Can they play together??
2006-06-14 19:49:02 · 12 answers · asked by Kevs 2 in Sports ➔ Basketball
Both simply are two of the best point guards in the NBA.. But their style of play simply wont blend.. Kidd earns his assists from spot up shooters and taller slashers.. Iverson likes to create a shot by himself.. it will be like mixing milk and pepsi..
2006-06-14 23:45:41 · answer #1 · answered by BLACKPOPE 2 · 2⤊ 1⤋
Kidd can play with anyone. Can Iverson accept the ball going to Kidd first every time and him finding the best option at the time?
2006-06-15 04:24:21 · answer #2 · answered by josh22n 2 · 0⤊ 0⤋
Don't think so! There's no room for Iverson on the Nets and I don't want Kidd to leave!
2006-06-15 12:12:54 · answer #3 · answered by Amber P 2 · 0⤊ 0⤋
Both are too inconsistent from long-range. Iverson likes to create his own shot; Kidd likes to create shots for others. Who would lead the break? Who would finish the break? I don't like it.
2006-06-15 02:55:03 · answer #4 · answered by HL 5 · 0⤊ 0⤋
hell no, that's two little men with the ball. It may seem good with a super passer and a ball hog, but there hasn't been a great combo of guards in a while. The best thing is to have the threat in the paint and one on the outside so there will be open looks and a spread out offense.
2006-06-15 03:37:07 · answer #5 · answered by carolinakres 3 · 0⤊ 0⤋
iverson likes to shoot, kidd will pass him the ball,
it will work out great
2006-06-15 16:28:56 · answer #6 · answered by Anonymous · 0⤊ 0⤋
It will work, but only because Kidd is an unselfish player, and he is used to playing with cocky ball-hogs (Vince Carter).
2006-06-15 02:55:03 · answer #7 · answered by chopsueybysoad 2 · 0⤊ 0⤋
they will be great if Iverson would learn yo pass the ball more
2006-06-15 02:57:01 · answer #8 · answered by bigbobrose 1 · 0⤊ 0⤋
It would be a Grand Prix
2006-06-15 03:43:13 · answer #9 · answered by ToErrIsHuman 2 · 0⤊ 0⤋
One will beat him like he beats his wife, the other will smoke him out.
2006-06-15 03:17:51 · answer #10 · answered by bendergarcia 2 · 0⤊ 0⤋