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Author Topic: Probabilities?  (Read 2140 times)
Shinrei
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« on: July 09, 2012, 06:20:24 AM »

Hi everyone,

As the title implies, i am interested in the probabilities of the Mistborn Gamemechanics. I was wondering if there's a comprehensive probability-table somewhere? Or if not, if the fine folks at Crafty Games are willing to put one up for us?

I always like to look at the distribution of a given dice-system and how it progresses. It gives me a feel for the capabilites of characters and of the in-world power-ladder, so to speak.

Thanks in advance.
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Baijo Gosum
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« Reply #1 on: July 09, 2012, 06:37:37 AM »

check page 454, there is a change of success by difficulty table
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JMobius
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« Reply #2 on: July 09, 2012, 08:09:41 AM »

I was really pleased that they included the table in the game, because it was evidence that they'd clearly thought about the matter. I've seen way too many 'the designers can't do math' RPGs.
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Crafty_Pat
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« Reply #3 on: July 09, 2012, 10:25:16 AM »

I was really pleased that they included the table in the game, because it was evidence that they'd clearly thought about the matter. I've seen way too many 'the designers can't do math' RPGs.

We actually brought on a statistician to double-check the core mechanic for us. He had to build a script to test it and it nearly killed his machine (the first set of calculations took a matter of minutes, while the last one was an overnight affair - and if we'd kept it going the next set would have taken just over a week to process).

Good times. Smiley
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oxinabox
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« Reply #4 on: July 11, 2012, 06:00:53 AM »

wow, I am seriously impressed.
This is really good.
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A New Dusk: Alloy of Law Adventure Game
http://www.crafty-games.com/forum/index.php?topic=6148
With twinborn, compounders, guns, explosives
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Shinrei
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« Reply #5 on: July 11, 2012, 02:18:19 PM »

Yeah, i should have known that CG had already included a table in the rulebook  Grin

Great stuff and another "thumbs up" for CG....and i think every rpg company should check their math this carefully. Something like: "The damned game runs on your game mechanics, dude!!!"  Wink
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Uysses
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« Reply #6 on: August 02, 2012, 08:04:56 PM »

Guess I learned generating functions for my own personal satisfaction working this out for myself.

On the other hand, at least I can check my work now!

Out of curiousity did the statistician use a straight monte carlo program to do it empirically or what? I sort of cheated and used Wolfram Alpha to do the binomial expansion and that was nice and fast. After that it was some simple RegEx.
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« Reply #7 on: August 02, 2012, 08:20:51 PM »

Out of curiousity did the statistician use a straight monte carlo program to do it empirically or what? I sort of cheated and used Wolfram Alpha to do the binomial expansion and that was nice and fast. After that it was some simple RegEx.

Uh, I don't understand anything you're asking after the phrase, "Out of curiosity..."  Embarrassed

For all I know, he sacrificed three chickens under the light of a harvest moon, danced naked around a faerie circle, did the hokey-pokey then waved his magic digital wand to make these numbers come together. Wink He's not active on this boards, else I'd suggest asking him directly, but I can say he has a Masters in mathematics and his numbers checked against my own rudimentary calculations up to 6 dice that I did by hand.
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Agent 333
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« Reply #8 on: August 02, 2012, 08:49:28 PM »

Out of curiousity did the statistician use a straight monte carlo program to do it empirically or what? I sort of cheated and used Wolfram Alpha to do the binomial expansion and that was nice and fast. After that it was some simple RegEx.

Uh, I don't understand anything you're asking after the phrase, "Out of curiosity..."  Embarrassed

For all I know, he sacrificed three chickens under the light of a harvest moon, danced naked around a faerie circle, did the hokey-pokey then waved his magic digital wand to make these numbers come together. Wink He's not active on this boards, else I'd suggest asking him directly, but I can say he has a Masters in mathematics and his numbers checked against my own rudimentary calculations up to 6 dice that I did by hand.

Let me see if I can translate for you:

Out of curiousity did the statistician use a straight monte carlo program to do it empirically or what?
A monte carlo program uses an algorithm to do a random calculation a bunch of times and average their results. Emperically checking the answers by just trying a lot and seeing what comes out.

