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Message: <blockquote data-quote="Vurt" data-source="post: 4438574" data-attributes="member: 1547">The question of how to calculate the expected result of rolling a d20 twice and choosing the best result came up a couple of times before, and here's what I worked out.  At first I thought it was 75% of the die, but when trying to work out the probabilities again I realized that this is technically incorrect.  What you get is actually a 75% chance to roll above average.Proof:  For each roll of a die, there is a 50% chance of rolling less than average.  Rolling twice, there is a 25% chance (0.5 x 0.5) of both rolls being below average, and thus a 75% chance that one of them will be above average.To find the expected result of this process, however, it looks like you need to add the sum of each outcome multiplied by its relative probability (i.e.  the weighted sum).  From my doodling, there doesn't appear to be an easy algebraic formula to do this.  The best I could come up with is  summation notation (i.e. sigma notation) or pseudocode.First the sigma notation:y = Sigma [x=1..n] (x(2x-1)/(n^2))Where y is the expected result, n is the number of sides on the die in question, and x is just a counter that runs from 1 to n.Now the pseudocode, using the same variables:y = 0; //initializefor x = 1 to n do {y = y + x * (2 * x - 1) / (n ^ 2)}Crunching the numbers, you get:Rolling a d2 twice and choosing the best result should average 1 and 3/4 (and not 75% of 2, or 1.5 as I had originally surmised).Rolling a d4 twice and choosing the best result should average 3 and 1/8th (or 3.125).Rolling a d6 twice and choosing the best result should average 4 and 19/36ths (or 4.52777...).Rolling a d8 twice and choosing the best result should average 5 and 13/16ths (or 5.8125).Rolling a d10 twice and choosing the best result should average 7 and 3/20ths (or 7.15).Rolling a d12 twice and choosing the best result should average 8 and 35/72 (or 8.486111...)....Rolling a d20 twice and choosing the best result should average 13 and 33/40ths (or 13.825).~~So to answer the OP's original question, take the average value of rolling twice, subtract the average value for just rolling normally, and round up or down as you see fit.  That should be the bonus you're looking for.</blockquote>

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