Numbers | Sum | Root |
|---|---|---|---|---|
| Dec 7, 2003 | Night | 1-0-5-9 | 15 | 6 |
| Dec 6, 2003 | Night | 7-0-3-7 | 17 | 8 |
| Dec 6, 2003 | Day | 0-7-2-1 | 10 | 1 |
| Dec 5, 2003 | Night | 2-8-5-2 | 17 | 8 |
| Dec 5, 2003 | Day | 0-9-8-1 | 18 | 9 |
| Dec 4, 2003 | Night | 6-9-8-5 | 28 | 1 |
| Dec 4, 2003 | Day | 7-8-8-9 | 32 | 5 |
| Dec 3, 2003 | Night | 5-1-3-9 | 18 | 9 |
| Dec 3, 2003 | Day | 5-4-5-4 | 18 | 9 |
| Dec 2, 2003 | Night | 0-7-9-2 | 18 | 9 |
| Dec 2, 2003 | Day | 5-7-5-2 | 19 | 1 |
| Dec 1, 2003 | Night | 4-7-4-1 | 16 | 7 |
| Dec 1, 2003 | Day | 1-6-5-5 | 17 | 8 |