3 Pancakes Problem 
You have 3 pancakes in a stack. 1 is burned on both sides, 1 burned on
  1 side, 1 burned on no sides. What is P(burned on other side — burned
  on the top)?

Intuition (which is always suspect in probability questions) tells me that the answer should be 1/2, as you only have 2 pancakes that are burned on 1 side, and of that 2, only 1 is both on both sides.  Googling tells me some ppl think it's 1/2, some think it's 2/3 (http://forums.xkcd.com/viewtopic.php?t=568)
I'm trying to apply Bayes rule to this problem but am finding it a bit tricky, can someone show me the working?
 A: If you are serevd a random pancake in a random orientation, then each of the burnt sides occurs with equal probability as top side. But in two out of these three cases, the other side is burnt as well, so the answer is $\frac 23$.
To argue with Bayes, note that $P(\text{burnt-top}|A)=0$, $P(\text{burnt-top}|B)=\frac12$, $P(\text{burnt-top}|C)=1$ where $A,B,C$ are the three pancakes.
If $P(A)=P(B)=P(C)=\frac13$, you obtain
$P(B|\text{burnt-top})=\frac13 $ and $P(C|\text{burnt-top})=\frac23 $ from Bayes:
$$ P(B|\text{burnt-top})=\frac{P(\text{burnt-top}|B)P(B)}{P(\text{burnt-top})}=\frac{P(\text{burnt-top}|B)P(B)}{(P(\text{burnt-top}|A)P(A)+P(\text{burnt-top}|B)P(B)+P(\text{burnt-top}|C)P(C))}=\frac{\frac12\cdot\frac13}{0+\frac16+\frac13}.$$
Since (given a burnt top) the distinction burnt vs. non-burnt bottom is precisely that between $C$ and $B$, the answer to the original question is - again - $\frac23$.
A: There are six possibilities: Three pancakes, and seeing the one of either side of that pancake. Since you saw one side was burned, that narrows it down to three possibilities: You're seeing the burnt side of the black/white pancake, you're seeing one burnt side of the black/black pancake, or you're seeing the other burnt side of the black/black pancake. The probability is then $\frac{2}{3}$ that the other side is also black.
