P(X > Y - 1/2), given X and Y are uniform(0,1) I'm trying to calculate this, using the law of total probability. My understanding is that the answer should be 1/2.
$\displaystyle P\left(X > Y - \frac{1}{2}\right) = \int_{0}^{1} P\left(X > Y - \frac{1}{2} \middle| Y = y\right) \cdot f_Y(y) \cdot dy$
Solving the first term on the integral:
$\displaystyle P\left(X > Y - \frac{1}{2} \middle| Y = y\right) = 1 - F_x\left(y-\frac{1}{2}\right) = 1 - \int_{0}^{y-\frac{1}{2}} dx = \frac{3}{2} - y$
And combining it with the second:
$\displaystyle P\left(X > Y - \frac{1}{2}\right) = \int_{0}^{1} \left(\frac{3}{2} - y\right) \cdot 1 \cdot dy = 1$
Which seems incorrect.. could someone please point out what I'm missing?

For $0<y<1$ we have
$$P\left(X>y-\frac{1}{2}\right)=
\begin{cases}
1 & \text{ if } y\le \frac{1}{2}\\
\frac{3}{2}-y & \text{ if } y>\frac{1}{2}
\end{cases}
$$
So
$$P\left(X>Y-\frac{1}{2}\right)=\int_0^{1/2}1\,dy+\int_{1/2}^1 \left(\frac{3}{2}-y\right) dy=\frac{1}{2}+\frac{3}{8}=\frac{7}{8}$$
