Now, the basic conflict resolution is:
"Roll xd6 and achieve at least y successes."
A success is the result of a 5 or 6 on a d6.
Characters get 0-6 dice for each skill, which may be modified by circumstances such as weapons, magic items, or monster evasion/toughness/etc.
I want to convert these numbers from a d6 system to the d20 system to get an idea of granting d20-like bonuses to rolls will do. But the percentages are really slowing me down, because I never learned much beyond the standard statistics integrated in my high school math class. =P
Now, the basic chance of success on a given roll is 1/3 or 33.3%, but the chance of overall success (assuming multiple rolls with only 1 success needed) increases, but how to determine this percentage increase without resorting to plotting it all out in a chart eludes me.
MULTIPLYING THE PERCENTAGES DOESN'T WORK.
Please don't reply just to tell me to multiply 33% by 33%, because that results in a LOWER chance of success. Your chance of success would actually DROP from 33% to 11%, just by getting to roll an extra die to try and succeed, and that MAKES NO SENSE.
The reason I'm posting this to the CO Boards is, I know that there's at least one math whiz (possibly lurking) that might know how to solve this, and it is related to D&D, after all.
I want to know how to convert over the success/failure percentages to the d20 System so I can compare bonuses from Base Attack Bonus, saving throws, and feats like Weapon Focus in order to see how they stack up. I may have to modify certain feats and classes in order to emulate the chances of success to get more of the "Arkham Horror" feeling.
Thanks all!
...
P.S. It occurred to me to provide an example of some numbers I would like to change over.
Let's say I'm fighting a monster, and my "Fight" skill gives me 4 dice. I have two weapons, one that adds 2 dice, and another that adds 3 dice. The monster is "physically resistant," and that means that my "+2 weapon" only actually adds +1 die instead (the other is magical and therefore unaffected). I need a total of 2 successes in order to beat the monster, and after totaling everything up, I have 8 dice to roll.
I want to convert all of that into percentage changes ... bonuses and penalties, and the like. I want to then be able to convert those numbers into d20 System equivalents ... Armor Class and attack bonus.
The probability of getting k or more successes on n dice is:
Sum{i=k..n,(1/3)^i*(2/3)^(n-i)*n!/i!/(n-i)!}
(I think.)
The odds of exactly X successes out of N attempts with a 33.3% chance of success is (the probability of X successes) times (the probability of N-X failures) times (the number of ways you can organize the N dice)
The first value is (1/3) ^ X = X successes
The second is (2/3) ^ N-X = N-X failures
The third is N choose X or, written out, (N!)/((X!)(N-X!))
And so to get the value for "at least" X successes, you have to calculate that value for each value from X to N and add the results. Though sometimes it's easier to do it for "not less than"
In your example of having 8 dice and needing "at least 2", that means you can't have 0 or 1 successes.
p(0 successes)=1*(1/3)^0*(2/3)^8 = 256/6561
p(1 success) = 8 * (1/3)*(2/3)^7=1024/6561
so your odds of failing are (256+1024=1280)/6561, so your odds of succeeding are 1-that = 5281/6561=about 80% = in d20 terms, a 5 or better succeeds.
I think there might be a way to generalize it, but it's escaping me at the moment. If I think of it, I'll update here
Technically, this is probability, not statistics.
The probability of getting k or more successes on n dice is:
Sum{i=k..n,(1/3)^i*(2/3)^(n-i)*n!/i!/(n-i)!}
(I think.)
Okay, I think I've got the differences between statistics and probability worked out now. Misnomer on my part.
Uh... okay. Lemme try and figure this out...
Sum{i=k..n,(1/3)^i*(2/3)^(n-i)*n!/i!/(n-i)!}
What's i supposed to be? My math is fuzzy.
hogarth is correct
The odds of exactly X successes out of N attempts with a 33.3% chance of success is (the probability of X successes) times (the probability of N-X failures) times (the number of ways you can organize the N dice)
The first value is (1/3) ^ X = X successes
The second is (2/3) ^ N-X = N-X failures
The third is N choose X or, written out, (N! * X!)/(N-X!)
And so to get the value for "at least" X successes, you have to calculate that value for each value from X to N and add the results. Though sometimes it's easier to do it for "not less than"
In your example of having 8 dice and needing "at least 2", that means you can't have 0 or 1 successes.
p(0 successes)=1*(1/3)^0*(2/3)^8 = 256/6561
p(1 success) = 8 * (1/3)*(2/3)^7=1024/6561
so your odds of failing are (256+1024=1280)/6561, so your odds of succeeding are 1-that = 5281/6561=about 80% = in d20 terms, a 17 or better succeeds.
I think there might be a way to generalize it, but it's escaping me at the moment. If I think of it, I'll update here
Wow, okay. So I'm wondering ... if a monster's toughness were higher (requiring more successes) then it would be "harder to kill" than if it had a higher defense (penalty to the number of dice rolled).
But then again, it seems that the whole thing would be based on however many dice you got to begin with.
edit: okay, I pondered over this a little bit.
