The Gaming Den

For those of you who don't know, there is errata published that "fixes" skill challenges. Namely, the DCs have been adjusted and there have been some slight tweaks to the system. The biggest change is that your number of failures allowed is always three, no matter what complexity skill challenge. The total number of successes, however, remains the same, which means that you're still rolling a billion dice.

http://www.wizards.com/dnd/files/UpdateDMG.pdf

That's a link to the errata.

Now, my question is: are the changes enough, are the changes too much, or are they just right? Personally, I'm assuming that the skill challenge system still sucks ass and that the devs have still managed to make it too hard. However, there has been no analysis of the math yet, so I'm going to look at it in this thread.

Keep in mind a couple things:

1. It's been awhile since I took a math course. Writers disdain math as being the lesbian sister of biology. Double-check my calculations.

2. It's been awhile since I took a stats course. Unlike the ENWorld thread, I'm going to be using conditional probability because I'm more comfortable with that.

3. I don’t know how this is going to work out. Again, I’m assuming fail, but the developers might have fixed it. I’m more than willing to be wrong, if the math checks out.

4. I’m rounding for the numbers I’m giving you, mostly to three decimal places. More than that is annoying. However, when I’m doing the actual calculations, I’ll be going out to whatever the calculator displays. I’m just going to give you the rounded numbers.

So here we go.

I'm going to examine a total of six skill challenges. A complexity one skill challenge at level one, a complexity five skill challenge at level one, a complexity one skill challenge at level fifteen, a complexity five skill challenge at level fifteen, a complexity one skill challenge at level thirty, and a complexity five skill challenge at level thirty.

All of these skill challenges will be of the “hard” difficulty—so DC 15, DC 23, and DC 33.

Here are some assumptions I’m going to make:

1. Characters will start with a +3 ability score modifier for the skill checks. At fifteenth level, this will be improved to a +5. At thirtieth level, this will be improved to +7.

2. Characters are trained in the appropriate skill, giving them a +5 bonus.

3. Characters do not take the Skill Focus feat, as it sucks and only a retard would take it.

4. Characters have a +1 racial bonus on the check. Most characters are probably going to have a +2 racial bonus on the skills they use, but we’ll just say +1 because it’s average. Also, it gives us a ¾ chance of success at level one, which is easy to work with.

Overall, this gives them a starting bonus of +9, equal to that of the characters in the original ENWorld thread.

Level 1; Complexity 1
Characters have a +9 bonus on the check. Thus, to hit a DC of 15, they have a 75% chance of success. A complexity one skill challenge requires four successes before three failures.

Four Successes
.75^4 = .316.

Four Successes, One Failure
(.75^4)*.25 = .079. Since this could happen in four different orders (FSSSS, SFSSS, SSFSS, SSSFS), we multiply this by four to get the percent chance, leaving us with .316.

Four Successes, Two Failures
(.75^4)*(.25^2) = .0198. Bah. Numbers wouldn’t make sense if I didn’t go out to four digits. Since this could happen in ten different orders (FFSSSS, FSFSSS, FSSFSS, FSSSFS, SFFSSS, SFSFSS, SFSSFS, SSFFSS, SSFSFS, SSSFFS), we multiply this by ten to get the percent chance, leaving us with .198.

A hard complexity one skill challenge at level one thus gives players about a 83% chance to succeed. My eyebrows are already raised.

At this point, I am no longer going to write out the combinations of successes and failures. There is a simple formula to determining the number of possible orders to the successful skill challenges. With one failure, there are n possible orders, where n is equal to the number of required successes. With two failures, there are 1+(2*n-1)+n-2+n-3…n-n possible orders (n again equal to the required successes).

I’m sure there’s a simpler way to express this, but again, maths is hard, and it’s been awhile since my last math course.

Level 1; Complexity 5
Twelve Successes
.75^12 = .0317.

Twelve Successes, One Failure
(.75^12)*.25 = .00792. Twelve total combinations: .095.

Twelve Successes, Two Failures
(.75^12)*(.25^2) = .00198. Seventy-eight total combinations: .154.

Wow. Fuck going farther than that. I had this all planned out, and we see fail within the first calculations: a complexity five skill challenge has about a 28% success rate whereas a difficult complexity one skill challenge has a 83% success rate.

