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Message: <blockquote data-quote="frisbeet" data-source="post: 1978654" data-attributes="member: 10287">DMG 3.5, p. 170.Therefore it’s a 3d6 world you’re all playing in.Question: how frequently on average would you expect to bump into an NPC with at least one 18?One answer, which many claim is a consensus opinion (stemming from the 1st ed. of D&D), is: virtually never; extremely rare.Rubbish!Either that or “rare” is an extremely relative term.Here’s why.There’s just 1 way of rolling an 18 with 3 d6 dice: all 3 dice are sixes.  Genius.There are 6^3 = 216 total ways of rolling 3 dice.  That’s 6 ways to roll the 1st die x 6 ways to roll the next die x 6 ways to roll the last die.The probability of rolling an 18 by 3d6 is therefore 1/216.  There is but 1 in 216 ways of rolling an 18 with 3 dice.Maybe that’s rare to some people.  I dunno, most villages a PC traipses through will have this many occupants.  Is one prodigy/village so rare?But I’m not done.  1/216 is really the probability of an NPC having an 18 Int, or 18 Str, or 18 in any one specified attribute.  That doesn’t answer the question.  The question, again: how frequently on average would you expect to bump into an NPC with at least one 18, without requiring it be one particular attribute?That probability is found by the simple trick of subtracting from 1 the probability of not having any 18s.  Probabilities are always between and including 0 and 1, incidentally; 1 represents the probability that the NPC will have 6 attributes, each between and including 3 to 18 (well, duh).The probability of not having any 18s = probability of rolling 17 or less for Str x probability of rolling 17 or less for Dex x … = P(17 or less) ^ 6.  That’s because a law of probability requires that if N independent things must happen to get a certain outcome, the P(outcome) = P(thing 1 happening) x P(thing 2 happening) x … x P(thing N happening).  The probability of rolling a 17 or less is 215/216.  There is 1 way to roll an 18 out of 216 total ways of rolling 3d6.  Therefore there’s 215 ways of not rolling an 18.  Then the probability of not having any 18s among 6 stats = P(17 or less) ^ 6 = (215/216)^6 = 0.972541746577387.The probability of having at least one 18 among 6 stats is, again, 1 minus this. 1-(215/216)^6 = 1 - 0.972541746577387 = 0.0274582534226134   If you’re accustomed to percentile, that’s 2.75% of the time.The answer to the question, the frequency, is defined as 1 over the probability: 1 / (1 – (215/216)^6) = 1 / 0.0274582534226134 = 36.418922376776So, a PC will bump into one NPC with at least one 18 in a crowd of 36 NPCs, on average.That’s rare????Ergo, you’d better give your PCs an advantage.  4d6-drop-lowest analysis shows the answer to be 1 / (1 – ((1296 – 21)/1296)^6) = 10.7103209606842.  A little fewer than 1 in 11 PCs (created by 4d6 drop method) will have at least one 18, on average.18 isn’t that spectacular, IMO.</blockquote>

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