The winning Connecticut Play4 numbers for December 6, 2003 were: night 1-1-6-6.
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 6, 2003 | Night | 1-1-6-6 | 14 | 5 |
| Dec 5, 2003 | Night | 6-7-9-4 | 26 | 8 |
| Dec 5, 2003 | Day | 7-0-2-0 | 9 | 9 |
| Dec 4, 2003 | Night | 3-6-4-3 | 16 | 7 |
| Dec 4, 2003 | Day | 7-4-5-6 | 22 | 4 |
| Dec 3, 2003 | Night | 0-5-3-6 | 14 | 5 |
| Dec 3, 2003 | Day | 5-0-3-6 | 14 | 5 |
| Dec 2, 2003 | Night | 9-3-0-9 | 21 | 3 |
| Dec 2, 2003 | Day | 6-9-1-9 | 25 | 7 |
| Dec 1, 2003 | Night | 4-3-4-3 | 14 | 5 |
| Dec 1, 2003 | Day | 3-2-5-0 | 10 | 1 |
| Nov 30, 2003 | Night | 6-2-8-0 | 16 | 7 |