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<blockquote data-quote="RogerShrubber" data-source="post: 903714" data-attributes="member: 12089">A fast and easy solutionI know this thread has been dead for several months, but Frisbeet pointed it out to me, and I thought I'd provide an easy way to accomplish this, without going into the discrete behind it.This works great if you have access to Matlab, a good TI, or some other 'calculator' with an expand function.Where:n: number of dices: number of sides per dieexpand((x+x^2+x^3...x^s)^n)For example, for s = 6 and n = 2, this results in:x^12+2x^11+3x^10+4x^9...6x^7+5x^6....The coefficient before each x is the number of ways to make the exponent.&nbsp; In this case, there is 1 way to make a 12, 2 ways to make an 11, 3 ways to make a 10, 6 ways to make a 7, 5 ways to make a 6, and so on.This can be very tedious when you have to multiply and recombine the terms by hand, but is quite fast when done by a calculator.When you combine these results with the total number of possible solutions (s^n), you can easily find the percent chance of each result.Further, this formula can be used to determine the results if several different dice are involved.Example: d4+d6 Expand((x+x^2+x^3+x^4)^1*(x+x^2+x^3+x^4+x^5+x^6)^1)x^10+2x^9+3x^8+4x^7+4x^6+4x^5+3x^4+2x^3+x^2I have found this very useful in my modification of Frisbeet's Weapon Analysis, and think you may appreciate them as well.Roger Shrubber</blockquote>

[QUOTE="RogerShrubber, post: 903714, member: 12089"] [b]A fast and easy solution[/b] I know this thread has been dead for several months, but Frisbeet pointed it out to me, and I thought I'd provide an easy way to accomplish this, without going into the discrete behind it. This works great if you have access to Matlab, a good TI, or some other 'calculator' with an expand function. Where: n: number of dice s: number of sides per die expand((x+x^2+x^3...x^s)^n) For example, for s = 6 and n = 2, this results in: x^12+2x^11+3x^10+4x^9...6x^7+5x^6.... The coefficient before each x is the number of ways to make the exponent. In this case, there is 1 way to make a 12, 2 ways to make an 11, 3 ways to make a 10, 6 ways to make a 7, 5 ways to make a 6, and so on. This can be very tedious when you have to multiply and recombine the terms by hand, but is quite fast when done by a calculator. When you combine these results with the total number of possible solutions (s^n), you can easily find the percent chance of each result. Further, this formula can be used to determine the results if several different dice are involved. Example: d4+d6 Expand((x+x^2+x^3+x^4)^1*(x+x^2+x^3+x^4+x^5+x^6)^1) x^10+2x^9+3x^8+4x^7+4x^6+4x^5+3x^4+2x^3+x^2 I have found this very useful in my modification of Frisbeet's Weapon Analysis, and think you may appreciate them as well. Roger Shrubber [/QUOTE]

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