I sort of cheated and used Wolfram Alpha to do the binomial expansion and that was nice and fast.
Ulysses used a freely available online tool to find the formula for the averages.

After that it was some simple RegEx.
He then used Regular Expressions to define the different parameters and check all the results.

 Grin
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Uysses
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« Reply #9 on: August 02, 2012, 09:21:11 PM »




"Ulysses used a freely available online tool to find the formula for the averages."

No, not really. I used a freely available tool to expand the generating functions. The entire problem here is that you can't figure out pairs by averages on the dice. The expected value on 2 dice is 7 and tells you nada about the odds of pairs.

Counting dice can be a pain, as I am sure you know, and while the dice game created here looks a lot like a formv of Yahtzee (descended from existing games) the odds are not the same. One of the easiest ways to get around this is to write a program to roll 10,000 dice, check against success, and figure that this is a good representation of the odds. When you brute force it like that it is sometimes called the "Monte Carlo" method for pretty obvious reasons. It's also nice and straight-forward, if not trivial, and can make a CPU run a little hot.

(Note: Pascal's triangle, the basis for modern combinatorics, was in fact created for a dice game as legend has it.)

I thought about Monte Carlo but I was obsessed with finding the mathematical way of doing it. Statistic books had sections on counting using combinatorics but that covered essentially permutations (n^6) and combinations in total, and not of sub-sets with in. So it didn't provide a way to count pairs.

Eventually, thanks to my brother and the Great Google (Ceremonial robe recommended but not required.), I found discrete math and generating functions:

http://stats.stackexchange.com/questions/3614/how-to-easily-determine-the-results-distribution-for-multiple-dice

You can get a formula that represents each number of times a specific combination is found.
So if you take, say. (A+B+C+D+E+F)^2 and expand it (which you can do on Wolfram Alpha per the link above) you get something like:
a^2 + 2 a b + 2 a c+2 a d+2 a e+2 a f+b^2+2 b c+2 b d+2 b e+2 b f+c^2+2 c d+2 c e+2 c f+d^2+2 d e+2 d f+e^2+2 e f+f^2

Each term is a combination of dice, but using letters for each face (A=1, B=2, etc). So [2 a b] means a 1 and a 2. The coefficient represents the number of times each combination occurs. So [2 a b] means there are two ways to roll a 1 and a 2 on two dice i.e. 1,2 or 2,1.. Everything to ^2 is a pair.

 This method expands out to N-dice. The downside is that there are 3003 combinations for 10 dice, and over 60m permutations. So while this is more manageable it's still a LOT of work. But since I know ^2 is a pair and ^3 is a triple, etc, it's possible to grab those terms using what amounts to a find function and then add up the coefficients using...well, Excel, because I wanted an answer more than I want to do the programming.

I wasn't questioning either his numbers or the wisdom of hiring a stats guy (thank you for that, getting it right does matter, and that's why we pay people with Masters.) (Hey, cool! I was right! I am genuinely a little surprised.....)

Here is what I worked out if you're curious. Alas, the expansion I embedded in the sheets didn't carry through, but I can always redo it to include them.

https://docs.google.com/spreadsheet/ccc?key=0AggHPTvQaNPydFMtN0w2eVg3OGZZVkxZU2N3S1JhaEE#gid=1

----
I confess, I worked all this out and was here to see if my numbers matched other peoples findings and then I find out there was a table...in the book...and the folks thinking about running it missed it too.....

I'd feel foolish but as I got the right answer and learned a very valuable tool I'm just gonna be happy, And finish my beer.
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MilitiaJim
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« Reply #10 on: August 02, 2012, 09:33:02 PM »

I'd feel foolish but as I got the right answer and learned a very valuable tool I'm just gonna be happy, And finish my beer.
Threat roll there, the beer is the activated AD for a crit.   Wink
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Bill Whitmore
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« Reply #11 on: August 02, 2012, 09:49:23 PM »




"Ulysses used a freely available online tool to find the formula for the averages."

No, not really. I used a freely available tool to expand the generating functions. The entire problem here is that you can't figure out pairs by averages on the dice. The expected value on 2 dice is 7 and tells you nada about the odds of pairs.

