The desired probability is $\frac{\displaystyle\sum_{n=0}^{505}\binom{2023}{4n}}{2^{2023}}$ - proof of this is left to the reader. Note that the numerator is the sum of the coefficients of the fourth powers of $(x+1)^{2023}$. A roots of unity filter gives that this value is \begin{align*} \frac{(1+1)^{2023}+(1+i)^{2023}+(1-1)^{2023}+(1-i)^{2023}}{4} &= \frac{2^{2023}+\left(\sqrt{2}e^{\frac{\pi i}{4}}\right)^{2023}+\left(\sqrt{2}e^{\frac{-\pi i}{4}}\right)^{2023}}{4} \\ &=\frac{2^{2023}+2^{1011}\sqrt{2}\left(e^{\frac{2023\pi i}{4}}+e^{\frac{-2023\pi i}{4}}\right)}{4} \\ &=\frac{2^{2023}+2^{1011}\sqrt{2}\left(e^{\frac{-\pi i}{4}}+e^{\frac{\pi i}{4}}\right)}{4} \\ &=\frac{2^{2023}+2^{1011}\left(\sqrt{2}e^{\frac{-\pi i}{4}}+\sqrt{2}e^{\frac{\pi i}{4}}\right)}{4}\\ &=\frac{2^{2023}+2^{1011}\left(1-i+1+i\right)}{4} \\ &=\frac{2^{2023}+2^{1012}}{4} \\ &=2^{2021}+2^{1010}, \end{align*}so our probability is $\frac{2^{2021}+2^{1010}}{2^{2023}}=\frac{1}{4}+2^{-1013}$. Extrapolating, $(x,y)=(2,1013)$, so our answer is $1013-20=\boxed{993}$.