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Old 05-27-2015, 02:51 AM   #11
vicky_molokh
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Default Re: Formulae for calculating Storyteller / Exalted d10 probabilities?

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Originally Posted by Anthony View Post
Probably because AnyDice understands what I'm doing and can optimize it well (it's not a hard optimization problem; erikthered's code could do with formatting and comments but is probably correct). Might I ask what you tried to give it?

Spreadsheet version: https://docs.google.com/spreadsheets...it?usp=sharing
I was using
output [count {7,8,9,10,10} in 9d10]
But it times out at higher pools. I'm guessing their ability to check whether a die result equals a given number is very slow . . . which seems odd to me, as we aren't parsing strings or something like that.
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Old 05-27-2015, 11:15 AM   #12
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Default Re: Formulae for calculating Storyteller / Exalted d10 probabilities?

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Originally Posted by vicky_molokh View Post
I was using
output [count {7,8,9,10,10} in 9d10]
But it times out at higher pools. I'm guessing their ability to check whether a die result equals a given number is very slow . . . which seems odd to me, as we aren't parsing strings or something like that.
No, the problem is likely that they have to actually generate the output of 9d10 when doing that, so the problem grows exponentially, as opposed to doing a simple expansion that grows quadratically.
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Old 05-28-2015, 11:47 PM   #13
jeff_wilson
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Default Re: Formulae for calculating Storyteller / Exalted d10 probabilities?

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Originally Posted by Anthony View Post
No, the problem is likely that they have to actually generate the output of 9d10 when doing that, so the problem grows exponentially, as opposed to doing a simple expansion that grows quadratically.
This sounds very likely; I have always needed to iteratively expand die roll results when ever working with more than a handful. The expansion allows work into the hundreds of dice even in environs like Excel.
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Old 05-31-2015, 10:02 PM   #14
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Default Re: Formulae for calculating Storyteller / Exalted d10 probabilities?

An alternative approach would be to use generating functions. Specifically, one can think of rolling n dice as you've described as the generating function (6+3x+x^2)^n and therefore the number of ways of getting exactly k successes is given by the coefficient of x^k in the expansion of the above polynomial. For instance, Wolfram Alpha will give the coefficients for 10 dice as:

Quote:
x^20+30 x^19+465 x^18+4860 x^17+38070 x^16+236196 x^15+1199610 x^14+5093280 x^13+18325845 x^12+56358990 x^11+148853781 x^10+338153940 x^9+659730420 x^8+1100148480 x^7+1554694560 x^6+1836660096 x^5+1776193920 x^4+1360488960 x^3+781021440 x^2+302330880 x+60466176
Which says that there is one way to get 20 successes, 465 ways to get 18 successes, and 302330880 ways to get only one success.
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