The winning Connecticut Play4 numbers for August 3, 2003 were: night 9-9-0-2.
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 |
|---|---|---|---|---|
| Aug 3, 2003 | Night | 9-9-0-2 | 20 | 2 |
| Aug 2, 2003 | Night | 4-3-2-8 | 17 | 8 |
| Aug 1, 2003 | Night | 0-0-4-0 | 4 | 4 |
| Aug 1, 2003 | Day | 8-6-7-8 | 29 | 2 |
| Jul 31, 2003 | Night | 4-6-9-9 | 28 | 1 |
| Jul 31, 2003 | Day | 9-5-6-4 | 24 | 6 |
| Jul 30, 2003 | Night | 5-0-6-7 | 18 | 9 |
| Jul 30, 2003 | Day | 6-7-8-8 | 29 | 2 |
| Jul 29, 2003 | Night | 4-0-1-6 | 11 | 2 |
| Jul 29, 2003 | Day | 4-2-3-2 | 11 | 2 |
| Jul 28, 2003 | Night | 9-6-5-9 | 29 | 2 |
| Jul 28, 2003 | Day | 6-7-7-7 | 27 | 9 |