A fair six-sided die is thrown n times. How to calculate certain probabilities of the sum S of these rolls? To name these probabilities:
the sum S being less than a threshold N
the sum S being at least the threshold N
the sum S being exactly the threshold N
the sum being in a certain interval $[N_1,N_2] $
e.g. after 100 rolls less than 367, at least 342, exactly 350 and in $ [350,351]$?
How could I approach that?
Addendum: Could I valuably do that with a Monte-Carlo-simulation if the probabilities should only be in a certain range of precision?
I wrote a R-script for the Monte-Carlo and checked it with a "manual" calculation. We lnow that with independent variables the expected value and the variance are linear, therefor i can calculate that and get the standard deviation $\sigma$. With that and the PDF of the standard normal distribution, which we know is usable here because of Irwin Hall, i can calculate the percentage.
Does that make sense? Is there a more precise approach?
 A: You are looking for
 $$ 
 \eqalign{ 
   & N(s,m) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ 
   {\rm 1} \le {\rm integer}\;x_{\,j}  \le 6 \hfill \cr  
   x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,m}  = s \hfill \cr}  \right. =   \cr  
   &  = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ 
   {\rm 0} \le {\rm integer}\;y_{\,j}  \le 5 \hfill \cr  
   y_{\,1}  + y_{\,2}  + \; \cdots \; + y_{\,m}  = s - m \hfill \cr}  \right. =   \cr  
   &  = N_b (s - m,5,m) \cr}  
 $$ 
where
 $$ 
 N_b (s,r,m) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ 
   {\rm 0} \le {\rm integer}\;x_{\,j}  \le r \hfill \cr  
   x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,m}  = s \hfill \cr}  \right. 
 $$ 
and is given by
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers  }s,m,r} \right.\quad  =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)} 
{\left( { - 1} \right)^k \binom{m}{k}
 \binom
 { s + m - 1 - k\left( {r + 1} \right) } 
 { s - k\left( {r + 1} \right)}\ }
$$
as thoroughfully described in this related post.
Note that the Cumulative Number of solutions for $s$ up to $S$ is given by
 $$ 
 \eqalign{ 
   & M_b (S,r,m) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ 
   {\rm 0} \le {\rm integer}\;x_{\,j}  \le r \hfill \cr  
   x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,m}  \le S \hfill \cr}  \right. =   \cr  
   &  = \sum\limits_{\left( {0\, \le } \right)\,\,s\,\, \le \,S\,} {N_b (s,r,m)} \quad  = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over {r + 1}}\, \le \,m} \right)} 
  {\left( { - 1} \right)^k \left( \matrix{ 
   m \hfill \cr  
   k \hfill \cr}  \right)\left( \matrix{ 
   S + m - k\left( {r + 1} \right) \cr  
   S - k\left( {r + 1} \right) \cr}  \right)}  \cr}  
 $$ 
which answers to your questions about getting  "no less than ..."  or "between $S_1$ and $S_2$".
We have that 
 $$ 
 \left( {r + 1} \right)^{\,m}  = \sum\limits_{\left( {0\, \le } \right)\,\,s\,\,\left( { \le \,r\,m} \right)\,} {N_b (s,r,m)}  
 $$ 
and therefore
 $$ 
 P_b (s,r,m) = {{N_b (s,r,m)} \over {\left( {r + 1} \right)^{\,m} }} 
$$
is the probability distribution of the sum $s$ of $m$ i.i.d. discrete variables, with support $[0,r]$.
For large values of throws ($m$), the Probability tends to the probability distribution of the
sum of $m$ continuous random variables uniformly distributed on $[-1/2,r+1/2]$, which is
known as Irwin Hall Distribution,  and which in turn becomes asymptotic to a normal distribution
 with mean and variance equal to $m$ times the mean and variance of the uniform random variable on $[-1/2,r+1/2]$, i.e.
 $$ 
 \eqalign{ 
   & P_{\,b} (s,r,m) = {{N_{\,b} (s,r,m)} \over {\left( {r + 1} \right)^{\,m} }} \approx   \cr  
  &  \approx {1 \over {\sqrt {2\pi m\sigma ^{\,2} } }}e^{\, - \,{{\left( {s - m\mu } \right)^{\,2} } \over {2m\sigma ^{\,2} }}}
  = {{\sqrt {6/\pi } } \over {\sqrt {m\left( {\left( {r + 1} \right)^{\,2} } \right)} }}e^{\, - \,6{{\left( {s - mr/2} \right)^{\,2} } \over {m\left( {\left( {r + 1} \right)^{\,2} } \right)}}}  \cr} 
$$
A: This is not an exact answer but an approximation using Central Limit Theorem
First we need to find $E(X)$ and $VAR(X)$ where $X$ is a random variable with Probability Mass Function:
$P(X=i)=1/6,$ $for$ $i=1,2,3,4,5,6$
So we have:
$E(X)=\frac{1}{6}(1+2+3+4+5+6)=\frac{21}{6}=\frac{7}{2}$
$E(X^2)=\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2)=\frac{91}{6}$
$VAR(X)=E(X^2)-(E(X))^2=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}$
We will use Central Limit Theorem to calculate the probability of the sum of 100 rolls being less than 367.(Your first example that is)
$$P(\sum_{i=1}^{100} X_i < 367)\simeq P(\sum_{i=1}^{100} X_i \leq 367.5)=P(\frac{\sum_{i=1}^{100} X_i-100*E(X)}{\sqrt{100*VAR(X)}} \leq \frac{367.5-100*E(X)}{\sqrt{100*VAR(X)}}) \simeq \Phi(\frac{367.5-100*E(X)}{\sqrt{100*VAR(X)}}) \simeq \Phi(1.0247) \simeq 0.847$$
You can work in a similar way to calculate the probability of the sum of $n$ rolls being greater than a certain number $A$ or between certain numbers $A$ and $B$. As for the $\Phi$ table you can find it easily on the internet. If you have any questions about how to use the Theorem for other similar problems ask in the comments
