D&D 4E Showing the Math: Proving that 4e’s Skill Challenge system is broken (math heavy)

[rabindranath72]

I love interdepartmental donnybrooks.

I'm on the side of pure math vs. the stats guy.

What about you?

I give pure math 3 to 1 over the Pascal variable.

Taking bets now!

LOL thanks for the humour

So, which would be "pure math"? Statistics is just an application of mathematics.

 

[Spatula]

- At 8th (party) level we see the complexity do a flip, and more complex challenges become easier, for 1st level skill challenge. Is this because of the level disparity: does the same thing happen for a 10th level party for a 3rd level skill challenge? If so maybe PCs should auto-win on skill challenges 5 below their level.

There's no such thing as a skill challenge that isn't your level, at least as far as the DMG skill challenge templates are concerned.

 

[pemerton]

rabindranath72:

On a simple probability tree I get that, out of 32 possible outcomes for 5 rolls at 50/50, there are 26 branches that have at least 2 failures before 4 successes. Those odds aren't going to change all that dramatically if I were to make the chance of succcess 55/45,

So you are saying that my use of a probability tree is in error. Why?

 

[vagabundo]

There's no such thing as a skill challenge that isn't your level, at least as far as the DMG skill challenge templates are concerned.

Well then I'm reading the OP wrong, which is possibly considering I get cross-eyed looking at that maths.

It is funny I do love maths.I do grinds for leaving cert students and I'm an engineer, but probability was something Ive always avoided.

 

[Eldorian]

P(X=x|r,p)=OVER(x+r-1, r-1)*p^r*(1-p)^x

You mean,

P(X=x|r,p)=OVER(x+r-1, x)*p^r*(1-p)^x ?

r= successes =4, p = probably of success = .55, x= number of failures = 2?

That's what I see on wikipedia. Never used this thing, myself.


Take a look at the cumulative distribution function for your guy. According to wiki,

http://en.wikipedia.org/wiki/Negative_binomial_distribution#Properties


Oh here we go. I think I've found what you're using. You did it wrong, tho.

The negative binomial random variable, denoted by X ~ nb(r, p) is a generalization of the geometric random variable. Suppose you have probability p of of succeeding on any one try. If you make independent attempts over and over, then X counts the number of attempts needed to obtain the rth success, for some designated r >=1. When r = 1, then X ~ geo(p).

We often let q = 1 - p be the probability of failure on any one attempt. Then the probability of having the rth success on the kth attempt, for k >= r, is given by

P(X = k) = C(k - 1, r - 1) * q^(k - r) * p^r .


See, you found the probability of getting the 4th success on the 6th attempt. This isn't even possible in the experiment at hand. Here k= x+r

What you want to calculate, using your method, is

P(X = 4) = C(4 - 1, 4 - 1) * (.45)^(4 - 4) * (.55)^4 = 0.09150625

plus

P(X = 5) = C(5 - 1, 4 - 1) * (.45)^(5 - 4) * (.55)^4 = 0.16471125


That is, the sum of the chance of getting the 4th success on 4th trial, and 4th success on 5th trial.

add those together and you get

0.2562175

Easily accounts for the OPs slight round off errors.

Source: http://www.wku.edu/~david.neal/statistics/discrete/negbinom.html

 

[rabindranath72]

Yes sorry, that's the correct form; I wrote it incorrectly.

The probability of failing the skill challenge would be:

P(X>=2)=P(X>1)=1-P(X<=1)=.84

What this means for the game, I have no idea at all. I just made a comment on the solution of the supposed problem (X failures before r successes).


Does the DMG state that you have a fixed number of total die rolls to succeed?

 

[Tervin]

Ehy, Mr. Funny, I am a mathematician.

/---/

So, the probability of getting AT MOST 2 failures is 0.26

Well this is back to front. We don't want to know the probability of getting at most 2 failures in 6 attempts. We want to know the probability of getting at most 1 failure in 5 attempts, as the second failure means that the skill challenge has failed.

The problem can be solved with several methods, but if we look for different things we will get different results.

 

[rabindranath72]

Well this is back to front. We don't want to know the probability of getting at most 2 failures in 6 attempts. We want to know the probability of getting at most 1 failure in 5 attempts, as the second failure means that the skill challenge has failed.

The problem can be solved with several methods, but if we look for different things we will get different results.

Yes, that was the cumulative. What we want is the survival function 1-P(X). See my previous post.

 

[Eldorian]

Does the DMG state that you have a fixed number of total die rolls to succeed?

Roll dice. Stop when you get two failures. If you have 4 (or more) successes, you win. Otherwise, lose.

Roll dice. If you get 4 successes, before 2 failures, you win. Otherwise...

Roll dice. If you get 4 (or more) successes on 5 attempts, you win. Otherwise....

Last one is what the OP did. They're all the same experiment.

 

[two]

LOL thanks for the humour

So, which would be "pure math"? Statistics is just an application of mathematics.

No sir, no.

Take that weak PR and keep it to yourself, or use it to amuse the children.

I have a math degree myself, and grew up in an academic household, so again let me repeat: no sir.

While it is true that set design is an art, it is also true that if you were to ask around, nobody's life goal is to be a set designer. They want to be the playwright, or at least the lead actor.

Again I say, no.

Statistics is a useful, valuable, difficult, etc. etc. {insert more fluff here} but it does not garner the same respect as other "more pure" forms of mathematical investigation.

Let's not argue, though. Let's walk arm-in-arm to the nearest pub and over a couple of warm pints, discuss... well, what have you?