Stealth Mathematics

[Illuminated One]

Well, since I've taken an interest in stealth I noticed that the probability table on p5 in the manual shouldn't work.
You still have 2 open ended dice rolled but these are d25.

So I was wondering if anyone had an easy formula to calculate that or even such a table already.

From what I'm thinking if it wasn't open ended it should be pretty easy to do that.
Formula is 40 + stealth + 2d25 >= destealth + 2d25.
That is 40 + stealth - destealth + 2d25 - 2d25 >= 0.
If I'm correct -2d25 = +2d25 - 52
So the formula can be expressed as 4d25 >= n which should be calculable.
The problem is it don't see how to bring in the open ended part on both sides here.

[Omnirizon]

according to RPG.net its the average of the dice without rerolls, divided by one minus the probability of a reroll:

OEbar = Xbar/(1-p)

= 25/(0.75)
= 29ish


however, the bigger your dice get the longer the tail on the probability curve gets and the more skewed your result is. it's rare to get a reroll, but when you do it makes a big difference.

[Illuminated One]

Thanks.

So this is the average number of eyes when throwing two open-ended dice?

Since I'm looking for the probability to beat the opposed roll against a fixed difference between stealth and destealth, I don't know if that helps though.

Hmm, I've tried to do it without oe and I came up with the following formula. The chance to roll x (as long as x < 52 afterwards) with 4 25 sided dice is (well, if I didn't make a mistake)

P(x) = ((x - 4) * (x - 2) * 3 * x - (x - 3) * (x - 2) * (2*x - 5) )/6*25^4

With that and excel one could make a table for non oe dice.

[Omnirizon]

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:

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

10/7.2801 = 1.37

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.

[Pehmyt]

I should mention that the statistics of d6 open ended dice were discussed in another thread, and many results apply as well to d25oe throws.

In particular, for X=24n+r with 1<=r<=24, the probabilities of getting X from 1d25oe and 2d25oe are

P(1d25oe = X) = 25^(-n-1) and
P(2d25oe = X) = [ (26-r)24n+r-1 ]/25^(n+2)

From the latter one can do the summation to get P(2d25oe > X), which gives the probability of throwing larger than X, and then one more summation to get a table like on p.5. The actual numbers are likely to ugly, as already the result for d6 involves some funny numbers. Maple/Mathematica is strongly recommmended.

The method of the previous post possibly also works for central values, but underestimates the tails, as the distribution is exponential instead of normal (Gaussian).

Somewhat unrelated, I'd like to point out that for any open-ended dice 1dNoe, the average value differs from that of corresponding dice by exactly 1/2: XbarOE = N/2+1 (for normal dice (N+1)/2). For N=25 this gives XbarOE=13.5, and for two independent dice <2d25oe> = 27.

[Taqwus]

Ran 2d25(oe) - 2d25(oe) for a million iterations.

FWIW --

487,798 trials < 0
609,032 trials < 5
720,064 trials < 10
812,051 trials < 15
881,251 trials < 20
928,169 trials < 25
957,732 trials < 30
962,028 trials < 31
965,965 trials < 32
969,564 trials < 33
972,647 trials < 34
975,562 trials < 35
977,937 trials < 36
980,157 trials < 37
982,168 trials < 38
983,949 trials < 39
985,496 trials < 40
991,444 trials < 45
995,048 trials < 50
997,217 trials < 55
998,457 trials < 60
999,120 trials < 65
999,484 trials < 70
999,709 trials < 75
999,826 trials < 80
999,916 trials < 85
999,949 trials < 90

[Illuminated One]

Thanks to all of you!