Taking damage, hit points, and attack bonus out of the equation, that means that if you needed one hit to kill it, and your average result on a d20 attack was 10.5, then that means if you were to hit it 80% of the time, it'd have an AC of, like 5.
More realistically, throwing the attack bonus back in there, it would probably have an AC of like, 15, and you'd have an attack bonus of like, +10.
...
I'm trying REALLY hard not to sound stupid about this, but my brain is fried just from trying to understand the equation, but I think this is what I've been looking for.
Thanks for your help so far.
EDIT: some other examples
If the monster let you roll 6 dice and you needed at least two to win, you'd have a 65% chance, meaning roll an 8 or better.
If the monster let you roll 8 dice but you needed at least four to win, you'd have a 25% chance to win, meaning roll a 16 or better.
EDIT: some other examples
If the monster let you roll 6 dice and you needed at least two to win, you'd have a 65% chance, meaning roll an 8 or better.
If the monster let you roll 8 dice but you needed at least four to win, you'd have a 25% chance to win, meaning roll a 16 or better.
Did you use the formula for those examples?
edit: Okay, I see where I'm getting hung up.
On the third part, that "N choose X or, written out, (N!)/((X!)(N-X!))"
would be "dice! / ((successes!
p(0 of 6) = 1*(1/3)^0*(2/3)^6 = 64/729
p(1 of 6) = 6*(1/3)^0*(2/3)^5 = 192/729
p(not 0 or 1 of 6) = (729-64-192=473)/729=.65
p(0 of 8)=1*(1/3)^0*(2/3)^8 = 256/6561
p(1 of 8) = 8 * (1/3)^1*(2/3)^7=1024/6561
p(2 of 8) = 28 * (1/3)^2*(2/3)^6=1792/6561
p(3 of 8) = 56 * (1/3)^3*(2/3)^5=1792/6561
p(4+ of 8) = (6561-256-1024-1792-1792=1697)/6561 = .25
edit: exactly right; N is dice and x is successes
yeah; sorry, I've always been bad at showing my work
edit: exactly right; N is dice and x is successes
I know the feeling. I *have* to show my work these days or I get so lost in what I'm doing. It feels like such a hassle.
...
So, to summarize what I understand so far, in order to figure out the probability of success, you have to figure out the probability of failure first (which you just subtract from 100% to get your prob. of success).
In order to figure out the chance of failure, you must use the formula to find out the individual prob. of EACH negative result. For example, if you need 3 successes, you must find out what the prob. of getting 0, 1, and 2 successes are, and add them together.
Now, the formula, as I understand it, has three parts, which I was able to clarify with that last post.
33.3% ^ success
times
66.6% ^ dice - success
times
dice! / ((successes!
N = # of dice.
So for 3 dice it would be 1 - (2/3) * (2/3) * (2/3) = 0.7 = 70%. Or on a d20 you'd need to roll at least a 21 - (70%/5%) = 7.
Except, you don't have to calculate the failures and subtract them, but in these particular cases, it's faster because there are fewer failure cases than success cases.
If you had 8 dice and needed 6 or more to win, you'd just add up the odds of having 6, having 7 and having 8 and be done. If i did that one with the failures, I'd have to add up the values for 0,1,2,3,4 and 5 and subtract them. Just do it whichever way is less math.
EDIT: Nathreet, that's the odds of having at least 1 success out of N dice; we're dealing with having at least X successes out of N dice
So, what you are trying to attempt here is to convert into d20 a board game that is based off an RPG that has already been converted into d20.
The d20 Call of Cthulu isn't bad. It does a decent job maintains the proper atmosphere, hopelessness, and kicked-in-the-nut feeling that the setting is famous for, and it does this while being mostly compatible with other d20 products and classes and such. It does add a much more combat-oriented feel, but as you are coming from Arkham Horror this is likely something you'll be comfortable with.
The best way to play with the Cthulu mythos is with Chaosium's latest revision, Call of Cthulu 6th Edition.
I hope this helps you get your wierd fiction and creepy horror game rolling.
EDIT: Nathreet, that's the odds of having at least 1 success out of N dice; we're dealing with having at least X successes out of N dice
Oh, then that's a bugger. I think there's an equation for this involving factorials, or that may apply to something else. Try wikipedia then.
EDIT:
Aha I know. The probability of getting X successes out of Y is the probability of getting 1 success out of Y times the probability of getting 1 success out of (Y - 1), etc.
So your chance of getting 3 successes out of 5 rolls:
(1-(1-P)^5)*(1-(1-P)^4)*(1-(1-P)^3).
Where P is the chance of succeeding on 1 roll.
EDIT 2:
Wait, that's not right.
EDIT 3:
Here it is:
http://en.allexperts.com/q/Probability-Statistics-2077/probability-problem-2.htm
n!/(x!(n-x)!)*P^x*(1-P)^(n-x)
n = number of dice rolled
x = number of successes needed
So chance of getting 3 good rolls out of 5 is:
5!/(3!2!)*P^3*(1-P)^2
=10*P^3*(1-P)^2
If P = 1/3 then the answer here is 16.5%.
