TITLE: Sum of $4$ dice rolls greater than the product
QUESTION [18 upvotes]: Let us roll a fair die $4$ independent times, and denote the outcomes as $X_1, X_2, X_3$ and $X_4$.
What is the probability of $X_1+X_2+ X_3+X_4 > X_1X_2 X_3X_4$?
My try:
I could get the answer for $2$ rolling case by enumerating possibilities, but couldn't get this larger problem. Can someone help me with this?
Thanks in advance for any help!

REPLY [4 votes]: Of course we can quickly simulate this, e.g. in R:
rolls = expand.grid(1:6, 1:6, 1:6, 1:6)
sum(apply(rolls, 1L, sum) > apply(rolls, 1L, prod))
# [1] 39

So there are 39 cases which fit the criterium out of $6^4$ in the sample space.
This approach allows us to easily generalize to $n$ rolls as long as $6^n * n * b$ fits in memory (where $b$ is the storage space for an integer), e.g. (*I've made some small adjustments to make it run more efficiently for large $n$):
p = sapply(2:10, function(n) {
  print(n)
  rolls = 
    as.matrix(do.call(expand.grid, lapply(integer(n), function(...) 1:6)))
  mean(rowSums(rolls) > apply(rolls, 1L, prod))
})
plot(2:10, p, type = 'o', lwd = 3L, xaxt = 'n', las = 1,
     xlab = '# of Dice', ylab = 'Pr[sum > product]',
     main = 'Extension of Dice Problem')
axis(side = 1L, at = 2:10)


To be clear, there's no inherent need to store all $n 6 ^ n$ dice rolls in memory, and this problem is trivial to stream so that the only RAM needed is $n * b+2 * b = O(n)$ storage; I only eschew the more memory-efficient approach for simplicity (the 2-line code here would be hard to match with explicitly arbitrarily nested loops)