OW describes the "Zeroing Effect" for Begginers

[Agema]

Okay, there is a small error. I'll repost the tables on Monday.

[Agema]

Originally Posted by Pehmyt

I think I should comment that as far as I can understand your spreadsheet, the results are ~10% off as soon as the open-ended rolls enter the picture. Did you take into account that 6 equals "5 +extra roll"? Not that it matters for the present discussion.

For example, the probability of getting 7 should be that of throwing one of (2;5), (3;4), (4;3), (5;2), (6,1;1), (1;6,1) (with ";" separating the two dice), that is 4/36 + 2/216 = 13/108 = 0.1204.

The results corresponding your column A can be computed exactly by writing the value of DRN roll as X=5k+r+1, with r between 0 and 4, and using

P(DRN = 5k+r+1) =  ((6-r)5k +r )/6^(k+2)

The chances of DRN being at least as large as a given number X=5k+r+1 can also be computed exactly, although this is not directly useful. It could be practical for generating a table similar to that in manual without losing too much precision. For 6-sided dice it reads

P(DRN >= 5k+r+1) = ( (48 - 13r + r^2)5k + 72 + r - r^2 )/( 2 * 6^(k+2) )

Yes, I did account for 6s being re-rolled, however there was an error in my working, I misplaced some columns in my working as if an extra roll was calculated as 6+d6, not 6-1+d6, which was where the 10% arrived. I've revised my tables accordingly, and I'll check them agains the book and repost them.

I had a look at those equations, and although my maths is pretty rusty these days I don't think they work for 2d6 which are the basis of DRNs in Dom 3. I'm not entirely clear what k and r represent - k = number of 6s rolled and [i]r[/r] = d6-1 excluding 6? It would work with something more like X=5k+r+r'+2.

[Pehmyt]

I'm not sure if I understand what you're saying, but naturally the formulas are not supposed to work for regular 2d6 (except for results below 6), the distribution is different as soon as we start rerolling.

In some approximate sense k represents the number of 6's rolled, at least for large values, but not exactly. See my example for getting DRN=7: there we have 7 = 5*1+1+1, so k=1 and r=1 and the P(DRN=7) = ((6-1)*5*1+1)/6^3 = 26/216 = 13/108, but among the 6 different ways of getting this there are 4 without any 6's and 2 with 1.

So k and r don't really depend on how you got your result, they are computed simply from the desired number X as
r = (X-1) % 5 (remainder)
k = (X-1)/5 (integer division, remainder dropped)

It might be possible to divide the result in two parts, each representing probabilities of getting the asked DRN with a given number of 6's. I have not looked into that, but if it is possible, it might lead to a more intuitive proof. Mine is using induction on k, and does not look very nice. On the other hand it works for any N-sided dice (N>1).