Dec 17 2004, 08:19 PM
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Moving Target ![]() ![]() Group: Members Posts: 556 Joined: 28-May 04 From: Moorhead, MN, USA Member No.: 6,367 |
Holy $hit I did it! Here's the formula as best I can write it without proper subscripts, superscripts, etc.
W = chances of success on the roll T = target number D = dice used S = successes needed e.g. threshold +1 C = (truncate (T-1)/6) + 1 (=1 for target numbers 2 through 6, 2 for 7 through 12, etc) B = 6C+1 A = 6^C-B K(n,S) = 1*(n-1)*(n-2)*(n-3)...*[n-(S-1)] .............1*(S-1)*(S-2)*(S-3)...*[S-(S-1)] W = Sum from n=S to D of K(n,S)*(T-A)^(n-S)*(B-T)^S/6^(C*n) I also wrote it up in QBasic. I'll edit that in below right after the spoiler test. It's in spoilers for length. [ Spoiler ] edit: fixed the last term C*n not 6*n! This post has been edited by Da9iel: Dec 17 2004, 08:52 PM |
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Da9iel The grand unified chances of success formula Dec 17 2004, 08:19 PM
Da9iel Yes it does work for all rational numbers of dice,... Dec 17 2004, 08:25 PM
BitBasher Um, but doesn't your chance of a sucess have l... Dec 17 2004, 09:28 PM
draco aardvark Or, for those of you out there who haven't use... Dec 18 2004, 12:19 AM
draco aardvark here's one to calculate the average number of ... Dec 18 2004, 01:16 AM
draco aardvark If I did it right, this one should calculate the l... Dec 18 2004, 01:57 AM
draco aardvark Hey Da9iel, can you explain how you got that formu... Dec 18 2004, 02:10 AM
Da9iel Actually its a purely empirical formula that exact... Dec 21 2004, 09:39 AM
draco aardvark I made the % chance of success because I have a ha... Dec 22 2004, 05:25 AM![]() ![]() |
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