4. [3] Three brothers Abel, Banach, and Gauss each have portable music players that can share music with each other. Initially, Abel has 9 songs, Banach has 6 songs, and Gauss has 3 songs, and none of these songs are the same. One day, Abel flips a coin to randomly choose one of his brothers and he adds all of that brother's songs to his collection. The next day, Banach flips a coin to randomly choose one of his brothers and he adds all of that brother's collection of songs to his collection. Finally, each brother randomly plays a song from his collection with each song in his collection being equally likely to be chosen. What is the probability that they all play the same song?
Answer: $\frac{1}{288}$. If Abel copies Banach's songs, this can never happen. Therefore, we consider only the cases where Abel copies Gauss's songs. Since all brothers have Gauss's set of songs, the probability that they play the same song is equivalent to the probability that they independently match whichever song Gauss chooses. Case 1: Abel copies Gauss and Banach copies Gauss (1/4 chance) - The probability of songs matching is then $1 / 12 \cdot 1 / 9$. Case 2: Abel copies Gauss and Banach copies Abel ( $1 / 4$ probability) - The probability of songs matching is then $1 / 12 \cdot 1 / 18$. We add the two probabilities together to get $1 / 4 \cdot 1 / 12 \cdot(1 / 9+1 / 18)=1 / 288$