The winning Connecticut Play4 numbers for October 7, 2002 were: day 2-2-7-3; night 9-2-8-4.
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).
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 |
|---|---|---|---|---|
| Oct 7, 2002 | Night | 9-2-8-4 | 23 | 5 |
| Oct 7, 2002 | Day | 2-2-7-3 | 14 | 5 |
| Oct 6, 2002 | Night | 6-7-6-5 | 24 | 6 |
| Oct 5, 2002 | Night | 8-9-3-4 | 24 | 6 |
| Oct 4, 2002 | Night | 3-2-2-7 | 14 | 5 |
| Oct 4, 2002 | Day | 9-3-1-6 | 19 | 1 |
| Oct 3, 2002 | Night | 0-3-4-8 | 15 | 6 |
| Oct 3, 2002 | Day | 7-6-8-2 | 23 | 5 |
| Oct 2, 2002 | Night | 4-5-9-4 | 22 | 4 |
| Oct 2, 2002 | Day | 7-2-7-6 | 22 | 4 |
| Oct 1, 2002 | Night | 1-1-6-4 | 12 | 3 |
| Oct 1, 2002 | Day | 8-5-0-7 | 20 | 2 |