for that you will need standard deviation:
i actually work with population samples, so discrete numbers a bit awkward to me because i'm not used to having them. I'm just going to stick with one dice for now, and it should work like this:
Code:
XbarOE = Xbar/(1-p)
XbarOE = 13/(0.75)
XbarOE = 17.3
25
s^2 = SIGMA(1/25)((i-Xbar)^2)
i=1
/and via a quick and dirty function hacked out in python/
s^2 = 80.4089
SD = sqrt(s^2)
SD = 8.9671
/the same process on the dice without rerolls yields/
SD = 7.2801
because i'm not too familiar with OE roll stuff and variance here, that number might not be trustworthy. because of this, let's work with the close-ended roll as an example, and you can try it with the open-ended result if you wish.
at any rate, take your SD and compare it with a z-distribution table.
http://www.statsoft.com/textbook/sttable.html#z
first, the number of SD's above or below your number of interest is the z-score:
so, if you want to know that chance to roll exactly 7.28 above or below a particular number, that's a z of 1. if you wanted to know 14.56, that's a z of two. you can do fractions of z also.
find your z-score on the table:
the rows are ones and tens place, the columns are hundredths place.
so for example, 1.96 would be row 1.9, column 0.06.
you are only interested in rather you roll above OR below a particular number. therefore you only use half the distribution. the result of the table will show the chance to roll within a certain deviation on the entire table. thus, if you want to know the chance to beat a roll that is 8 more than the average, you can't look at the table and calculate a result as if you could roll within a space of eight around the mean.
to account for this you will need to double the result, and subtract it from 100. thus, if you are interested in whether or not you could beat a roll that is 10 higher than the mean, you would first divide that number by the standard deviation, getting your z-score
now find 1.37 on the table. it is a .4147
double it for .8294
subtract that from 100 for about .18
you would have a chance of 18 percent of beating a roll of 10 above the mean (23 in this case).
you can try that with the OE rolls with have a mean of 17.33 and an SD of 8.9671 if you like, but i'm not sure the math would be correct because i'm not sure if the variance formula would work the same for those rolls. however, it might because we already know the mean and there's nothing i've seen about variance that cares how that mean is obtained.