can some on tell me were to buy a wii
2006-12-10 12:05:56 · 12 answers · asked by Anonymous in Games & Recreation ➔ Other - Games & Recreation
Keep an eye on the Top Toys for Christmas page at Alex's Coupons:
http://www.alexscoupons.com/top-toys-christmas-2006.html
They're tracking when the Wii, PS3 and a bunch of other toys are available. Right now (1:30am PST 12/12/06) Walmart has the Wii Holiday Bundle IN-STOCK for $583.56. Hope this helps...
-- T
2006-12-11 20:39:20 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Wal-Mart obviously is a good choice. Or Best Buy, Toys-R-Us, Game stop, Game Crazy, the internet, Target, and probably any other department store. Although, at this time of year, seeing as they were only recently released, it will be hard to get one. If you dont care about the price, eBay could work for you.
2006-12-10 20:09:37 · answer #2 · answered by Rafael Santiago 1 · 0⤊ 0⤋
Wal-mart
2006-12-10 20:08:03 · answer #3 · answered by ? 7 · 0⤊ 0⤋
The best place would be EBAY they are the only place that will have it. It might be way more than in the stores but its your best bet.
2006-12-10 20:09:29 · answer #4 · answered by Pdubbs 2 · 0⤊ 0⤋
ebay, amazon.com, overstock.com, best buy, circuit city, walmart, GameBuzz, Fred Meyer, Fry's, Sears, Costco, some kmarts have it, ebgames, gamestop, Target
2006-12-10 20:15:40 · answer #5 · answered by Geek Girl 3 · 1⤊ 0⤋
Ebay!
2006-12-10 20:42:15 · answer #6 · answered by michael 1 · 0⤊ 1⤋
i went shopping today, and they dont have any available anywhere! everyone should have some about next sunday so try then. you can also try ebay, but it will be 2x the price!
2006-12-10 20:11:21 · answer #7 · answered by Daugz 2 · 0⤊ 0⤋
ebay...but buy fr:a seller who has a pic of the receipt and 100% pos feedback...
2006-12-10 20:08:06 · answer #8 · answered by Anonymous · 0⤊ 0⤋
you could try going to ebay,or trying searching on the internet
2006-12-10 20:11:13 · answer #9 · answered by hpz ftw 4 · 0⤊ 0⤋
ebay got a bunch of them
2006-12-10 20:13:37 · answer #10 · answered by 0six_mix 2 · 0⤊ 1⤋