The Bottom Line
Skill challenge errata fails. Everything is either a cakewalk or really fucking hard. Given a character who isn't trained in a particular skill, the success rate for a high-complexity skill challenge drops precipitously, while characters who have a modicum of competence can breeze through lower-complexity skill challenges. The only way that these can balance out is if your low-complexity skill challenges involve groups of retarded, shit-covered chimpanzees for players, whereas your high-complexity skill challenges require bomb-squad levels of specialization to succeed.

As a final note, feel free to correct my math. Call me a fucking moron. Tell me how my calculations could have been done better by the aforementioned apes. Maybe I'm completely wrong about all of this. Again, I'm not a math major, and I did all this on the computer, so I don't have any of my work to double-check.

Edited for "math = hard."

Psychic Robot wrote:groups of retarded, shit-covered chimpanzees for players

Nice.

CatharzGodfoot wrote:

Psychic Robot wrote:groups of retarded, shit-covered chimpanzees for players

Nice.

I don't suppose that means you agree with the mathematical analysis, does it?

I've only got as far as skimming the errata. Easy/Moderate = autopass regardless of complexity at low levels was the first thing that struck me, followed by Hard+Complex = likely fail at any level. I can't get my brain in gear sufficiently to figure out the stats right now; like you, it has been a LONG time since I've had to deploy anything other than basic arithmetic in anger.

But writers *love* lesbians. As a general rule.

Psychic Robot wrote:

CatharzGodfoot wrote:

Psychic Robot wrote:groups of retarded, shit-covered chimpanzees for players

Nice.

I don't suppose that means you agree with the mathematical analysis, does it?

No, it just means that I was impressed by your phrase construction.

It figures that my first real post here would be all mathy...

The math in your second case is off (should be .316, making grand total 83%), and I'm pretty sure your state counting in the 12 success example is off, but the underlying work looks fine.

Since counting states blows, here's a more streamlined way to do the whole thing for those at home who want to play along . Beware, math derivation stuff is in the spoiler, results are below.

Ok, so if you need 4 successes before 3 failures, you need to make at most 6 rolls. From the combinatorics PoV, SSSSFF, SSSSSF, SSSSFS, and SSSSSS are all individual results. Using your old DC 15 example, the odds for each of these are 0.1780, 0.0593, 0.0593, and 0.0198, which sum to 0.3164 just like you had in your first example (after account for rounding errors anyway). No one would ever roll these cases of course, they'd stop after the first 4 cause the last two don't matter, but including them in makes the math easy. And since all we care about are cases where you get at least 4 successes and at most 2 failures on your 6 rolls, I'm going to bring them along.

Case A, No Failures: There is exactly 1 case in which each roll is a successes. The of getting it is S^6, where S is the probability of a success.

Case B, 1 Failure: There are exactly 6 cases where one of the rolls is a failure and the other 5 are successes. We don't care about 2 of these cases in practice, but we have to include them. The probability of getting one of these outcomes is S^5*(1-S). 1-S is just the probability of failure, and you can just use that if you want.

Case C, 2 Failures: There are exactly 5 cases where both of our two allowed failures happen right next to each other: 1 at the front of 4 successes, 1 at the end of 4 successes, and 3 breaking them up. There are also 10 cases where the failures happen with successes in between them. Total of 15 cases with two failures. Probability of getting one of these is S^4*(1-S)^2.

So we've got 22 outcomes that we actually care about. Going back to the OP for a second, this is the same number you would get if you added in the extra cases that we don't actually care about, so we're still in agreement.

So, after weighting each prob by its number of cases, we get this obnoxious thing:

[1 * S^6] + [6 * S^5 * (1-S)] + [15 * S^4 * (1-S)^2].

Doing all that math gets us an 83% probability, exactly what you got after correcting your second case.

The reason to do it this way is because you can generalize your numbers of cases much more easily, and then just multiply through instead of worrying about counting states. If you need n successes, then you have 1 case with no failures where the probability is S^(n+2), n cases with 1 failure where the probability is S^(n-1)*(1-S), and two groups of 2 fail cases: n-1 cases of fail/fail + n*(n+1)/2 cases of fail/stuff/fail where the probability is S^(n-2)*(1-S)^2.

