Two walkers returning home with one always behind

Two independent simple symmetric random walks run on the integers: Tort starts at $0$ and Bort starts at $4$, each moving $\pm1$ with equal probability every step. Find the probability that after $10$ steps both are back at their starting points and Tort has been strictly to the left of Bort after every one of the $10$ steps. Writing the answer as

$\frac{\binom{a}{b}^{p}-\binom{c}{d}^{p}}{2^{r}}$

with $b\le a/2$ and $d\le c/2$, find the sum of the squares of the six integers $a,b,c,d,p,r$.

$638$. View the pair as one walk in the plane with diagonal steps; the gap $Y-X$ moves by $0$ or $\pm2$, so it can only reach $0$ by landing on the diagonal. Reflecting across $x=y$, bad paths from $(0,4)$ back to $(0,4)$ match paths to $(4,0)$, giving $\frac{\binom{10}{5}^2-\binom{10}{3}^2}{2^{20}}$. Then $100+25+100+9+4+400=638$.