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April 15th, 2010, 04:41 AM  #1 
Newbie Joined: Apr 2010 Posts: 2 Thanks: 0  d20 probabilities
So I have this homebrew Pen and Paper system, and I need to calculate some probabilities. It is an opposed roll with d20s, so my d20 minus your d20. I need to calculate the probability of the outcome being: 10 or greater, 5 or greater, 1 or greater, 1 or less, 5 or less, 10 or less. Now, this could be easily done with a matrix, but I am genuinly interested in knowing the math. Also, it is important that I know how to modify the results easily, so that I might for instance calculate these probablities for a (d20+5)d20. Any help would be appreciated. 
April 15th, 2010, 07:14 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: d20 probabilities
Let's say you want to get 15 or higher on an opposed roll 1d20  1d20. If the first die is a 20, then there are 5 possibilities, 1 through 5. If the first die is a 19, then there are 4; etc. So the total number of ways to get at least 15 is 5 + 4 + 3 + 2 + 1 = 5(5 + 1)/2 = 15. There are 20^2 = 400 possibilities overall, so the chance of getting 15 or better is 15/400 = 3.75%. Notice that trying to get T on 1d20 + t  1d20 is the same as trying to get Tt on 1d20  1d20 (and 1d20  t  1d20 is the same as getting T+t), so that doesn't introduce any more complexity. 
April 15th, 2010, 01:30 PM  #3 
Newbie Joined: Apr 2010 Posts: 2 Thanks: 0  Re: d20 probabilities
Oh right. I had stared myself blind at the problem. I had made myself think it was much more complicated than that. Thank you very much, sir. 
April 15th, 2010, 09:17 PM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: d20 probabilities Quote:
Notice that if you don't need to be as high as the other roll you should calculate the number of times you fail rather than the times you succeed (subtracting from 400, naturally) to be able to use that nice sum formula.  

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