World of Darkness Dice and Probability

[Baalthazaq]

I've got a bit of a problem, and I thought this forum might be a cool place to post it, saying as y'all seem to like math and (I'm guessing) probably have some experience with roleplaying systems in general.

I'm trying to break down the system in order to get an understanding of what odds my players have in any given situation, I'm trying to find an equation which represents WoD dice rolls. It seems like it should be trivial, but I'm clearly missing a step somewhere.

WoD dice work in this way:
You roll a number of dice (X), and you are attempting to get at least a specific number of "successful rolls" (Y).
A successful roll is anything >7. So 8, 9 and 10 are considered successes.

I.e. Bob needs to get 1 success in order to accomplish his task, he has 3 dice to roll which represents his skill. If any of them succeed, he succeeds. If he needed 2 successes, then 2 of the dice would need to roll successes in order for him to succeed.

Easy, 30% chance of success, X attempts, Y successes.

Here is where I'm getting thrown however. "10 again".
In the WoD system there is another factor called "10 again". Where if the player rolls a 10 on any previous dice, he gets to roll another dice for each 10, technically meaning you can get 10 successes from a single dice (it's just extremely unlikely), and I simply cannot factor that into the solution.

For example. Bob has 3 dice to roll. He needs 5 successes. A seemingly hopeless task that should be beyond his skill to accomplish under normal circumstances. Luck is on his side however, and he rolls 10, 10, 8.
3 successes, and 2 more rolls!
He rolls his 2 dice, one falls short at 3, but the other is another 10, giving him another success and another roll.
His final roll comes up 9 and he succeeds in his hopeless task, the bullet he was firing ricocheting off the floor and catching the thug in the scrotum in an amazing stroke of luck.

Shouldn't be the hardest question in the world, but I've spotted some easier, so enjoy.

(Mods: Feel free to move this into Games or Mathematics, I wasn't sure exactly where in the 3 forums it should go. I felt this was most fitting but if you feel otherwise, I apologize.)

[Madge]

I'm GMing a Werewolf: the Forsaken campaign at the moment, but I am not as crazy about probability - I just get the players to roll whichever stats look about right and let the chips fall where they may. I've got a bit of an instinct as to what is a reasonable amount of dice for them to be rolling for any task, so I think after a while you tend to internalise this sort of thing.

I do know that the "10 again" rule basically makes the chance exactly 1/3 - because you have a 30% of getting a 10, and then a 3% chance of getting 2 10s, and then a 0.3% chance of a third ten, and so on. So one die gives you a 1/3 of a success, on average.

[Goldstein]

For the actual probabilities,

Spoiler:

let P(r,n) be the probability of getting r successes with n die.

Consider the rolls one at a time. '10 again' essentially means that you get another roll on top of your allotted n rolls.

P(r,n)
= 70% P(r-1,n) [Rolling 1-7 reduces your available dice without reducing the number of successes you still require]
+ 20% P(r-1,n-1) [Rolling 8-9 reduces both your available dice and the number of succcesses remaining]
+ 10% P(r,n-1) [Rolling 10 reduces the number of successes you still require, but doesn't cost you a die]

Now we have a recurrence relation and, trivially, P(1,1) = 30%, P(1,0) = 0%, P(0,1) = 100%. Whack it into a spreadsheet and away you go. Bob had just over 0.2% chance of making that scrotum-shot.

[EricH]

Actually, I made up such a spreadsheet some time ago to do just that, back when we were playing a lot of White Wolf games. Let me see...ah, here we go; looks like 'some time ago' = eight years ago. But it's small, so I'll just attach it right here. Have fun.

Edit: Oh, FYI, the dice you mention will be the kind I labeled 'Specialty Dice', because some game or other made the distinction; just don't use any other kind, and you're all set for that game.
Edit again: New version of the file...20 months later...because the old one became unavailable for some reason. The default target number is 8 in this version, so it matches what we all seem to expect.

[a1s]

that formula seems very complicated (in the spreadsheet). and it doesn't match Goldstei's numbers...

[EricH]

The complexity of the formula is because, through various games, White Wolf kept changing how its dice behaved. The difference in outcome is because I left the default where I had it so many years ago, where 7 is a success and 6 is a failure. Change the number in cell C6 to an 8, and you'll see the same results. Sorry about that.

[Goldstein]

I like that, EricH. I was drawing tables like that while hoping to come up with some sort of formula, but I decided it wouldn't be practical to do it by hand and took it to every powergamer's favourite tool, Excel.

[Baalthazaq]

Perfect Goldstein, thanks.

I was going wrong by doing it as a 30% success, 70% failure, then trying to tag on the 10 again rule after that, as opposed to 70:20:10.

The spreadsheet isn't exactly what I want but it's trivial to build a data table out of it now, which again is perfect for what I need.

