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Message: <blockquote data-quote="Eldorian" data-source="post: 4508573" data-attributes="member: 10504">Rolling twice squares the chance of failure.  This is a good thing, since the chance of failure is between 0 and 1.Let me introduce you to a concept called a binomial distribution.Lets say p is a number between 0 and 1 representing the chance of success.  (think of it as a percentage chance of success, where 1=1.00=100%).  Then 1-p=q is the chance of failure.The chance of getting at least one success on one roll is of course p.The chance of getting at least one success on two rolls is p^2 + 2p(1-p)= 2p-p^2=(2-p)p=(2-(1-q))(1-q)=(1+q)(1-q)=1-q^2.This, however, is an annoying way to think about it.  Usually when determining odds with dice type questions, it's easier to calculate the odds of failure, and then success is 1- that.  So when would you fail on two rolls?  Two failures, i.e., q^2.  So success has probability 1-q^2.Basically, take the chance of failure, square it, and if that number is lower than the chance of failure with some sort of bonus, then a reroll is better.  A reroll is only worse than a static bonus when q^2 is close to q, ie, q is big.  So when it's a long shot, it might be better to have a bonus instead of a reroll.Now, when there isn't a target number, like in initiative, then a reroll of a d20 gives expected value of 13.825, which is less than the 14.5 you get from improved initiative.  However, there is less variance in the reroll, i.e., you're less likely to roll mega low or mega high when you have a reroll.  This makes your initiative more predictable, which is better for you as a player (if you're designing a character to win fights.  If you're designing him to have fun playing then maybe having a lot of swing factor is cool for you).To do this is a bit harder.  There are 400 possible rolls of 2d20.  So basically you count how many ways of getting 1 there are, multiply that by 1, and add that to the ways of getting 2 multiplied by 2, etc.1(1+0) +2(2+1) + 3(3+2) etc, ie the sum of k(2k-1)=2k^2 -k which requires some slightly arcane formulas to evaluate, n(n+1)(2n+1)/3 - n(n+1)/2  where n is the number of sides on the die.  You divide this by the total number of possible rolls, 400, and you get the magic number 13.825.  This gives the expected value of rerolling a die, picking the best of both rolls.If there is an easier way of doing this last bit, I don't know it.</blockquote>

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