Probability that number of heads exceeds sum of die rolls

Let X denote the sum of dots we see in 100 die rolls, and let Y denote the number of heads in 600 coin flips. How can I compute 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 1 since E(X) = 350, E(Y) = 300, \text{Var}(X) \approx 292, \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.

Answer

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

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.]

Attribution
Source : Link , Question Author : Community , Answer Author : BruceET

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