You, your parents, your sister, go to visit grandma for her birthday. Grandma made a cake for the party. If she puts $20$ raisins in the cake at random in the cake, and she divides the cake into $5$ equal pieces, what's the probability that you get one more than than your sister?
Let X be the number of raisins you have and Y be the number of raisins she has.
$X,Y \sim Bin(20, 1/5)$
Either $P[X-Y = 1]$ or $P[X-Y \neq 1]$.
$P[X-Y > 1] + P[X-Y=1]+ P[X-Y<1]=1$.
It's true that $X, Y$ are each individually binomially distributed... but I don't think that's quite as helpful as you want it to be, because $X$ and $Y$ are quite dependent on one another. (Example: if you know that $X = 15$, then this forces $Y \leq 5$.)
I'd consider some cases separately and compute the probability of each case directly:
To compute any one of these, we can use a multinomial distribution. For instance, the probability that you get 3 raisins and your sister gets 2 (hence, 15 go to the unused portion of the cake) is $$\binom{20}{3, 2, 15}(1/5)^3 \cdot (1/5)^2 \cdot (3/5)^{15} = \left( \frac{20!}{3! \cdot 2! \cdot 15!} \right) (1/5)^3 \cdot (1/5)^2 \cdot (3/5)^{15}$$ which can be simplified, of course. You then have $10$ such terms which represent disjoint events, and whose union is the event that you want, so you can add their probabilities to get your answer.
(It's possible that there's a simpler way to do this, but this was the simplest one I could think of right now.)