First, a small error: When you say P(X=1)=(d-2)/10, you mean P(X=0)=(d-2)/10.
Secondly, your formula for the standard deviation of old WoD is not correct. It should be sqrt((12-d)/10 - ((10-d)/10)^2) = sqrt(20+10d-d^2)/10. You have only the first term of the formula.
Thirdly, regarding botch, I analysed this in http://www.rpg.net/news+reviews/colu...ly18jun02.html (which seems to be temporarily unavailable, probably as an effect of the machine update). I didn't derive a general formula, but looked at the probabilities up to 3 dice in order to show that the botch risk increases with increasing skill.
Fourthly, for the probability of rolling at least two identical dice in k dice, it is 1 minus the probability of all being different. The probability of all k dice being different is 10/10*9/10*8/10*...*(11-k)/10 = 10!/(10-k)!/10^k. So the probability of a match is 1-10!/(10-k)!/10^k. This gives 0.0, 0.1, 0.28, 0.496, 0.6976, 0.8488, 0.93952, 0.981856, 0.996371, 0.999637 for k=0 to 10. This doesn't quite match your numbers.
The first term of the polynomial is obviously n*d^(n-1), where n is the number of dice. The second term appears to be (n^3-n)/2*d^(n-2), but after that I can't offhand find a system.
I got bored over winter break and worked out some of the probabilities for white wolf. Here's a list of the probabilities of success, failure, and botching for certain difficulties and certain numbers of dice. I didn't figure out an explicit formula, but used recursion based on the probabilities of a single die being a botch, success, or neither of the two. This does assume that botch means more 1s than successes. The forum doesn't seem to want me to be able to post the actual spreadsheet, so this is just the list of overall probability of success, failure, and botch (treating the last two separately), instead of the probabilities of getting X successes on Y dice at Z difficulty.
Hopefully this is understandable. First column is number of dice, 2nd is prob. of success, 3rd is prob. of fail, 4th is prob. of botching. Hope you enjoy.
Like avgotts, I didn't make a formula, but used a program to calculate the probabilities. I used my own dice-roll calculator "Roll" to do it. In this, you just describ ethe roll in a simple "programming" language and it will calculate the probability distribution. For Old WoD, the "program" is:
let c=N#d10 in
(count >=T c)-(count =1 c)
N is the number of dice and T is the target number.
I get the same numbers as avgotts.
For the new World of Darkness (with reroll on 10), it gets slightly more complicated:
count N# >7
let x = d10 in
repeat if =10 x then d10 else 0#0
I found the problem with Paul's calculation of the chance of getting a pair. His formula is correct, but when he calculates P(3), he uses k=3 in one of the places where he should use k=2, i.e., P(3) = p(2)+(1-P(2))*3/10 instead of p(2)+(1-P(2))*2/10, so he gets 0.28 instead of 0.19.
Since all the subsequent values depend on this, these are wrong as well.
It seems somewhat unlikely that someone will read this, due to the thread's age, but I'd still like to try.
I'm looking to generalize the formula for matching dice in a few ways. I'm looking (for starters) to compute the probabilities for matching triples, quadruples, etc. I've been working at this for a day or two now, but I just can't seem to get it right.
Further I'd also like to be able to compute chances for 2 pairs, pair and triple, 3 pairs, pair and quadruple, etc. This last bit is likely going to be fairly difficult, so no priority there. It's quite doable to brute-force this last bit by just generating enormous amounts of dice rolls, but I'd prefer to have mathematically approximated probabilities.
So far I've worked out by hand the first couple of probabilities:
prob of triples with
3 dice = 0.01
4 dice = 0.036
5 dice = 0.081
and ofcourse the probability of a triple with 21 dice should be 1.
Oh well, I've decided to brute-force it after all. I can still compute the exact probabilities ofcourse by just generating all different rolls. Here's the first part. The numbers before the arrow are number of pairs, number of triples, number of quadruples, etc. After the arrow is the exact probability, or as close as Java can approximate it without using BigDecimals. Note that the entries are sorted by ascending WotG's-usefulness (smallest sets first, largest sets last).
Re: #15: More Probability Examples
Linear Mode
