Resolution Mechanic Query

[Calithena]

After some initial experimentation I really like the method of opposed dice pool resolution in Sorcerer and Donjon. One thing I like about it is that I don't know exactly how much another die is worth. However, it occurs to me that this information is actually known, or should be knowable, and that while true ignorance is a glorious thing (in the case of RP mechanics at least, on the assumption that you like enforcing actor stance, or at least ruling out 'calculator stance', which is a questionable assumption, but I digress), ignorance which merely stems from the fact that someone else knows something I don't is not. So what I'm wondering is: let's say you've got two opposed die pools of variable size and you want to know

(a) the chances for each side to win

and

(b) what the average number of successes are for the victor

or best of all

(c) given m dice vs. n dice, say m,n=1 to 20, what are the chances for each possible result ranging from m victories for the first player to n victories for the second? This would require a three-dimensional graph to represent visually but spreadsheets will do this sort of thing too.

Actually, there's a fourth dimension. You need an answer to (c) for each 'sidedness' of dice. So actually you need 5 3d graphs, plus extras if you want to use d7's or d30's or other odd resolution methods.

Does anyone know this information in part or whole? If so, I'd love it if you sent it to me at calithena@juno.com, or posted some of it here if it doesn't whelm the site with data. Or provided a link if it's up somewhere else already.

I will graciously provide the answer for 1 on 1 myself, with d10's: there is a 45% chance for a single victory for either side, and a 10% chance for a push. (In Sorcerer of course you reroll or find a different way to interpret when m=n and there are ties on every die.)

Thanks in advance for any help.

[Mike Holmes]

I had, at one point, written up a spreadsheet that calculated these things, but have lost it since. It wasn't easy to make, IIRC. How much is it worth to you? :-)

More approrpiately, why do you need to know? Because there are some trends that can be stated that might suffice instead of actual figures.

Mike

[Calithena]

I think I've figured out most of the trends myself from scrupulous home analysis and obsessive reading of various discussions in Sorcerer and Donjon and on this site, etc., though I suppose there are always more things I might learn. What I want is the actual numbers.

I'd pay say $20 for a full answer to (c) with all five types of dice, provided that the algorithms were provided along with the numbers so I could check the math. It's worth more than that in the sense that it would probably take me a day or so to DIY it and my time is worth more than $2 an hour, but I'm not a wealthy man. Actually, I'm not 100% sure I could do it at all except by 'dartboard' analysis (have the computer roll 20 against 12 say 1,000,000 times and back-engineer the percentages from the results) because some of the probability stuff involved seems to get pretty hairy to crunch directly. But I'm not sure. I'd like to avoid becoming sure on the basis of my own statistical analysis if someone else has already done the work.

I am such an old fogey. I bet there are calculators and programs you can just punch this stuff into now to get answers to questions like these. When I was an undergraduate I used to invert 4x4 matrices of differential equations in my head just for practice; now there's no point in any of that because the computers do it for us. And what's worse is that not only can I not use the computers, I can't do the stuff I used to do when I was a math whiz any more either. If somebody makes a quick $20 by giving me an answer to this from a home computer program I'd like a copy of the program too so I can check the work...