Author: Eric Finley (---.micralyne.com)
Date: 06-19-2002 14:51
I've been working on this, on and off (mostly the latter) for a while. It's related to, but not directly equivalent to, your (excellent!) work on the program itself.
Can you find an algebraic solution for the average value of a roll of NkM in the 7th Sea/L5R dierolling mechanic? That is, roll N open-ended d10s, summing the M highest? Note that I do not say a specific average; I can (and have) done that myself. The challenge, and I keep thinking I'm close, is to come up with an algebraic solution to the puzzle. It'll probably NOT be something you can just intuit during a game, the way one can more straightforward systems, but it'd be nice to have it and see, anyway. [In practice I find that assuming 6 per kept plus 2 per unkept works remarkably well. Then, of course, I go and do drastic things based on that and eat my hat... :)]
As a starting point, the average roll on a normal dA is, of course, (A+1)/2; less trivially, the average roll on an *open-ended* dA is (A/(A-1)) of this value, such that where a normal d10 averages to 5.5, the open-ended one averages to six-point-one-repeating, 6.1111..., which I can't figure out how to represent here. :)
Even less trivially, and more usefully toward any eventual solution, is that the average roll on an open-ended d10 which you know is showing at least a value of B turns out to be
xbar(B) = [1/(11-B)] * [(-1/2)B^2 + (1/2)B + 61.1111...]
As such, if your highest unkept die is showing a B, then each of your kept dice can be counted as having a value of xbar(B). What gets me is an algebraic solution for what P(B) is, the probability that your highest unkept die shows a B, given N and M; if I could get that, then
xbar = Sum(B=1 to 10) [P(B)*xbar(B)]
would be trivial.
I can envision a geometric solution to P(B) for simple cases... especially when there's only one *unkept* die, it's easy (and neat - winds up being nested squares, or cubes). But a general solution to P(B) continues to elude...
Some of us have waaay too much education and free time. :)
So, I challenge you. Beat me to a solution. I won't be installing or compiling roll(), so you'd have the edge in being able to destruct-test algebraic models; I also have no formal stats training, just a good head for figures. :) If you win, I'll send you, I don't know, something cool, maybe my friend's misprinted d6 which shows 0,1,2,3,4,7, which some GMs let him use anyway. Open challenge, too... not just the author. You know how it is once you get your teeth in something...
- Eric
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