Given the symmetry of the RR/WW card distribution, you can (and Olaf did) legitimately rephrase the original question from:
'The face is white. What are the chances that the other side of the card is also white?'
to
'What are the chances of drawing a card with the same colour on both sides'
...at which point you're just counting cards.
This solution scales so long as the distribution of RR/WW cards is identical.
ie. given 3 RW, 5 RR, and 5 WW cards (13 total), the correct answer to both the original question and the 'rephrased question' is now 10/13.
Equivalency proof?
R = Number of Red/Red Cards
W = Number of White/White Cards
M = Number of Red/White Cards
1) given white, 2nd face also white. p(wg,w) = 2W/(2W+M)
2) given white, 2nd face red. p(wg,r) = M/(2W+M)
3) given red, 2nd face also red. p(rg,r) = 2R/(2R+M)
4) given red, 2nd face white. p(rg,w) = M/(2R+M)
The chances of seeing a white face in the first side are : (2R+M)/(2R+2W+2M)
For R=W, this simplifies to (2R+M)/(4R+2M) = 1/2
(and likewise for red)
p(w,w) = 2W/(4W+2M)
p(w,r) = M/(4W+2M)
p(r,r) = 2W/(4W+2M)
p(r,w) = M/(4W+2M)
5) p(same) = p(r,r)+p(w,w) = 4W/(4W+2M) = 2W/(2W+M)
6) p(diff) = p(r,w)+p(w,r) = 2M/(4W+2M) = M/(2W+M)
The terms 1) and 5) are identical for R=W