Painted cube probability 
Each face of a cube is painted either red or blue, each with
  probability 1/2. The color of each face is determined independently.
  What is the probability that the painted cube can be placed on a
  horizontal surface so that the four vertical faces are all the same
  color?

The correct answer is $5/16$. I don't understand why my solution (below) is incorrect.

I will compute the probability that the four vertical faces are painted red and then multiply by two to cover the symmetric blue case. To compute this value, I will condition on the number of faces painted red.
Case 1: All faces are painted red. This occurs with probability $1/64$ and guarantees the four vertical faces to be red. 
Case 2: Five faces are painted red. This occurs with probability $1/32$ and given this event occurs, there is a ${5\choose 4}/{6\choose 4} = 1/3$ chance of choosing four red-painted sides to be the vertical sides.
Case 3: Four faces are painted red. This occurs with probability $1/16$ and given this event occurs, there is a ${4\choose 4}/{6 \choose 4} = 1/15$ chance of choosing four red-painted sides to be the vertical sides. 
Note that I don't need to consider the cases where there are less than $4$ red-painted sides, since their probabilities will equal $0$. Therefore, my final answer is
$2 \cdot \left((\frac{1}{64} \cdot 1) + (\frac{1}{32} \cdot \frac{1}{3}) + (\frac{1}{16} \cdot \frac{1}{15})\right) = 0.0604166666 \neq 5/16.$

 A: There are various mistakes in your attempt.
A pair of opposite faces is equally colored with probability ${1\over2}$, and the number of such single-color pairs  is binomially distributed on $[0..3]$. With probability ${3\over8}$ we have exactly two such pairs, and with probability ${1\over2}$ the color of these two pairs is the same. With probability ${1\over8}$ we have exactly three such pairs, among them at least two of the same color. It follows that the required  probability $p$ is given by
$$p={3\over8}\cdot{1\over2}+{1\over8}\cdot1={5\over16}\ .$$
A: There are $\binom66$ possibilities to have six sides red, so there are six possibilities to have four vertical faces of the same color — any placement will do.
There are $\binom65$ possibilities to have five sides red, so there are two possibilities to have four vertical faces of the same color — blue must be on top or on bottom.
There are $\binom64$ possibilities to have four sides red, so there are two possibilities again to have four vertical faces of the same color — the two blue faces must be on bottom and on top.
There are $\binom63$ possibilities to have three sides red, so there is no possibility to have four vertical faces of the same color.
Two red faces correspond to four blue ones, one red face to five blue ones and zero red faces to six blue ones.
So we have $6+2+2+0+2+2+6=20$ possibilities to have four vertical faces out of $64$ possibilities in total, which gives a probability of $5/16$.
