smc

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 $|x-y|>\frac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality: 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: $\left(\frac{1}{4}\right)\cdot \left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}\right)=\boxed{\frac{7}{16}}$.