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
$2 \cdot \left((\frac{1}{64} \cdot 1) + (\frac{1}{32} \cdot \frac{1}{3}) + (\frac{1}{16} \cdot \frac{1}{15})\right) = 0.0604166666 \neq 5/16.$
Best Answer
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 $$p={3\over8}\cdot{1\over2}+{1\over8}\cdot1={5\over16}\ .$$