Another way is by simulating a million match-offs between $X$ and $Y$
to approximate $P(X > Y) = 0.9907\pm 0.0002.$ [Simulation in R.]

    set.seed(825)
    d = replicate(10^6, sum(sample(1:6,100,rep=T))-rbinom(1,600,.5))
    mean(d > 0)
    [1] 0.990736
    2*sd(d > 0)/1000
    [1] 0.0001916057   # aprx 95% margin of simulation error

[![enter image description here][1]][1]

_Notes_ per @AntoniParellada's Comment:

In R, the function `sample(1:6, 100, rep=T)` simulates 100 rolls a fair die;
the sum of this simulates $X$. Also `rbinom` is R code for simulating
a binomial random variable; here it's $Y.$ The difference is $D = X - Y.$
The procedure `replicate` makes a vector of a million differences `d`.
Then `(d > 0)` is a logical vector of a million `TRUE`s and `FALSE`s, the `mean` of which is its proportion of `TRUE`s--our Answer. Finally, the last statement
gives the margin of error of a 95% confidence interval of the proportion
of `TRUE`s (using 2 instead of 1.96), as a reality check on the accuracy
of the simulated Answer. [With a million iterations one ordinarily expects
2 or 3 decimal paces of accuracy for probabilities--sometimes more for
probabilities so far from 1/2.] 


  [1]: https://i.sstatic.net/YqRwm.png