You've got it.
Except, you don't have to calculate the failures and subtract them, but in these particular cases, it's faster because there are fewer failure cases than success cases.
If you had 8 dice and needed 6 or more to win, you'd just add up the odds of having 6, having 7 and having 8 and be done. If i did that one with the failures, I'd have to add up the values for 0,1,2,3,4 and 5 and subtract them. Just do it whichever way is less math.
About the success/failure thing: right, duh. I chock that one up to fried brain.
hope that table helps, Dither
That looks right now; you started fixing it right as I started to leave a comment about it; quick work
hope that table helps, Dither
Thanks playswithfire and Nathreet, I hereby declare you both made of win.
...
Actually, I'm not sure which values to add to determine *more* successes. Oh, wait ... you add horizontally? No... I'm not sure what you mean about adding the chance of getting more successes?
He went back and did the adding for you; it now shows the odds of getting at least that many successes; that's what he meant by 'fixed the table to reflec this'; you just never saw the broken version of the table
Oh, okay. I get it now. I was thinking to myself that no matter how many dice rolls you got, the chance of success should never be 100%, let alone MORE than 100%. :D
edit: Okay, so now the process of converting some of the rolls and difficulties in the board game to reflect d20 bonuses and DCs ... this should make for an interesting project.
Thanks for your help guys. I'd love to post some of the results of my own findings / discoveries, I suppose I'll just update this thread when I find something interesting.
You "add" probabilities by multiplying their opposite probabilities: 0.33 "add" 0.33 = 0.66 * 0.66 = 0.4356 = 44%
*skims thread*
You "add" probabilities by multiplying their opposite probabilities: 0.33 "add" 0.33 = 0.66 * 0.66 = 0.4356 = 44%
That makes sense. I think I remember that now from my math classes of yore. I'll have to keep that in mind for less... complicated... stuff that I attempt in the future.
Assuming 33% chance of success per roll.
If you take the chance for success, 1/3, and multiply it by 1/3, then what you're actually determining is your chance to get two successful dice. Instead, to find the chance to succeed on X dice, you don't want to find how likely you are to succeed, but rather, how likely you are not to fail. In other words, you'd find the chance to fail and multiply that.
On two dice, you'd have .66x.66 (~44%) chance to still fail. On 3 dice it'd be .66x.66x.66 (~29%) and so on.
Similarly, to figure out the chances to get exactly one success on, lets say, 5 dice, you'd just need to use .66 for 4 dice, and .33 for the 5th. So, .66x.66x.66x.66x.33 (~6%).
As I said, I know other people already did the math for you, but yeah...now if you need to tweak something you know the "easy way."
On two dice, you'd have .66x.66 (~44%) chance to still fail. On 3 dice it'd be .66x.66x.66 (~29%) and so on.
Similarly, to figure out the chances to get exactly one success on, lets say, 5 dice, you'd just need to use .66 for 4 dice, and .33 for the 5th. So, .66x.66x.66x.66x.33 (~6%).
No quite; .66x.66x.66x.66x.33 is the odds that the first four you roll fail and the fifth one is the success. Since order doesn't matter, you have to multiply by 5c1 = 5!/(4!*1!)=5 to get the real value of the odds of getting exactly one out of five. So, it's around 30% as shown in the unmodded table supplied by Ubasrawr.
The formula for exactly X successes on N rolls, as has already been stated, is
N!/((N-X)!*X!) * pSuccess^X*(1-pSuccess)^(N-X) or, for the specific 1/3 case
N!/((N-X)!*X!) * (1/3)^X*(2/3)^(N-X)
You dropped the first term from the function in your example, which you can do for the exactly 0 case, but not any of the others.
I think its been covered, but I wanted to give you a brief explanation for why what you initially thought (multiply the chances) didn't work and how to make it work.
Thanks, I actually figured that out on my own; I just wanted to throw a warning out there to anyone that thought the answer was simpler than it looked.
Cmon people. Am I going to have to write your answers in LaTeX and upload it to my photobucket so the poor fella can understand what you are writing. This SUM{i = blabalbal} is appalling looking and any mathematician would be ashamed of such poor typography.
Please. If you were really serious, you'd:
1) use LaTeX2HTML and use [html][/html] tags
2) have done it already
That makes sense. I think I remember that now from my math classes of yore. I'll have to keep that in mind for less... complicated... stuff that I attempt in the future.
Wait! I got it wrong. You still need to go back to the original probability. The correct equation:
0.33 OR 0.33 = 1 - 0.66 * 0.66 = 1 - 0.44 = 0.66
0.1 OR 0.1 OR 0.1 = 1 - 0.9 * 0.9 * 0.9 = 1 - 0.729 = 0.271
The "addition" is actually an "or" operation - the result is the probability that at least one event is true (at least one die meets the target number).