Wednesday, June 20, 2012

Using Coins to Simulate Dice

Being a Role playing game fan I worry too much about dice.  One of my more "useful" musings was how to represent Dice with coins.  This will represent a good enough solution, I came up with a fix to make the probabiliy work out correctly but it is unwieldy.

So you are going to need a ton of coins of one variety and a single coin of another variety.  I use pennies and quarters myself.  While this could be performed with several successive coin flips I normally use a hand full of coins to speed it up.  As a note if you really dont care about probablility you can use pente stones as well, you just need two colors.  One will represent the pennies and the other the quarter.

Ok the first step in this is to devide the dice you are wishing to simulate by two.  A nice thing about this method is it can represent any even dice so you can do the whole RPG dice set, but also weird stuff like a d14. Next you will take that many pennies.  Then you will add your quarter.  Shake em all up and toss them out just like you would in AD&D.  From here assign 1 to heads and 2 to tails.  For the quarter Heads is 0 and tails is -x.  In this case x is the number of pennies thrown minus 1.  Now you just add them up.

You will see this is not correct as per a single dice roll.  but taking a quick look at the rolls possible for a D6 you will see it creates a bell curve not represented on dice.

Penny 1: Penny 2: Penny 3:  Quarter 0 / -2
2         2       2 =      6     4
2         2       1 =      5     3
2         1       1 =      4     2
1         1       1 =      3     1

This makes the break down of roll probablility something like the following

1= 12.5% 2= 12.5% 3= 25% 4= 25% 5= 12.5% 6= 12.5%

This is opposed to a six sided die which all six have an even chance of hitting.  There is a solution but it gets a bit goofy. Basically if you notice my number distribution is the two numbers around half of the dice sides that get the extra percentange.  So is 3/4 on a D6, 5/6 on a D10 and 10/11 on a D20.  YOu you need to do to fix this is use a coin flips to nomralize this percentage hiccup.

What I came up was sort of a logical coin flip.  here you are introducing another coin type so lets call it the dime.  you are going to a number of dimes equal to Y where Y is the number of pennies minus 1.  Here you only use the dimes if the original throw results in half the die type or half the dice type plus 1.  Basically those are the two numbers that are most probably, the 3 and 4 on the standard six sided dice. As I have said the dimes are logical and have instructions rather than numbers associated with them.

If the Dime is heads you do nothing if the dime is tails you shift the result of the original role by one.  The direction of the shift is dependent on the number originally rolled.  Numbers equal to half the number of dice sides you subtract 1.  Number equal to half the number of dice sides plus 1, well here you add one.  Keep in mind you are probably throwing more than one die here.  But only for previous results of 3 and 4.

So on our D6 form above

Original Throw Dime one heads Dime one tails Dime two heads Dime two tails
3         3                 2         2                1
4         4                 5         5                2

Again I am just some roleplayer and the math is beyond my basic understanding of probability I wish someone with a math background could take a look and tell me if this actually works.

NOTE: In looking at this i am question the the second dime flip.  I know when I added these I was basically pushing the extra percentage to different numbers.  So in by basic system i am ending up with 3 and 4 as overly represented numbers.  After the first dime flip the more common numbers push to 2 and 5.  I almost need to add something that then splits the extra percentage in to maybe taking the first dime flip into account and saying if that was tails and the second was tails then when you have two results of tails then shift as we did in the previous example.  But here you would need to do it on results of 2 and 5 since those are the most frequent numbers.

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