Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$
There are several cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$.
The four cases are:
Case 1: $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$.
Case 2: $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$.
Case 3: $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$.
Case 4: $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$.
Each case has a $\frac{1}{4}$ chance of occurring (as it requires two coin flips).
For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\frac{1}{2}$.
For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\frac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \frac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\frac{1}{2}$.
By symmetry, Case 3 has the same success rate as Case 2.
For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \tfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:
[asy] filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1)); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); [/asy]
The shaded area is $\frac{1}{4}$, which means the probability for success for case 4 is $\frac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$).
Adding up the success rates from each case, we get:
$\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\right) = \boxed{\textbf{(B) }\frac{7}{16}}$.