Quote from: Age of Fable;283672Do you count the 2d6 as two seperate totals, or as a single total?

Sorry for the delay in answering.

Two separate totals. That is, he rolls 1d12+2d6 and scores 2 (1d12), 5(1d6) and 3 (1d6), so he gets a 5, the highest roll.

No worries.

OK: your chance of getting n as your best result, out of a roll of three dice, of size a, b and c.

To get n, you have to get

i) no results of more than n. and

ii) at least one roll of exactly n.

Your chance of getting a result of more than n is based on the number of faces on the dice which show more than n.

If the dice is less than or equal to n, then this will be 0 (your chance of getting more than 4 on a d4 is 0 for example).

If the dice size is n or more, the chance will be (x-n)/x, where x is the size of the dice.

So I'm going to write this as max(0,(x-n)/x), 'max' meaning 'the higher of', because I can't find the mathematical symbol for that.

So your chance of not getting a result of more than n is 1-max(0,(x-n)/x).

So your chance of rolling three dice, of sizes a, b and c, and not getting any result of more than n, is

(1-max(0,(a-n)/a))*(1-max(0,(b-n)/b))*(1-max(0,(c-n)/c))

On to part two.

If you've rolled three dice and gotten no result of more than n, what's the chance that you got at least one result of exactly n?

Looking at the chance that a single dice will not show n:

We can disregard all numbers greater than n, because we're only looking at cases where no such number was rolled.

If the dice size is less than n, the chance will be 1.

If the dice size is greater than or equal to n, the chance will be (n-1)/n.

So the chance that all dice will show a number other than n can be written as

((n-1)/n)^z , where z is the number of dice whose size is greater than or equal to n.

Thus the chance that at least one dice will show n, will be 1-(((n-1)/n)^z)

and therefore, the chance of rolling three dice, of sizes a, b and c, and getting the highest single result of n, is

(1-max(0,(a-n)/x))*(1-max(0,(b-n)/x))*(1-max(0,(c-n)/x))*(1-(((n-1)/n)^z)), where z is as above.

So, plugging this in to Excel, I got the following for 1d6 or 1d6 or 1d12:

1   0.002314815
2   0.016203704
3   0.043981481
4   0.085648148
5   0.141203704
6   0.210648148
7   0.083333333
8   0.083333333
9   0.083333333
10   0.083333333
11   0.083333333
12   0.083333333

The first number is a possible best result, the second number is the chance of that result. So the most common result by far is 6.

For d10 or d8 or d6, the chances are:

1   0.002083333
2   0.014583333
3   0.039583333
4   0.077083333
5   0.127083333
6   0.189583333
7   0.1625
8   0.1875
9   0.1
10   0.1

Given the chance of each individual result for either player, it's easy to find the chance of any given combination (just multiply the two).

So I did a grid in Excel that does that.

Then it's pretty easy to just add up the chance of getting any case where the first player wins, and the same for the second player winning and for a tie.

If you have player A rolling a d12, d6, and another d6, and player B rolling a d10, d8 and d6,

player A should win about 47.6% of the time,
player B should win about 41.1% of the time,
and they should draw about 11.2% of the time.

I've attached a spreadsheet to this (in a zip file because you can't upload Excel files to this forum) - you can change the dice sizes (up to 30) and it'll update the chances of various results.

QuoteOriginally Posted by Age of Fable
((n/d)^Y) * (1-(((n-1)/n)^Y))

I did some calculations with this, but not being terribly adept at mathematics, the results don't really mean anything to me. Any chance you could give me an idea as to how much your chances increase on successive additions of d10s, take the highest? I'm trying to make a skill system based on that, and I want to know how high I should go with the number of dice I allow. Thanks.

For game designers who need to toy around probabilities for their new improved secret mechanism there's a nice tool called troll. I'm not going to pretend it's user friendly. If you're allergic to math formulas, it's not for you. If you're scared of running an executable with a command line, it's not for you. But if you're willing to try and spend 20 minutes to understand the software you'll be rewarded.

Example:
Say I want to get probabilities for a pool of dice to have at least one die superior or equal to a target number. In troll language the formula could look like this :
   count N < x d y
which can be deciphered as : count the number of dice that exceeds N in a pool of x y-sided dice. Basically the roll is a success for any case where the count is superior to 0.
The command line to get a result looks like this :
   troll 0 filename.txt N=8 x=5 y=10
which means I demand an evaluation of probability for a succes on a 5d10 pool against a target of 8. Here is the result :
  Value    % =                 % >=
     0 :   32.768             100.0
     1 :   40.96               67.232
     2 :   20.48               26.272
     3 :    5.12                5.792
     4 :    0.64                0.672
     5 :    0.032               0.032

Average = 1.0    Spread = 0.894427191  Mean deviation = 0.65536
The case 0 (roll failed) is rated with a 32.7% chance. So basically it's a success with sth like 67% chance.

Spread good troll lovin' !

Quote from: The Worid;290807I did some calculations with this, but not being terribly adept at mathematics, the results don't really mean anything to me. Any chance you could give me an idea as to how much your chances increase on successive additions of d10s, take the highest? I'm trying to make a skill system based on that, and I want to know how high I should go with the number of dice I allow. Thanks.

That formula is only for 3 dice.

I'll work this out ASAP - but chances of what?

The idea is that you add your attribute+1d10, trying to exceed a target number. Skill levels allow you to roll more d10s, taking the highest to add to your attribute. My question is, how much does the average result go up by for each successive d10? So, if I needed a 6 or more (assume no bonus), I have a 50% chance at 1d10, and then higher chances at 2d10, 3d10, and so on.

Thank you again.

Rolling X d10s, your chance of getting at least one result of Y or more:

Your chance of getting less than Y on a single roll is (Y-1)/10

Your chance of getting less than Y on all of X rolls is ((Y-1)/10)^X

So your chance of getting Y or more at least once on X rolls is 1-(((Y-1)/10)^X)

I see. Many thanks; sorry for the delay in response.

No worries.