[Math] How to calculate the sum of combinations of 1000 dice rolls

probability

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?

Best Answer

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

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

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

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

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 $m\sigma ^2$, 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.

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