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Thread: Help with dice averages.

20130301, 07:38 AM (ISO 8601)
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 Mar 2007
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 Knoxville Tennessee
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Help with dice averages.
Me and my brother are working on a game system with d6s where a pool is rolled and 56 are succeses. But were hitting a stump trying to work out averages, i can work them out a little, but he is just hopeless. Does anyone know if there is a tool out there or a tutorial for working out averages like that?
:EDIT: I can find stuff for normal avarages, but averages for rolls of 56 or rolls of 56 with how often a 1 might come up along the rest of that and all that are a bit beyond me.Last edited by Dragonus45; 20130301 at 07:44 AM.
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20130301, 07:57 AM (ISO 8601)
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 Mar 2009
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 Long Shiny Cloudland
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Re: Help with dice averages.
Try http://anydice.com/
Depending on what you mean by 'average', a little programming will get your problem done for you. If you have any more specific problems I can see what I can cobble together.
EDIT: Rolling a pool of each size from one to ten, counting fives and sixes and the same, counting ones sound like what you wanted. The Summary tab provides the mean (i.e. the average count). The At Least tab will tell you the probability of getting at least a given number (obviously).Last edited by Fortuna; 20130301 at 08:04 AM.
If I creep into your house in the dead of night and strangle you while you sleep, you probably messed up your grammar.
I'm always extremely careful to hedge myself against absolute statements.

20130301, 08:07 AM (ISO 8601)
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 Mar 2007
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 Knoxville Tennessee
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Re: Help with dice averages.
Thanks, i had actually found that but i couldn't program it to make it do what i wanted.
Thanks to Linklele for my new avatar!
If i had superpowers. I would go to conventions dressed as myself, and see if i got complimented on my authenticity.

20130301, 08:07 AM (ISO 8601)
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 Mar 2010
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Re: Help with dice averages.
Not sure what you mean by average, but I'll try to be useful...
Chance of a given number being rolled on a single d6 is ~18.89% or 1/6.
Chance or a 5 or a 6 is therefore ~33.33% or 1/3.
So, an average (not to be confused with typical) roll of [n]d6 will yield n/6 1's and n/3 5's or 6's.
So, figure out what the human average is for a given dice pool in your game (I'll call that x). A "typical" (mildly challenging, but not too tough) challenge should probably have a target number of successes equal to x/3. If 1's cancel out successes, then it should be x/3x/6=x/6. Anything less than that should be considered easy, anything more should be exceptionally tough.
Or I'm just completely wrong about all of this. I like to keep that option open.Last edited by Jack of Spades; 20130301 at 08:09 AM.

20130301, 09:43 AM (ISO 8601)
 Join Date
 Feb 2013
Re: Help with dice averages.
Not sure what you're trying to work out, exactly. Is it the chance to get any X successes (56 on d6) when rolling any N d6s?
That's your basic binomial distribution where p = 0.5, n is the number of dice rolled, and i is the number of successes you're calculating the probability for.
Easiest thing to do is to plug this into Excel / OpenOffice Calc:
=(COMBIN(n;i))*(POWER(p;i))*(POWER(1p;ni))
You can make a table of it, with n and i values on the rows and columns (or the other way around), if you know Excel / Calc spreadsheet basics.
Plugging that in, for instance, I can see that, rolling 6 dice, the chance of 0 successes is 8.8%, 1 success is 26.3%, 2 successes is 32.9%, 3 successes is 22.0%, 4 successes is 8.2%, 5 successes is 1.7%, and 6 successes is 0.1%.
NB: If my math is wrong, somebody spank me over it. Haven't done binomial distribution in a while.

20130301, 10:27 AM (ISO 8601)
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 Oct 2010
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 Dallas, TX
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Re: Help with dice averages.
If you will ask a specific question, we'll answer it, but we need to know the question.
What, exactly, do you want to calculate?

20130301, 10:41 AM (ISO 8601)
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 Jun 2011
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Re: Help with dice averages.
Make custom dice. If your d6 rolls "1 success on a 5 or 6" and "1 success on a 1", you want a die that looks like this: {1,0,0,0,1,1}
It ends more or less like this: http://anydice.com/program/1e98

20130301, 12:25 PM (ISO 8601)
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 Mar 2007
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 Knoxville Tennessee
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Re: Help with dice averages.
Thanks to Linklele for my new avatar!
If i had superpowers. I would go to conventions dressed as myself, and see if i got complimented on my authenticity.