The Question
What is the probability, rolling $n$ six-sided dice twice, that their sum each time totals to the same amount? For example, if $n = 4$, and we roll $1,3,4,6$ and $2,2,5,5$, adding them gives
$ 1+3+4+6 = 14 = 2+2+5+5 $
What is the probability this happens as a function of $n$?
Early Investigation
This problem is not too hard for $n = 1$ or $n = 2$ via brute force...
For $n = 2$:
Tie at a total of $2$: $ \frac{1}{36} * \frac{1}{36} = \frac{1}{1296} $
Tie at a total of $3$: $ \frac{2}{36} * \frac{2}{36} = \frac{4}{1296} $
etc.
so the answer is $ \frac{1^2 + 2^2 + 3^2 + ... + 6^6 + 5^2 + ... + 1^2}{1296} = \frac{\frac{(6)(7)(13)}{6} + \frac{(5)(6)(11)}{6}}{1296} = \frac{146}{1296} $
Note that I use the formula: $\sum_{k=1}^{n}k^2=\frac{(n)(n+1)(2n+1)}{6}$.
Is there a way to do this in general for $n$ dice? Or at least a process for coming up with a reasonably fast brute force formula?
The Difficulty
The problem arises that the sum of squares is not so simple when we get to three dice.
Using a spreadsheet, I figured out we need to sum these squares for 3 dice:
$ 1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1 $
For a brute force answer of $\frac{4332}{46656}$. Note how we can no longer use the sum of squares formula, as the squares we need to sum are no longer linear.
Some Thoughts
I am no closer to figuring out an answer for $n$ dice, and obviously the question becomes increasingly more difficult for more dice.
One thing I noticed: I see a resemblance to Pascal's Triangle here, except we start with the first row being six $1$, not one $1$. Se we have:
1 1 1 1 1 1 1 2 3 4 5 6 5 4 3 2 1 1 3 6 10 15 21 25 27 27 25 21 15 10 6 3 1 1 4 9 16 25 36 46 52 54 52 46 36 25 16 9 4 1 ...
but that's still a process, not a formula. And still not practical for $n = 200$.
I know how to prove the formula for any cell in Pascal's Triangle to be $C(n,r) = \frac{n!}{r!(n-r)!}$... using induction; that doesn't really give me any hints to deterministically figuring out a similar formula for my modified triangle. Also there is no immediately obvious sum for a row of this triangle like there is (powers of 2) in Pascal's Triangle.
Any insight would be appreciated. Thanks in advance!