Hi Whymme,

Thank you for the response. I'm still not convinced. You slightly
misinterpreted what I said: prob. 2/3 is total prob. for both colors,
(i.e. prob. that the color on the other side is the same as the one you see, and it is 2/3 just because there are 2 out of 3 cakes colored this way. In case of 3 colors there would be AA,BB,CC,AB,AC,BC
6 cakes and the prob. of the other side being the same 1/2 or 3 out of 6 for all the colors combined). So for a specific color the total prob. should be devided by number of colors, for they "OR"ed into the
total prob. because of the symmetry.

If you look at the Bayes probabilities for the 3 cakes

P(BB)=1/3 prob to get Brown/Brown cake
P( not BB)=2/3 prob of not BB cake

P(side B/BB)=1 you are looking at Brown side
P(side B/not BB)=1 you are looking at Brown side

Then

P(BB/side B)=P(side B/BB)* P(BB) / ( P(side B/BB)*P(BB)+P(side B/not BB)*P( not BB) )= 1*1/3 / (1*1/3+1*2/3) = 1/3

It seems the trick is in the fact that once you picked the cake and
see a side there is no flexibility on choosing the other: it is already chosen and the prob. is 1 out of 3 for 2 colors or 1 out of 6 for 3 colors.

Please correct me if I'm still wrong.

Thanks.

BG