<s>[INST] Sum of $4$ dice rolls greater than the product 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!
 [/INST] A solution to the equation has at least one $1$.
When you roll all 2s, the equation is false.
$2 + 2 + 2 + 2 > 2 \times 2 \times 2 \times 2$
$8 > 16 \implies false$
Adding one to any of the numbers always maintains the falsehood of this equation, or the truth of the reverse.  Assuming the opposite equation:
$x_1 + x_2 + x_3 + x_4 \le x_1x_2x_3x_4$
Adding $1$ to one of the die results in the product results in a value that is $x_2x_3x_4$ higher.
$(x_1 + 1)x_2x_3x_4 = x_1x_2x_3x_4 + x_2x_3x_4$
Adding $1$ to one of the die results in the sum makes the sum $1$ higher, and $x_2x_3x_4 >= 1$, so we can say:


*

*If $x_1 + x_2 + x_3 + x_4 \le x_1x_2x_3x_4$, then $(x_1 + 1) + x_2 + x_3 + x_4 \le (x_1 + 1)x_2x_3x_4$.


Therefore, any solution to $x_1 + x_2 + x_3 + x_4 > x_1x_2x_3x_4$ must have at least one $1$.
The lowest possible dice roll with exactly one $1$ goes like this:
$1 + 2 + 2 + 2 > 1 \times 2 \times 2 \times 2$
$7 > 8 \implies false$
Therefore, the solution must require at least two $1$s.
Three or more $1$s.
If there are at least 3 $1$s, the equation is always true.  Note this includes the case of all four $1$s.
$1 + 1 + 1 + x > 1 \times 1 \times 1 \times x$
$x + 3 > x \implies true$
Exactly two $1$s.
With exactly two $1$s, let's explore the other two values.
If one of the other values is a $2$, then that restricts the other value to $3$ or less.
$1 + 1 + 2 + 2 > 1 \times 1 \times 2 \times 2$
$6 > 4 \implies true$
$1 + 1 + 2 + 3 > 1 \times 1 \times 2 \times 3$
$7 > 6 \implies true$
$1 + 1 + 2 + 4 > 1 \times 1 \times 2 \times 4$
$8 > 8 \implies false$
If both the other values are at least $3$, then we don't have a solution.
$1 + 1 + 3 + 3 > 1 \times 1 \times 3 \times 3$
$8 > 9 \implies false$
All solutions


*

*All four $1$s.  1 solution.

*Three $1$s and we don't care what the last value is (this excludes the four $1$s case): 5 possible choices for the last value, 4 choices as to on which die it occurs: 20 solutions

*Exactly two $1$s: The other two values are both $2$.  There are $4\choose2$ or 6 choices to distribute the two $2$s.

*Exactly two $1$s: The other two values are $2$ and $3$.  There are four choices for the $2$, leaving three choices for the $3$, for a total of 12 solutions here.


I count 39 possible solutions, out of $6^4$ possibilities.
$39\div1296 = 13\div432 \approx0.0301 $
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