AMC12 2017 B

Answer (D): The set of all possible ordered pairs $(x,y)$ is bounded by the unit square in the coordinate plane with vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. For each positive integer $n$, $\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor = -n$ if and only if $\frac{1}{2^n} \le x < \frac{1}{2^{n-1}}$ and $\frac{1}{2^n} \le y < \frac{1}{2^{n-1}}$. Thus the set of ordered pairs $(x,y)$ such that $\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor = -n$ is bounded by a square with side length $\frac{1}{2^n}$ and therefore area $\frac{1}{4^n}$. The union of these squares over all positive integers $n$ has area $$ \sum_{n=1}^{\infty}\frac{1}{4^n}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}, $$ and therefore the requested probability is $\frac{1}{3}$. (It is also clear from the diagram that one third of the square is shaded.)