The winning New York Numbers numbers for October 11, 1999 were: evening 9-7-5.
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 11, 1999 | Evening | 9-7-5 | 21 | 3 |
| Oct 10, 1999 | Evening | 0-9-9 | 18 | 9 |
| Oct 9, 1999 | Evening | 0-2-6 | 8 | 8 |
| Oct 8, 1999 | Evening | 8-9-3 | 20 | 2 |
| Oct 7, 1999 | Evening | 4-6-6 | 16 | 7 |
| Oct 6, 1999 | Evening | 7-4-0 | 11 | 2 |
| Oct 5, 1999 | Evening | 8-5-2 | 15 | 6 |