Counting dice can be a pain, as I am sure you know, and while the dice game created here looks a lot like a formv of Yahtzee (descended from existing games) the odds are not the same. One of the easiest ways to get around this is to write a program to roll 10,000 dice, check against success, and figure that this is a good representation of the odds. When you brute force it like that it is sometimes called the "Monte Carlo" method for pretty obvious reasons. It's also nice and straight-forward, if not trivial, and can make a CPU run a little hot.

(Note: Pascal's triangle, the basis for modern combinatorics, was in fact created for a dice game as legend has it.)

I thought about Monte Carlo but I was obsessed with finding the mathematical way of doing it. Statistic books had sections on counting using combinatorics but that covered essentially permutations (n^6) and combinations in total, and not of sub-sets with in. So it didn't provide a way to count pairs.

Eventually, thanks to my brother and the Great Google (Ceremonial robe recommended but not required.), I found discrete math and generating functions:

http://stats.stackexchange.com/questions/3614/how-to-easily-determine-the-results-distribution-for-multiple-dice

You can get a formula that represents each number of times a specific combination is found.
So if you take, say. (A+B+C+D+E+F)^2 and expand it (which you can do on Wolfram Alpha per the link above) you get something like:
a^2 + 2 a b + 2 a c+2 a d+2 a e+2 a f+b^2+2 b c+2 b d+2 b e+2 b f+c^2+2 c d+2 c e+2 c f+d^2+2 d e+2 d f+e^2+2 e f+f^2

Each term is a combination of dice, but using letters for each face (A=1, B=2, etc). So [2 a b] means a 1 and a 2. The coefficient represents the number of times each combination occurs. So [2 a b] means there are two ways to roll a 1 and a 2 on two dice i.e. 1,2 or 2,1.. Everything to ^2 is a pair.

 This method expands out to N-dice. The downside is that there are 3003 combinations for 10 dice, and over 60m permutations. So while this is more manageable it's still a LOT of work. But since I know ^2 is a pair and ^3 is a triple, etc, it's possible to grab those terms using what amounts to a find function and then add up the coefficients using...well, Excel, because I wanted an answer more than I want to do the programming.

I wasn't questioning either his numbers or the wisdom of hiring a stats guy (thank you for that, getting it right does matter, and that's why we pay people with Masters.) (Hey, cool! I was right! I am genuinely a little surprised.....)

Here is what I worked out if you're curious. Alas, the expansion I embedded in the sheets didn't carry through, but I can always redo it to include them.

https://docs.google.com/spreadsheet/ccc?key=0AggHPTvQaNPydFMtN0w2eVg3OGZZVkxZU2N3S1JhaEE#gid=1

----
I confess, I worked all this out and was here to see if my numbers matched other peoples findings and then I find out there was a table...in the book...and the folks thinking about running it missed it too.....

I'd feel foolish but as I got the right answer and learned a very valuable tool I'm just gonna be happy, And finish my beer.

Holy good god of awesomeness, we have another one on the forums! Cheesy

Welcome aboard Uysses!
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Uysses
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« Reply #12 on: August 02, 2012, 09:56:00 PM »

D'oh! Sure looks that way.

Exploding dice, on a roll and keep system (L5R), *still* make my head.....shutdown. Not explode. Definitely not explode.
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Bill Whitmore
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« Reply #13 on: August 02, 2012, 09:58:21 PM »

Yea, I tried to get a formula for Roll and Keep calculations...didn't go so well.

As a matter of fact, I think I am getting sick again just from thinking about it.  Lips Sealed
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Uysses
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« Reply #14 on: August 03, 2012, 06:21:23 AM »

There were some very long posts by L5R math geeks on making to work. Monte Carlo/Brute force seems the easiest method, although I suppose a generating function would be possible.

Fortunately, both Troll and Dicelab have worked it out for the days you want an answer and not a problem to solve.

http://www.diku.dk/~torbenm/Troll/
http://semistable.com/dicelab/

Or R and the Dice package:
http://www.r-project.org/
http://cran.r-project.org/web/packages/dice/dice.pdf
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