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The following PDF gives an explanation on page 11. Unfortunately I do not know how to reproduce it here.

http://web.mit.edu/~qchu/Public/TopicsInGF.pdf

In short, I am not sure how the symmetry with 18 and 10 arises and the other less trivial parts of the proof (why for example are we looking for coefficients of x6 and not x18)? I have a feeling I am missing something simple with this approach.

user73041
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Not that it really matters, but in actual dice the $1$ is opposite the $6$, the $2$ is opposite the $5$, and the $3$ is opposite the $4$. Thus if the dice are thrown onto a glass coffee table, the short gambler below the table and looking up sees $7-k$ whenever the person looking from above sees $k$.

So if from above the dice show $a,b,c,d$, then from below they show $7-a,7-b,7-c,7-d$. Thus from below they show sum $28-(a+b+c+d)$.

It follows with $4$ dice, a sum of $x$ and $28-x$ are always equally likely. More generally, with $d$ dice a sum of $x$ and $7d-x$ are equally likely.

André Nicolas
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  • Thank you, but for example here we have 4 boxes and 18 values to distribute among these boxes with no restriction - if I am correct. Why then is n = 28 and not n + k - 1 = 21? – user73041 Apr 20 '13 at 00:39
  • The coefficient of $x^6$ in $(1-x)^{-4}$ is $\binom{6+4-1}{3}$. This is unconnected to the $28$ stuff, whose only use was to change the problem from finding the coefficient of $x^{18-4}$ to finding the coefficient of $x^{10-4}$, which is simpler-looking. – André Nicolas Apr 20 '13 at 00:56
  • I understand now, thanks again, you've been a great help on several questions today. My only remaining question is about the simplification resulting in 1-x in the denominator. This happens often it seems in evaluating generating functions. Is it a simple algebraic manipulation? – user73041 Apr 20 '13 at 01:10
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    @user73041: Yes, it is a simple manipulation, and one you know. We started with $x+x^2+\cdots+x^k$, where $k=6$. This is $x(1+x+\cdots+x^{k-1}$). The thing in parentheses is a finite geometric series. For $x\ne 1$, its sum is $\frac{1-x^k}{1-x}$. In the unlikely event that you do not know this, you can verify it by calculating the product $(1-x)(1+x+\cdots+x^{k-1})$. There is a whole lot of cancellation. – André Nicolas Apr 20 '13 at 01:18
  • I had a feeling that's what it was but the extra certainty is always helpful, thanks! – user73041 Apr 20 '13 at 01:42