# An extension of a game with two dice

This question is an extension of a previous question already correctly answered: A game with two dice. In that question, we had two dice: the first one, when rolled, determined the number of times that the second die would be rolled. The goal was to quantify, before any dice was rolled, the probabilities of the sum of the second die rolls (the sum is always between 1 and 36). Now we have two imaginary dice, both with infinite faces. Again, we roll the first die that will determine how many times we’ll roll the second die. The probability functions for each die are power functions:

For the first die $P(x) = ax^{-k}$ and for the second die $P(x) = bx^{-q}$, where $a, b, k$ and $q$ are constants, $x\in{N^+}$, $k, q>1$, and $0<a, b<1$

We want to calculate the probabilities of the sum of the second die, before any dice is rolled.

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What have you tried? Specifically, have you tried to understand and generalize the generating function approach to the previous question? – Greg Martin Mar 10 '14 at 17:20
Also note that you must have $k,q>1$, not $k,q>0$, for convergence, and that necessarily $a=1/\zeta(k)$ and $b=1/\zeta(q)$. (I'm assuming the infinitely many faces are labeled with $x$ in the positive integers.) – Greg Martin Mar 10 '14 at 17:21
@GregMartin. IMHO, the function also converge for k,q>0. I can do the PGF but the problem is to group all of them by the same sum – Luis Gonilho Mar 10 '14 at 17:33
I'd say the PGF will be $$ab\sum_{i=1}^{n}{i^{-(k+iq)}}\$$ The next question is how to group all the combinations – Luis Gonilho Mar 10 '14 at 17:57
For a single die, if the probability of the positive integer $x$ being chosen is $ax^{-k}$, then the total probability is $\sum_{x=1}^\infty ax^{-k}$. This sum does not converge if $x\le1$. – Greg Martin Mar 10 '14 at 21:34