2007-02-01 22:57:10 · 12 answers · asked by Anonymous in Family & Relationships ➔ Singles & Dating
both at the same time...
2007-02-01 23:00:54 · answer #1 · answered by Angel 2 · 0⤊ 0⤋
Since i have blonde, brunette, and a redhead in my life i would tend to think color of hair has nothing to do with it. It's the girls personal character traits that makes up each of them separately.
All the women in my life are fun i couldn't pick which is more fun to be with as they are just so nice.
2007-02-02 07:04:12 · answer #2 · answered by michael_trussell 4 · 1⤊ 0⤋
It kind of depends on what you definition of fun is. If you are a party animal, lol, blondes may be more fun. I think brunettes seem to be more sensible when it comes to partying and being reckless. Then again I may just be biased because I'm a brunette. Oh and I don't mean to offend anyone who is blonde. This is just my basic outlook. You may be different.
2007-02-02 07:02:57 · answer #3 · answered by Anonymous · 0⤊ 1⤋
It's not the hair color that makes a girl fun to be with!
2007-02-02 07:02:10 · answer #4 · answered by lanisoderberg69 4 · 1⤊ 0⤋
you can't know how much fun a person is by the colour of thier hair, its just a stereotypical view, besides, who keeps thier natural hair colour anymore anyway???
2007-02-02 07:01:23 · answer #5 · answered by *â??*à Æ?Æ?Ÿ*â??* 2 · 1⤊ 0⤋
haahaa...thats a funny question, because in my country there are no blonds...if you say blonds have more fun, nobody in our country has enough fun ..!!! lol
2007-02-02 07:05:43 · answer #6 · answered by ellllow 1 · 1⤊ 0⤋
very racist question... what is wrong with the women who aren't blonde or brunette???
2007-02-02 07:01:13 · answer #7 · answered by Forlorn Hope 7 · 1⤊ 1⤋
Brunette...........I am one and we have alot of fun all the time...yahooo.
2007-02-02 07:01:55 · answer #8 · answered by bette69 5 · 0⤊ 0⤋
Every hair color has exciting ones and duds.
2007-02-02 07:09:11 · answer #9 · answered by mary 2 · 0⤊ 0⤋
bolnd men
2007-02-02 07:01:39 · answer #10 · answered by gay pharoah tha don 1 · 1⤊ 0⤋