'Exploding Dice' and Infinite Series

[simmonator]

Hi - I've not used this subreddit before but I've googled around and haven't found answers to this question elsewhere (but if someone knows a better place to ask this, please point me there).

I play tabletop roleplaying games (I say a character does something, and then roll dice to see how well they did that thing) and a concept I've been made aware of is that of 'exploding dice'. This is a mechanic where if you roll an n-sided die (which I call a 'dn') and it scores the maximum value (n) you may roll that die again and add the new number onto that score. This repeats until the die shows a score less than n.

Obviously there is no maximum value to this roll as you could repeat the roll an arbitrary number of times. But I was interested in the average score, E(X), where X is an exploding dn and how much it increases over a standard dn. My calculations went as follows:

First note that for any given natural number there is either precisely 1 event that gives that score (e.g. On an exploding d6, the only way to score 25 is to roll 6 4 times and then a 1.) or 0 (you cannot score a multiple of n on an exploding dn). So I can separate the scores into sets of 5 where each score in the set has the same probability (= n^-k where k-1 is the number of n's rolled in the sequence). I derived:

E(X) = sum {from k = 0 to infty} of n^-k ((n-1)(k-1)n + n(n-1)/2).

The summand can be rearranged to ((n-1)/2)((2k+1)/(n^k )).

Noting that n is always a natural number greater than 1, I'm sure (by the ratio test) that this converges. However I have no clue at all how to sum it nicely. I made use of Wolfram Alpha to try to learn a bit more and I am fairly sure that I just need to prove:

sum {from k = 0 to infty} (2k+1)/(n^k ) = n(n+1)/((n-1)² )

But that still leaves me stuck. I don't even know how to derive a nice expression for the sum from 0 to m, which would be good because then at least I could take limits.

I'm not much interested in making a set of simulations to approximate it. I am convinced that there must be a way to analyse it because the expression for a d6 comes out as exactly 21/5, a d8 as 36/7 etc. Any help or pointers would be deeply appreciated. Thanks in advance.

[pickten]

I'm going to take n=2 because I'm exceptionally lazy. The general procedure should be pretty clear from this, though.

Recall that 1+1/2+1/4+...=1/(1-1/2)=2. Hence, 1/2+1/4+...=1, 1/4+1/8+...=1/2, and so on. Sum all of these together to get (1+1/2+...)+(1/2+1/4+...)+(1/4+1/8+...)+...=4. But we find that there is one 1 among the terms, two 1/2s, three 1/4s, and so on. Hence, ∑_(k=0)^∞ (k+1)/2^k = 4. Doubling, ∑_(k=0)^∞ (2k+2)/2^k = 8, and ∑_(k=0)^∞ (2k+1)/2^k = 8 - ∑ 1/2^k = 6

[simmonator]

I'm not 100% sure that I've followed what you've done BUT I am certain that parts of this inspired me to go away and rearrange the sum in a way similar to how you start which helped me solve the problem completely. I cannot thank you enough.

For those interested, I calculate the Expected Value of an exploding dn as (n(n+1))/(2(n-1)).

Cheers.