The Philadelphia 76ers have the luxury of possessing not one, but two lottery picks in the upcoming NBA draft on June 26. With one of the deepest classes in recent memory set to enter the league, two picks is quite a valuable commodity to have.
Not only does the team possess the No. 3 and 10 picks in the first round, they’ll also be working with five picks in the second round (No. 32, 39, 47, 52 and 54) as
In a recent live chat over at ESPN.com concerning the draft, insider Chad Ford discussed the possibility of Philadelphia packaging their first round picks together to entice the Cleveland Cavaliers to give away their No. 1 overall selection.
Q: Chad – Do you think the Cavs and 76ers would both agree to a 1 for 3+10 swap? Seems like it would allow the Cavs to load up on picks they could move in tandem with someone like Waiters/Bennett/Thompson for an established star. Whereas the 76ers guarantee themselves Wiggins.
A: I’m not sure the Sixers would do this. They need talent and that No. 10 pick is very valuable to them. I think they’d offer Thaddeus Young and 3 for No. 1, but including the 10th pick would be too much for them. I don’t blame them. There’s a good chance, if Joel Embiid’s back checks out that he’s the No. 1 pick this year. If he goes 1, there’s a very good chance the Sixers top target, Andrew Wiggins is there at No. 3
Ford makes a valid point. The 76ers finished the regular season with a record of 19-63, second-worst in the NBA behind the 15-67 Milwaukee Bucks. Not only is there a high chance that Wiggins, the player currently atop their draft board, will be available when they make their No. 3 pick, but their No. 10 pick could also end up being a valuable contributor in their rotation in 2014-15.
It’s simply to0 high of an asking price to simply move up two spots. If the Cavaliers go with Embiid as they’re expected to do, there will be no reason for Philadelphia to even consider such a move.