Here's Why the Cavaliers Will Not Trade the #1 Pick They Got in the Kyrie Irving Deal

Flashback to 2014 and the NBA draft. On June 26th, the Cavaliers drafted Andrew Wiggins 1st overall. The idea of pairing him with budding star Kyrie Irving was exciting.

But more exciting than that was the possible return of LeBron James to Cleveland. LeBron had opted-out of his Miami contract as expected, and rumors were flying. The Heat were trying so hard to keep him, they made a draft-night trade to acquire a player LeBron was fond of: Shabazz Napier.

LeBron picked Cleveland, promised a title, and delivered one. He also got the Cavs to trade Wiggins for Kevin Love, and the trio of Irving, Love and LeBron has been to three straight finals.

[Aside: Would the Cavs have gone to three straight Finals with Wiggins instead of Love? I believe so. That’s how dominant LeBron has been.]

How does that scenario apply to this season’s Cavs, LeBron and the #1 pick they acquired in the Kyrie Irving deal? It’s simple: LeBron won’t commit, and Cleveland’s best hope of holding onto LeBron beyond this season is keeping that pick and drafting a star player in 2018.

No, that player may not play with LeBron – but that high draft pick – Michael Porter, Marvin Bagley, whomever – could be flipped the way the Cavs flipped Wiggins in 2014, in exchange for a star.

But the reason you can’t trade the pick before LeBron makes a decision is simple: You could trade for someone like Kristaps Porzingis, Marc Gasol, whomever – have it not work out, and then LeBron leaves. And you’re now out the #1 pick to grow the team, and you’ve got a guy who could leave via free agency.

It would be business suicide to trade the pick before LeBron makes a free agent decision. Hey LeBron, stay with us, we’re flip this pick and pieces for whatever we can. And let’s be honest – there isn’t a player in the NBA that the Cavs could realistically trade for at the deadline who is going to put the Cavs over the Warriors.

If LeBron won’t commit, you must keep the pick.