Ok, for character creation in my game, I came up with a mechanic where you take 8 d6 and roll them. You then group like numbers and count each dice as a bonus to skills. So, if I rolled 3 sixes, 2 ones, 2 fours and 1 two, I'd have a skill at 3, another at 2, etc. I wanted to figure out the probability of getting different values, though. Essentially, I want to know what's the likelyhood of getting 0, 1, 2.... 7 or 8  of the same number. I got these two, but I don't know how to do the math on the others. Thanks for any help you can assist me with!

0 = .225

8 = .0000004

There are plenty of utilities on the web that can help you with this. Google "Smallroller" for one that I use. Just tool around a bit, and you'll find others.

If this isn't resolved come Tuesday, PM me; I'm going to a weekend convention and don't have time right now, but that's a pretty simple probability problem.

David,

I'm getting:

0 of a specific digit = (5/6)^8 ~= .232568  (I'm using Excel.  Maybe different due to rounding?)
0 of any digit = 1 - (8*7*6*6*5*5 / 6^8) ~= .97

8 of a specific digit = (1/6)^8 ~= .0000006
8 of any digit = (1/6)^7 ~= .000004 

Correct me if I'm wrong.  I might be a bit rusty.

-Jason

The difficulty lies in the "extra" dice.

Like, the probability of getting 5 out of 5 is 1/6^5 

But the probability of getting 5 out of 8? I remember enough of my math to know it ISN'T the same as the above.

The small roller program can't do that, thanks for the advice, though. I tried poking around but aren't seeing any that do what I'm describing.

Actually, I think I did the math wrong on "0 of any digit."  I thought 97% was a bit much.  It should be:

0 of any digit = 1 - (8*7*6*6*5*19 / 6^8) ~= .886

How I did the math on this:

The easy way, as I saw it, to calculate in what cases you'd have all zeroes, was to find out the negative case, when you had at least one of every digit.  Since there are 8 dice and 6 digits, and each die must have at least one digit, this leaves us with two possibilities:

(a) One digit has 3 dice with that result
(b) Two digits have 2 dice with that result

Part (a) is easiest, so we do that first.  There are six possible digits that could have 3 as a result, so 6 is a multiplier for everything.  For the five digits left, their possible positions among 8 dice equal 8*7*6*5*4.  So for part (a), we have 8*7*6*6*5*4.

In part (b), we know that there are 15 possible combinations of any 2 digits (6*5/2).  So 15 is our multiplier.  The possible positions of the other four digits are 8*7*6*5.  Among the 2 doubled digits in the four positions remaining, there are 6 possible orderings of how they could go (4*3/2).  So for part (b), we get 8*7*6*6*5*15.

Add these two together, and you get 8*7*6*6*5*19.  As you can probably guess from my last post, I forgot about the 15 multiplier in part (b) and posted a bit hastily.  Divide this by 6^8 which is the number of total combinations, and there you are.

As for the probability of getting any digits with 5, 6, or 7 combos; this should be actually easier than calculating 1, 2, 3, or 4.  Give me a while and I'll try to post some guidelines for doing that.  Also, if the above explanation of how I got the chance of getting any zeroes isn't clear, let me know.

-JT

David,

7 of any digit: 6 possible digits multiplied by 8 possible positions that are not the digit multiplied by 5 possibilities for the other digit.  8*6*5 / (6^8) ~= .00014

6 of any digit: 6 possible digits multiplied by 28 possible combinations of positions that are not the digit (8*7/2) multiplied by 5 possibilities for the first other digit multiplied by 5 possibilities for the second other digit.  28*6*5*5 / (6^8) ~= .0025

5 of any digit: 6 possibile digits multiplied by 56 possible combinations of positions that are not the digit (8*7*6/6) multiplied by 5 possibilities cubed for the 3 remaining digits.  56*6*(5^3)/(6^8) ~= .025

So even getting 5 of any digit is still fairly remote.  Just barely over 1 in 40 characters will have exactly a 5 in anything, and having exactly a 6, 7, or 8 in any skill is rare indeed.

I think you might be able to go on from here using this math, but remember when doing "4 of any digit" or lower, there is a possibility for overlap, either with the calculations already done or within and of itself.  Perhaps the way around this is just subtracting out the specific cases as many times as the overlap in calculation would occur.

Good luck. ^^  I like mathy mechanics which don't require the user to be mathy.

-JT