Lego Friends ?

[stewpette]

My daughter wants Olivia's House for Christmas and I see on Walmart.com there is a bundle for Olivia's House and Heartlake Vet for $79 - I think this is a good deal, but if anyone is knowledgeable aboout Lego Friends, can you confirm? Thanks!

 

[belle22]

Looking for Lego Friends as well (not the house).. any sales on the other pieces?

 

[kalliyan1]

My daughter wants Olivia's House for Christmas and I see on Walmart.com there is a bundle for Olivia's House and Heartlake Vet for $79 - I think this is a good deal, but if anyone is knowledgeable aboout Lego Friends, can you confirm? Thanks!

Yes that seems like a good price. I think the house itself is $69 and the vet is like $40

 

[stewpette]

Yes that seems like a good price. I think the house itself is $69 and the vet is like $40

Ok great, that is what I thought. But I have been going crazy with all the sales and couldn't recall if I had seen these two items separately at cheaper prices. I am going to get them and am glad to get 2 bigger sets for just a bit over what I would have been spending for one!

 

[MAH4546]

That's a good deal.

Also, LEGO Friends are becoming scarce. LEGO.com is starting to sell out, as is toysrus.com, and stores carrying them are running out. It's a hot holiday gift - buy LEGO Friends sets now because they won't be plentiful again until January!

Some of the smaller sets are also being retired - most notably Heartlake Dog Show and Butterfly Beauty Shop - as a new batch of sets will trickle into stores soon. So what's left in stores (both those examples are already sold out directly from LEGO) is all that's left.

Toys R Us has Buy One, Get One Half Off on Friends right now, but they so grossly overprice most Friends sets, it's not much of a deal, with one exception. Combine the Groupon $10 for $20 Toys R Us gift certificate (still available) with the sale, and you can get Olivia's Treehouse and Heartlake Pet Show for $24 (retail is $40; TRU retail is $44, because TRU doesn't inflate the price on these as much as other sets for whatever reason).