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The formula doesn't make sense for k = 0, that is right. If you want a formula that makes sense for this case too, let l be the lowest die that won't be dropped. This is always well-defined. To define 0^0 = 1 makes algebraically perfectly sense. And it should be defined this way as the...

I have associated the formula with Gauß as there is an anecdote about him: when he was a 9-year-old pupil in school, the teacher was a little bit lazy and wanted to give his pupils an exercise which to solve would have taken some time for them. He asked them to add the first 100 positive...

Re: Re: Re: Re: Analytical solution Ah, yeah, of course. I have been obviously to sleepy recently... Have you understood how to derive the summation formula for the 1 + 2 + ... + n? Nathan

Re: Re: Analytical solution Yes, you're right. And this has already been right in the formula. Again, my mistake: I wanted to say the number of B's is u. It's value is l, of course. Yes! Formula by Gauß: 1 + 2 + 3 + 4 + ... + N = N * (N + 1) / 2. Proof for this? Look at: (1 + N) + (2 + (N...

Thank you! One note: you should write the binomial coefficients (n over r) without the line between the n and the r, otherwise they look like fractions...

I don't have a programme to hack my formula into at the moment, so I still hope that someone else will be doing this. The initial poster mentioned that he calculated an average of 13.4 by brute force for 5d6, drop lowest 2. As your method gives a 13 here, we see that your formula can only be an...

Your method won't lead to the correct results but should give a good approximation. With respect to my last comment on how to get a formula for the likelihood of a certain total N: First you have to find a formula for C(l, r, u), the number of ways to add up n - u - r numbers from 1...d to N -...

Re: Re: Analytical solution No, I haven't. In my formula, there are always only three summations, independent of the numbers n, k, and d. And this isn't bad, and quite normal for an analytic formula calculating an average. However, just counting each possible result would lead to a growing...

Analytical solution An approach to an analytic solution: Let us assume we have n dice with d sides each, numbered from 1...d. (In your case n = 5, d = 6). We want to drop the lowest k dice (k = 2 in your case). How many ways are there to roll these dice? In total, there are n^d = n * ... * n...

Re: Hmm I will be in Cambridge (near Castle Hill, doing a Postdoc year there) until the end of this september. However, I am not quite sure how much time I will have during the rest of my stay in lovely Cambridge. I don't have a gaming group in the moment, too. Nathan

So when? Let us all know! ;)

Social skills Are there any changes/clarifications on how the social skills like Bluff and Diplomacy work? There have been long debates on these skills (role-play vs. roll-play, etc.), so I wonder if there is something new on this in 3.5? Or are these things in the 3.5 PHB? Thank you!

Bardic music Hiya! I have some questions concering bardic music... Have they been answered yet? I don't think. 1) Has Fascinate changed? 2) Does a bard need to fascinate a subject first if they want to use Suggestion on this subject afterwards? 3) Has the duration changed in any way? Do...

Leadership Maximum level for cohorts now PC-level minus 2. Source: Andy Collins (www.andycollins.net).

I you really want to prove that the MTh is/isn't too powerful, the restriction to ask for a Clr5/Wiz5/MTh10 combo is too narrow. Something like Clr3/Wiz7/MTh10 or Clr3/Wiz3/MTh10/PrC4 should also be allowed. Or any other combination involving the MTh10. Nathan

CRGreathouse, didn't we debate about such a question (4d6, drop lowest, etc., drop crap characters) on these boards some months ago? I cannot search the boards but maybe someone else could dig deep into this board and recover the former thread. We ended up with some nice tables and formulas.

The definition of the binomial coefficient (n over s) for n any number and s any integer with s >= 0 is given by: (n over s) := (n * (n - 1) * (n - 2) * ... * (n - s + 1))/(1 * 2 * ... * s). As one can easily check, the nominator includes a factor of zero (and therefore vanishes) if s >= n.

Re: More than one die type... Note that the binomial coefficient (n over s) is per definition zero, if s is an integer greater than n (and n positive). Therefore, you don't need the constraints m1 <= n1 and m2 <= n2.

With respect to druids: They should also change the spell reincarnation so that it is compatible with Savage Species and the LA/ECL system.

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