Probability of sum of 6 picked integers from [1..36] How we would calculate probability that sum is bigger than 92.5 if we pick 6 random numbers from [1..36] ?
Not putting them back (if took 1, then we can pick [2..36] etc.
I cannot think of how to put this because there are lot of variations of how this can fall. Unless to write some script.
 A: Personally I only know about generating function which produces the answer with aid of computer. It may be easier to calculate the answer using script, without using any combinatorical or generatingfunctionological technique. The number of ways to choose six distinct numbers whose sum is equal to or lower than 92 is denoted as $S(92)$:
$$S(92)=[x^{92}y^6]\dfrac{1}{1-x}\prod_{k=1}^{36}(1+x^ky)$$
where $[x^{92}y^6]$ indicates the coefficient of $x^{92}y^6$, and the total number of ways to choose six distinct numbers is
$$N={36 \choose 6} $$
Thus, the desired probability is $$\dfrac{N-S(92)}{N}$$
This particular question is quite complex yet widely known. I found a past post where more general question was answered. This is the link:
The number of ways to write a positive integer as the sum of distinct parts with a fixed length
The following generating function is probably more useful, so I quoted it from the above link. 

Let $q(n,k)$ be the number of partitions of $n$ into $k$ distinct
  parts. The generating function is $$Q_k(x)=\sum_{n\ge
> 0}q(n,k)x^n=\frac{x^{k+\binom{k}2}}{(1-x)(1-x^2)\dots(1-x^k)}\;.$$

