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Message: <blockquote data-quote="ggroy" data-source="post: 6011722" data-attributes="member: 83805">OK.  All the calculations in my previous two posts are wrong !!!It turns out for a dN "exploding dice" scenario, in order for the probabilities to make sense (ie. the total probability of all dice roll possibilities have to add up to 1), it requires a crucial assumption.When one rolls an "N" on a dN die, one immediately has to roll the dN die again and add the result to the previous result.  This is a crucial assumption for the math to work.Let's look at the case of a d4 "exploding dice".If one rolls a 1, 2 or 3 on the d4, that's the damage.If one rolls a "4" on the d4, then one rolls the d4 again and adds it to the previous result of 4, resulting in damage possibilities of 5, 6, 7, or 8.   If one rolls another "4" on the second d4 roll, then one rolls the d4 again and adds it to the previous result of 8, resulting in damage possibilities of 9, 10, 11, or 12.  (Ad infinitum).So requiring the d4 to be rolled again after rolling a "4", this means the damage possibilities are:- 1, 2, 3 -> each case individually with a probability 1/4- 5, 6, 7 -> each case individually with a probability (1/4)^2- 9, 10 , 11 -> each case individually with a probability (1/4)^3- 13, 14, 15 -> each case individually with a probability (1/4)^4- etc ...In this d4 "exploding dice" system with the above mentioned crucial assumption, we can never roll a damage score of 4, 8, 12, 16, etc ...If we add up all the probabilities, we have:3*(1/4) + 3*(1/4)^2 + 3*(1/4)^3 + 3*(1/4)^4 + ... Using the formula a/(1-x) = a + a x + a x^2 + a x^3 + a x^4 + ... for |x|< 1, this infinite sum adds up exactly equal to (3/4)/[1-1/4] = 1.  (This is just a simple infinite geometric series).This can be easily generalized to a dN "exploding dice" system, where one rolls the dN die again after rolling an "N".  So we will never see a damage score of N, 2N, 3N, 4N, etc ....  The damage possibilities are:- 1, ..., (N-1) -> each damage case individually with a probability 1/N- N+1, ..., N + (N-1) -> each damage case individually with a probability 1/N^2- 2N+1, ..., 2N + (N-1) -> each damage case individually with a probability 1/N^3- 3N+1, ..., 3N + (N-1) -> each damage case individually with a probability 1/N^4- etc ...(An exercise for the reader:  show that the dN "exploding dice" probabilities all sum up to 1).To get the average damage of this dN exploding dice system, we just take the weighted sum of the probabilities with each damage case value:[1 + ... + (N-1)]/N + { [N+1] + ... + [N+(N-1)] }/N^2+ { [2N+1] + ... + [2N+(N-1)] }/N^3+ { [3N+1] + ... + [3N+(N-1)] }/N^4+ ...Though not entirely mathematically rigorous, we will rewrite this sum (handwaved in a cavalier manner) by grouping the terms in a more suggestive form:[1 + ... + (N-1)]*[1/N + 1/N^2 + 1/N^3 + 1/N^4 + ...]+ (N-1)*[1/N + 2/N^2 + 3/N^3 + ...]As an exercise for the reader, this infinite sum becomes quite simple:N(N+1)/[2(N-1)]for the average damage of this dN "exploding dice" system.For various dN die, we have the average "exploding dice" damage N(N+1)/[2(N-1)]: d4 -> 3.33d6 -> 4.2d8 -> 5.14d10 -> 6.11d12 -> 7.09d20 -> 11.05 in contrast to the average damage of a single dN die (N+1)/2: d4 -> 2.5d6 -> 3.5d8 -> 4.5d10 -> 5.5d12 -> 6.5d20 -> 10.5</blockquote>

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