2006-08-29 12:58:10 · 13 answers · asked by mitteycole 3 in Sports ➔ Boxing
Leonard wins. Ray is naturally bigger. he started his career as a welter. Oscar started as a super feather. Both are skilled but Leonard had faster feet and hands.
2006-08-30 05:09:04 · answer #1 · answered by Brent 5 · 0⤊ 0⤋
I have to go with Ray Leonard on this one though De La Hoya is a skilled fighter. I believe that Leonard had too much speed and power and would dominate De La Hoya before stopping him in the 10th round.
2006-08-29 18:44:38 · answer #2 · answered by toughguy2 7 · 0⤊ 0⤋
Sugar Ray Leonard for sure by KO
2006-08-29 13:14:12 · answer #3 · answered by kleber 2 · 0⤊ 0⤋
that isn't even a contest the talent of a great like sugar ray leonard against Delahoye. Leonard would beat tyson's record in the olympics for the fastest knock out
2006-08-29 13:08:31 · answer #4 · answered by kinky couple 2 · 0⤊ 0⤋
Leonard. Too big naturally. Remember, Leonard was able to win a title at 175. De La Hoya was a bust at 160.
Leonard was also quicker, with more skill, and a much bigger banger.
2006-08-29 15:35:04 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Ray Leonard by a knockout.
2006-08-29 13:03:12 · answer #6 · answered by Anonymous · 1⤊ 0⤋
Tough question. If Sugar dances and moves, he'd win 9/10 times. However, if he fought toe to toe and if De La Hoya fought like he fought against Ike Qourtey, then De La Hoya would stand a good chance.
2006-08-29 15:37:24 · answer #7 · answered by eprometheus 2 · 0⤊ 0⤋
I would say Oscar but he wouldn't knock Sugar out.It would be a split decision. They are both well schooled & took on all challengers. Sugar in his later years fought in a tougher division(middle weight) & beat Marvin Hagler who dominated his weight class for over a decade. It's to close to call but i'll stick with Oscar.
2006-08-29 13:13:55 · answer #8 · answered by guvner_46 3 · 0⤊ 0⤋
sugar ray leonard!
2006-08-30 11:20:12 · answer #9 · answered by helenh 3 · 0⤊ 0⤋
ray leonard
2006-08-29 22:21:34 · answer #10 · answered by rizwano 7 · 0⤊ 0⤋