[Caesar]

The probability that a die generates zero successes is 7/10. The probability that a die generates one success is 2/10+1/10*7/10=27/100 (eight or nine OR ten and zero successes on the following roll). The probability of a die generating two successes is the probability of rolling a ten times the probability of the die generating one success. So the probability of the die generating n>0 successes is 27/10*1/10^n. With this we can calculate the probability of not rolling enough successes. By example, the probability of not rolling two successes with two dice is (I will call the probability of rolling n successess with a single die p(n).) p(1)p(0)+p(0)p(1)+p(0)+p(0). Or, to solve your example the probability of rolling at least 5 successes with 3 dice is 1-(3p(4)p(0)²+6p(3)p(1)p(0)+3p(3)p(0)²+3p(2)²p(0)+3p(2)p(1)²+6p(2)p(1)p(0)+3p(2)p(0)²+p(1)³+3p(1)²p(0)+3p(1)p(0)²+p(0)³). That is roughly 0.002 - a highly unlikely scenario.

[DRWS]

Caesar's equation is correct, and illustrates the huge amount of work necessary for calculating each probability of an outcome. I started to calculate each possible (within reason) outcome for up to ten dice, but eventually cheated and used averages of tens of thousands of generated rolls in Excel to create a table with a little bit of error.
The earlier posted .xls file is no longer accessible, but I would be interested in seeing an efficient spreadsheet that calculates each probability accurately. You can see that there are 10 parts to Caesar's equation for 3 dice (not counting the "1-"). For 7 dice there are 34 parts to the equation for each target number of successes. For ten dice there are 61 parts to the equation.

[Pjotr]

I've been working on probabilities in nWoD on and off for several months, and so far came up with the following formulae:

Code: Select all

P(s=0 | d=1) = 0.7                                                          (1)   the probability of getting no successes on a single die (not a luck die)
                                       
P(s=0 | d=n) = 0.7^n                                                 n>0    (2)   the probability of getting no successes on n dice - generalization of eq. (1)
                                       
P(s=n | d=1) = ((0.1^n)*(0.7))+((0.1^(n-1))*(0.2))                   n>0    (3)   the probability of getting exactly n successes on a single die (not a luck die)
equivalent   = (2.7)*(10^-n)                  
                                       
Es(d=n) = n*SUM [i*P(s=i | d=1)]                   over i=1 to inf   n>=0   (4)   the expected (average) value of the number of successes obtained from n dice                  
   Another way of approaching the expected value is this:                                    
   Let x be the number of successes obtained from a single die. Let r be the reroll chance (r= .1 for '10-again').                                    
   Then x = .7*0+(.3-r)*1+(1+x)*r                                    
   Solving for x gives gives x= .3/(1-r)                                    
   Since this is the expected value for a single die, we can simply multiply by n to get:                                    
                                       
Es(d=n) = n*(.3/(1-r))                                                      (4a)  the expected (average) value of the number of successes obtained from n dice

P(s<=n | d=1) = SUM [P(s=i | d=1)]                 over i=0 to n     n>=0   (5)   the probability of getting at most n successes on a single die (not a luck die)
                                       
P(s>=n | d=1) =1- SUM [P(s=i | d=1)]               over i=0 to n-1   n>0    (6)   the probability of getting at least n successes on a single die (not a luck die)
equivalent   = 0.1^(n-1)*0.3   = 3*10^-n                                    
                                       
P(s=1 | d=n) = n*P(s=1 | d=1)*[P(s=0 | d=1)]^n-1                     n>=0   (7)   the probability of getting exactly 1 success on n dice                  
                              
P(s<=n | d=m) = SUM [P(s=i | d=m)]                 over i=0 to n            (8)   the probability of getting at most n successes on m dice         
                              
P(s>=n | d=m) = 1- SUM [P(s=i | d=m)]              over i=0 to n-1          (9)   the probability of getting at least n successes on m dice         


Besides that, I've been using Excel to run simulations and plot nifty graphs based on them, but as far as the OP is concerned we need only look at equation (9) above. For small dice pools and successes needed, Ceasar's formula works quite well, but it becomes quite a monster when you start thinking about realistic dice pools and target numbers.
For example, a character with 4 dex and 5 firearms (plus specialization in assault rifles) firing a long burst from an AK will roll 4+5+1+3+4 = 17 dice. If she wants to down a character in one go, she'll likely need 7 or 8 successes (let's say 8 ).
The problem is that even P(s=7 | d=17) - the first term of the sum in eq. (9) - is quite difficult, because those 7 successes can come from any combination of the 17 dice. In theory, you could roll 7 successes on 1 die and zero on the other 16, or any combination of successes from a single die that sums to 7, as illustrated here:
(7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(17 permutations)
(6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(272 permutations)
(5,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(272 permutations)
(5,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(2040 permutations)
(4,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(272 permutations)
(4,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(4080 permutations)
(4,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(9520 permutations)
(3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(2040 permutations)
(3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(2040 permutations)
(3,2,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(28560 permutations)
(3,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0) =>(30940 permutations)
(2,2,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(9520 permutations)
(2,2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0) =>(61880 permutations)
(2,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0) =>(74256 permutations)
(1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0) =>(19448 permutations)
The probabilities for each of the above sequences can be calculated using eq. (2) and (3), and then multiplied by the number of permutations and added up.. but that's only the first term for use with eq. (9)! A similar exercise will have to be done for s=6 through 0 to find the other terms, which, while getting easier as s gets lower, is still quite a feat.

I have an Excel sheet which does the calculations up to n=5 and m=20 (including interaction with the 9-again and 8-again rules) but that may be too large to post. If you want to take a look at it, drop me a pm with your email.