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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

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 TRUEs and FALSEs, the mean of which is its proportion of TRUEs--our Answer. Finally, the last statement gives the margin of error of a 95% confidence interval of the proportion of TRUEs (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.]

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

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 TRUEs and FALSEs, the mean of which is its proportion of TRUEs--our Answer. Finally, the last statement gives the margin of error of a 95% confidence interval of the proportion of TRUEs (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.]

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

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 TRUEs and FALSEs, the mean of which is its proportion of TRUEs--our Answer. Finally, the last statement gives the margin of error of a 95% confidence interval of the proportion of TRUEs (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--maybe more for probabilities so far from 1/2.]

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

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 TRUEs and FALSEs, the mean of which is its proportion of TRUEs--our Answer. Finally, the last statement gives the margin of error of a 95% confidence interval of the proportion of TRUEs (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.]

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

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

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

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 TRUEs and FALSEs, the mean of which is its proportion of TRUEs--our Answer. Finally, the last statement gives the margin of error of a 95% confidence interval of the proportion of TRUEs (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--maybe more for probabilities so far from 1/2.]