Roll twice, take best

[Edivad]

Something I found on my computer and felt like posting here before deleting. Basically, time ago I was horribly bored and wondered - how good is rolling twice and taking the best result, mechanically? Here's a short explanation.


P(X), that is, the chance of obtaining X as a result by rolling a d20 twice and taking the best result is:
1/400 + (X-1)/20 * (1/20) * 2 = [1 + 2(X-1)]/400


(This is because you can obtain X by either getting X from both d20, or from getting a number lower than X with one dice and X with the other – you have to multiply this by 2 because this can happen in two separate orders, that is, you can roll the lower number before or after the X)

P(1): 1/400 or 0.25%
P(2): 3/400 or 0.75%
P(3): 5/400 or 1.25%
P(4): 7/400 or 1.75%
P(5): 9/400 or 2.25%
P(6): 11/400 or 2.75%
P(7): 13/400 or 3.25%
P(8): 15/400 or 3.75%
P(9): 17/400 or 4.25%
P(10): 19/400 or 4.75%
P(11): 21/400 or 5.25%
P(12): 23/400 or 5.75%
P(13): 25/400 or 6.25%
P(14): 27/400 or 6.75%
P(15): 29/400 or 7.25%
P(16): 31/400 or 7.75%
P(17): 33/400 or 8.25
P(18): 35/400 or 8.75%
P(19): 37/400 or 9.25%
P(20): 39/400 or 9.75%

The average is the sum of X * P(X) for X = 1, ..., 20
1/400 + 2 * 3/ 400 + ... + 20 * 39/400 = (1 + 2*3 + ... * 20*39)/400
= 13.825


Which is quite a benefit compared to the standard average of 10.5.