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For two dice rolls we can calculate the number of combinations for each summed total:

1. Rolling a 2: one chance (1&1)


2. Rolling a 3: two chances (2&1)(1&2)


3. Rolling a 4: three chances (3&1)(1&3)(2&2)


4. Rolling a 5: four chances (4&1)(1&4)(3&2)(2&3)


5. Rolling a 6: five chances (5&1)(1&5)(4&2)(2&4)(2&2)


6. Rolling a 7: six chances (6&1)(1&6)(5&2)(2&5)(4&3)(3&4)


7. Rolling an 8: five chances (6&2)(2&5)(5&3)(3&5)(4&4)


8. Rolling a 9: four chances (6&3)(3&6)(5&4)(4&5)


9. Rolling a 10: three chances (6&4)(4&6)(5&5)


10. Rolling an 11: two chances (6&5)(5&6)


11. Rolling a 12: one chance (6,6)

How do we go about this for n dice rolls?
For example how do we find the total number of values which will sum to 150 if we roll 100 die?

Note that the answer to your problem in general, can be given as follows.

Let us define:
$$
eqalign{
& {rm No}{rm .},{rm of},{rm solutions},{rm to};left{ matrix{
{rm 1} le {rm integer};y_{,j} le r + 1 hfill cr
y_{,1} + y_{,2} + ; cdots ; + y_{,m} = s + m hfill cr} right. = cr
& = {rm No}{rm .},{rm of},{rm solutions},{rm to};left{ matrix{
{rm 0} le {rm integer};x_{,j} le r hfill cr
x_{,1} + x_{,2} + ; cdots ; + x_{,m} = s hfill cr} right. = cr
& = N_b (s,r,m) cr}
$$
then we have the formula

$$
N_b (s,r,m)quad left| {;0 le {rm integers };s,m,r} right.quad = sumlimits_{left( {0, le } right),,k,,left( { le ,{s over r}, le ,m} right)} {left( { - 1} right)^k left( matrix{
m hfill cr
k hfill cr} right)left( matrix{
s + m - 1 - kleft( {r + 1} right) cr
s - kleft( {r + 1} right) cr} right)}
$$

Also consider that the o.g.f. on the parameter $s$ is:

$$
F_b (x,r,m) = sumlimits_{0, le ,,s,,left( { le ,mr} right)} {N_b (s,r,m);x^{,s} } = left( {1 + x^{,1} + x^{,2} + ; cdots ; + x^{,r} } right)^{,m} = left( {{{1 - x^{,r + 1} } over {1 - x}}} right)^{,m}
$$

For more details have a look to the answers to this other post.

When the number of rolls ($m$) takes large values, the formula above becomes impractical and we shall resort to an asymptotic approximation.

To this regard note that each variable $x_j$ is a discrete uniform variable with support $[0,r]$, therefore with mean and variance given by $$
mu = {r over 2}quad sigma ^{,2} = {{left( {r + 1} right)^{,2} - 1} over {12}}
$$
The sum of $m$ such variables tends very quickly to be Normally distributed with mean $m mu$ and variance $msigma ^2$, that is

$$
p(s,r,m) = {{N_{,b} (s,r,m)} over {left( {r + 1} right)^{,m} }};; to ;{cal N}left( {m{r over 2},;m{{left( {r + 1} right)^{,2} - 1} over {12}}} right)
$$

Refer also to this related post.

This topic is called 'convolutions' in probability and computer science.
Because of the intensive and repetitive computation necessary, finding
exact probabilities of sums on n > 2 dice is usually done by computer
algorithm and several examples are available by googling 'convolutions discrete probability dice'; some have nice pictures, even if you ignore the code. Also, I found one well-written book chapter that shows formulas in mathematical rather than program form:
www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf

The 'envelope' of the PDF for two dice is formed by two straight lines. For three
dice the envelope is made of pieces of three parabolas, one for each tail, and an inverted
one in the center, with a result that is already suggestive of the shape of a normal density.
As n increases, envelopes very rapidly become very close to a normal PDF. As pointed out in comments, convergence is to be expected because of
the Central Limit Theorem. But the rate of convergence is astonishingly fast.

So for $n = 1000$, a normal distribution with the appropriate mean 1000(7/2) and variance 1000(35/12)
should give an extremely good approximations to normal, even reasonably far into the tails (remembering, of course, that 1000 rolls of a die cannot give values less than 1000 or
greater than 6000).

Dice with any number of sides could be treated in the same way for $n ge 2$.
A coin is also a 'two-sided die', and multiple coin tosses are binomial, which converges to normal. Even unfair dice would work, but some configurations of unfairness might converge more slowly to normal. Convolution algorithms work for unfair dice as well as fair ones, but results may be more difficult to summarize.