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<blockquote data-quote="nsruf" data-source="post: 762625" data-attributes="member: 872">I am too lazy to work out the general formula, but you can write a quick-and-dirty algorithm for getting the exact probabilities easily:If you want n p-sided dice rolled, generate all possible outcomes (vectors of length n with entries 1..p), sum over the entries in each outcome, and count how often you get each result. Divide this count by the total number of outcome vectors to get the probability to roll exactly this number. To get the probability of rolling equal or less, add up all probabilities for rolling equal or less.Example, using 2d6: you have 36 outcome vectors(1,1), (1,2), ..., (6,6)and possible results ranging from 1+1=2 to 6+6=12. If you count how often you get each result, you get2 x1, 3 x2, ..., 7 x6, 8 x5, ..., 11 x2, 12 x1.As you started with 36 outcome vectors with equal probability, your chance to roll 2 is 1/36, to roll 3 is 2/36, etc. Resulting in1/36 to roll 2 or less(1+2)/36 to roll 3 or less...(1+2+3+4+5+6)/36 to roll 7 or less(1+2+3+4+5+6+5)/36 to roll 8 or less...(1+2+3+4+5+6+5+4+3+2+1)/36 to roll 12 or lessNote that the last is really 36/36 or 100 %.You can easily generalize this method to more dice of different types.</blockquote>

[QUOTE="nsruf, post: 762625, member: 872"] I am too lazy to work out the general formula, but you can write a quick-and-dirty algorithm for getting the exact probabilities easily: If you want n p-sided dice rolled, generate all possible outcomes (vectors of length n with entries 1..p), sum over the entries in each outcome, and count how often you get each result. Divide this count by the total number of outcome vectors to get the probability to roll exactly this number. To get the probability of rolling equal or less, add up all probabilities for rolling equal or less. Example, using 2d6: you have 36 outcome vectors (1,1), (1,2), ..., (6,6) and possible results ranging from 1+1=2 to 6+6=12. If you count how often you get each result, you get 2 x1, 3 x2, ..., 7 x6, 8 x5, ..., 11 x2, 12 x1. As you started with 36 outcome vectors with equal probability, your chance to roll 2 is 1/36, to roll 3 is 2/36, etc. Resulting in 1/36 to roll 2 or less (1+2)/36 to roll 3 or less ... (1+2+3+4+5+6)/36 to roll 7 or less (1+2+3+4+5+6+5)/36 to roll 8 or less ... (1+2+3+4+5+6+5+4+3+2+1)/36 to roll 12 or less Note that the last is really 36/36 or 100 %. You can easily generalize this method to more dice of different types. [/QUOTE]

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