Air Jordan 1 X Dior

I saw that. But, no bot and it looks like more than half have already been passed out to celebrities, friends and family. So, it looks like a no to me
 
I saw that. But, no bot and it looks like more than half have already been passed out to celebrities, friends and family. So, it looks like a no to me

On top of that, stores have started to reach out to clients for pairs (fingers crossed on my 42). After all is said and done, I don't see many pairs being available for a supposed "general release".
 
Price is $2,200, not $2000 flat FYI.

As for the rumored raffle, I highly doubt 7500 pairs are up for grabs through that channel.
 
Oh and these are stupid fire. :{ I need now. The collection is too fire. Of course it's only gonna' be celebrities that are able to obtain them.
 
wait...what raffle? I want these bad. I just know its impossible. any help though would be greatly appreciated
 
wait...what raffle? I want these bad. I just know its impossible. any help though would be greatly appreciated

I wouldn't expect many pairs to be available through Dior after the seeded pairs to celebrities and VICs get their hands on them. Dior started reaching out to clients last week, and I'm sure a bulk of the pairs available have already been allotted to loyal Dior customers. It's in their best interest to keep clients happy, as opposed to giving someone who is not a frequent buyer a shot at them.
 
I wouldn't expect many pairs to be available through Dior after the seeded pairs to celebrities and VICs get their hands on them. Dior started reaching out to clients last week, and I'm sure a bulk of the pairs available have already been allotted to loyal Dior customers. It's in their best interest to keep clients happy, as opposed to giving someone who is not a frequent buyer a shot at them.
thank u
 
how many of the lows did they make? those don't seem as limited as the highs?
 
Eventually, if my dear client will pull through with it then I'll be one of the few lucky who may potentially snag a pair.

Fingers X'd
 
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