2007-06-11 07:11:41 · 11 answers · asked by Jack M 1 in Entertainment & Music ➔ Celebrities
Beyonce
2007-06-11 07:14:12 · answer #1 · answered by JAQUELINE Z 2 · 1⤊ 0⤋
Definitely Beyonce. She is such a natural curvaceous beauty, while Pamela Anderson is a big plastic thing. Plus, Pamela lets it all hang out there while Beyonce makes elegant and sexy work together for her.
2007-06-11 14:40:19 · answer #2 · answered by writeaway 4 · 0⤊ 0⤋
pamela anderson 15 years ago maybe but i think beyonce is sexier because she has that natural look and she has nice even cures but pamelas boobs is like not even with her body and then nowadays she puts on too much make up which gives her an older look so i think beyonce is sexier
2007-06-11 14:19:59 · answer #3 · answered by Anonymous · 0⤊ 1⤋
Pamela Anderson disgusts me and is fake. Beyonce is hot but cant sing at all. so yeah, its Beyonce.
2007-06-11 14:19:13 · answer #4 · answered by Anonymous · 0⤊ 1⤋
lol Pamila?
Pamela Anderson have HIV
And Beyonce isn't that hot...
I will go with Julia Volkova from t.A.T.u band lol
2007-06-11 14:15:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
personally pamela anderson . she still is a sex symbol at 40 (now how often can u say that )and i don't think beyonce will.
http://photos.ivillage.com/entertainment/starsnapshots/2007/06/pamela_anderson.html
look at this i mean hey come on!
2007-06-11 14:18:24 · answer #6 · answered by cute-goddess 5 · 0⤊ 2⤋
Catherine Zeta Jones is hotter than both of them!
2007-06-11 14:21:49 · answer #7 · answered by Carla P 4 · 1⤊ 1⤋
they're two totally different types of women in class and talent. Beyonce has them, Pamela doesn't......
2007-06-11 14:14:45 · answer #8 · answered by caramelchocolatecamilla 3 · 1⤊ 0⤋
Beyonce by a mile, and talented too.
2007-06-11 14:14:53 · answer #9 · answered by godron_wookie 4 · 3⤊ 0⤋
I heard they got the same tattoo.
A bulls-eye on there sitting eye!!
Do not care. there's starving children out there too and they look horrible
2007-06-11 17:38:03 · answer #10 · answered by NEMESIS 3 · 0⤊ 1⤋