A friend once told me that even ignoring the size of the scoops, a "spindown" D20 (Where neighboring faces are numbered sequentially, for easy look-ups.) produces different odds than a normal D20, which has more varied neighboring faces. Is this true?
Solved – Does the distribution of dice numbers per se matter
dice
Related Solutions
You should not do a calculation of probability for an event deemed surprising post hoc as if it were an event specified before it was rolled (observed).
It's very difficult to to do a proper calculation of post hoc probability, because what other events would have been deemed at least as surprising depends on what the context is, and also on the person doing the deeming.
Would three ones twice in a row at an earlier or later stage of the game have been as surprising? Would you rolling three ones have been as surprising as him rolling them? Would three sixes be as surprising as three ones? and so on... What is the totality of all the events would have been surprising enough to generate a post like this one?
To take an extreme example, imagine a wheelbarrow-full-of-dice (ten thousand, say), each with a tiny individualized serial number. We tip the barrow out and exclaim "Whoah, what are the chances of getting this?" -- and if we work it out, $P(d_1=3)\cdot P(d_2=6)\cdot \ldots P(d_{10000}=2)$ is $6^{-10000}$. Astronomically small. If we repeat the experiment, we get an equally unusual event. In fact, every single time we do it, we get an event so astronomically unbelievably small that we could almost power a starship with it. The problem is that the calculation is meaningless, because we specified the event post-hoc.
(Even if it were legitimate to do the calculation as if it were a pre-specified event, it looks like you have that calculation incorrect. Specifically, the probability (for an event specified before the roll) of taking three dice and rolling $(1,1,1)$ is $(1/6)^3 = 1/216$, because the three rolls are independent, not $1/56$, and the probability of doing it twice out of a total of two rolls is the square of that - but neither the condition of being pre-specified nor the "out of two rolls" actually hold)
Exact solutions
The number of combinations in $n$ throws is of course $6^n$.
These calculations are most readily done using the probability generating function for one die,
$$p(x) = x + x^2 + x^3 + x^4 + x^5 + x^6 = x \frac{1-x^6}{1-x}.$$
(Actually this is $6$ times the pgf--I'll take care of the factor of $6$ at the end.)
The pgf for $n$ rolls is $p(x)^n$. We can calculate this fairly directly--it's not a closed form but it's a useful one--using the Binomial Theorem:
$$p(x)^n = x^n (1 - x^6)^n (1 - x)^{-n}$$
$$= x^n \left( \sum_{k=0}^{n} {n \choose k} (-1)^k x^{6k} \right) \left( \sum_{j=0}^{\infty} {-n \choose j} (-1)^j x^j\right).$$
The number of ways to obtain a sum equal to $m$ on the dice is the coefficient of $x^m$ in this product, which we can isolate as
$$\sum_{6k + j = m - n} {n \choose k}{-n \choose j}(-1)^{k+j}.$$
The sum is over all nonnegative $k$ and $j$ for which $6k + j = m - n$; it therefore is finite and has only about $(m-n)/6$ terms. For example, the number of ways to total $m = 14$ in $n = 3$ throws is a sum of just two terms, because $11 = 14-3$ can be written only as $6 \cdot 0 + 11$ and $6 \cdot 1 + 5$:
$$-{3 \choose 0} {-3 \choose 11} + {3 \choose 1}{-3 \choose 5}$$
$$= 1 \frac{(-3)(-4)\cdots(-13)}{11!} + 3 \frac{(-3)(-4)\cdots(-7)}{5!}$$
$$= \frac{1}{2} 12 \cdot 13 - \frac{3}{2} 6 \cdot 7 = 15.$$
(You can also be clever and note that the answer will be the same for $m = 7$ by the symmetry 1 <--> 6, 2 <--> 5, and 3 <--> 4 and there's only one way to expand $7 - 3$ as $6 k + j$; namely, with $k = 0$ and $j = 4$, giving
$$ {3 \choose 0}{-3 \choose 4} = 15 \text{.}$$
The probability therefore equals $15/6^3$ = $5/36$, about 14%.
By the time this gets painful, the Central Limit Theorem provides good approximations (at least to the central terms where $m$ is between $\frac{7 n}{2} - 3 \sqrt{n}$ and $\frac{7 n}{2} + 3 \sqrt{n}$: on a relative basis, the approximations it affords for the tail values get worse and worse as $n$ grows large).
I see that this formula is given in the Wikipedia article Srikant references but no justification is supplied nor are examples given. If perchance this approach looks too abstract, fire up your favorite computer algebra system and ask it to expand the $n^{\text{th}}$ power of $x + x^2 + \cdots + x^6$: you can read the whole set of values right off. E.g., a Mathematica one-liner is
With[{n=3}, CoefficientList[Expand[(x + x^2 + x^3 + x^4 + x^5 + x^6)^n], x]]
Best Answer
In theory: No; each side of a perfect n-sided die has a 1 in n chance of appearing.
In the real world: Yes; die shapes are imperfect, often favoring opposing sides, hence the tradition of making sure opposing sides are equal on average. As explained in this answer. Having all high numbers on one side also makes it more easy to fake fair rolls, like merely spinning or skittering the die.