[Math] Sum of $4$ dice rolls greater than the product

combinatoricsdiceprobability

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!

Best Answer

Case work:

  • If 3 or 4 of the rolls come up 1, it's straightforward to see that the desired inequality holds.
  • If 0 or 1 of the rolls come up 1, then we can show that the inequality never holds. (The product of three integers, all $>1$, is always at least 2 greater than their sum.)
  • The remaining case is when exactly 2 of the rolls come up 1. Empirically, you can find that the only case when the inequality is satisfied is when the rolls are some permutation of $3,2,1,1$ or $2,2,1,1$.

There is only 1 combination of all 1's, $5\times 4=20$ combinations with three 1's, $4\times 3=12$ permutations of $3,2,1,1$, and $\binom{4}{2}=6$ permutations of $2,2,1,1$. This adds up to a total of 39 rolls that satisfy the inequality. There are $6^4=1296$ total rolls, so the probability that the inequality holds is $39/1296 \approx .03009$.

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