# Probability that number of heads exceeds sum of die rolls

Let $$XX$$ denote the sum of dots we see in $$100100$$ die rolls, and let $$YY$$ denote the number of heads in $$600600$$ coin flips. How can I compute $$P(X > Y)?P(X > Y)?$$

Intuitively, I don’t think there’s a nice way to compute the probability; however, I think that we can say $$P(X > Y) \approx 1P(X > Y) \approx 1$$ since $$E(X) = 350E(X) = 350$$, $$E(Y) = 300E(Y) = 300$$, $$\text{Var}(X) \approx 292\text{Var}(X) \approx 292$$, $$\text{Var}(Y) = 150\text{Var}(Y) = 150$$, which means that the standard deviations are pretty small.

Is there a better way to approach this problem? My explanation seems pretty hand-wavy, and I’d like to understand a better approach.

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


In R, the function sample(1:6, 100, rep=T) simulates 100 rolls a fair die;
the sum of this simulates $$XX$$. Also rbinom is R code for simulating
a binomial random variable; here it’s $$Y.Y.$$ The difference is $$D = X – Y.D = X - Y.$$
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.]