And here's the fantastic general result:

Probability of success = S^(n+2) + n*S^(n+1)*(1-S) + [n*(n+1)/2 + n - 1]*S^(n)*(1-S)^2

where n is the number of success you need, and S is the chance of success.

This gives the 83% you should have gotten for your first example, and a slightly lower 26.5% chance of success for your 12 success example because there are 4 less successful states than you counted.

Successes are actually worse than you were calculating. Hooray for system features!

Edit - Fixed equation.

Oh. Yeah. Yes... I mean, I was totally about to say that, but then the phone rang

Welcome aboard. Sincerely.

L33t math skillz r0x0r, but the ability to explain them well to the lay person is even better. 8)

EDIT: I meant to ask... what is the expression for the general case in which the number of permitted failures is a variable?

If you have n total items, and you want to pick k of them in an order-independent way, then the total number of combinations is called "n-choose-k" (or the binomial coefficient) and is calculated:

n! / [ k! * (n-k)! ]

So if you want to count all the ways you can roll 8 times with exactly 3 failures, the number of permutations is 8-choose-3, or 8! / (5!*3!) = 56.

So if you've got probability S of succeeding on any given roll and you need to get N successes before you get F failures, your overall probability of success should look something like this:

Summation from k = 0 to (F-1) of:
S^(N+F-1-k) * (1-S)^k * (N+F-1)-choose-k

The intuition here is that you're rolling N+F-1 times (the most you would ever need to roll to get N successes before F failures), and k is the number of failures rolled, so you're summing over all possible number of failures the probability of rolling that number of failures in a specific order times the number of orders.

Wow, that looks ugly in text form. Can you read that? I may have to make a pretty picture in an equation editor...

Amra wrote: I meant to ask... what is the expression for the general case in which the number of permitted failures is a variable?

Short answer:

There isn't an easy one. Sorry :-/

Long answer:

What I've got so far is the counting rules for 3 distinct states, and the probability of landing in any one of them. Multiply and add them all up, and that's the total probability of landing in any of them. Every additional failure you allow is an extra state that you could land in, you need to add another term to the equation. Which you can do of course, it's just uglier. A lot uglier.

Maths ahead. Feel free if you want the really really long answer, or just trust me that it gets bad

Right, if your still reading, you really want to do this, so here's the general probability of success :

Sum, g=1 to f, of Ng(n)*S^(n+f-g)*(1-S)^(g-1)

Where Ng(n) is the counting function for the state with n+f-1 rolls and g-1 failures, f is the total number of failures ever allowed, and g is just a glorified for loop counting from 1 to f...

You can expand out the first 3 terms, which I will do here cause I did them already, and then the general probability of success becomes:

S^(n+f-1) + n*S^(n+f-2)*(1-S) + [n*(n+1)/2 + n - 1]*S^(n+f-3)*(1-S)^2 + ...

Where "..." are the additional terms of the form Ng(n)*S^(n+f-g)*(1-S)^g for all g from 3 on up to f-1. For 3 failures, this reduces to the formula I posted before.

I'm not sure if there is a general case of Ng(n) or not, and don't have the time to get into it / research it. Luckily, Ng(n) is the gth box of the (n+f)th row of pascal's triangle, so someone may have done it already and I'm just not finding / remembering it. More importantly, it allows you to cheat like so:

1) Get a copy of Pascal's Triangle. If you have lots of possible rolls (n+f > 16), you'll need a bigger one than the one I've linked..
2) Go to row n+f on your triangle.
3) The first box in the row is the number of states with 0 failures. The second box is the number of states with 1 failures. Etc.
4) Determine the probability of each allowed group and multiply by it's box. So multiply the no failures probability by the 1st box, 1 failure prob by the second box, 2 failures by the 3rd, 3 by the 4th, etc. Stop after you do whichever box represents your max allowed failures (so box max fails +1).
5) Add all that shit together if you didn't do it while you were going along.
6) Repeat as desired and masochistic tendencies allow.

And if anyone knows how to calculate the yth box of the xth row of pascal's triangle, that would be helpful for the general case. I've blown work off enough for now though.

Or what Manxome said. It's much more elegant when you can remember the functions you need....

The new Skill Challenge System is exceedingly easy to evaluate for any set of inputs. Mathematically it reduces down to a system where you roll the dice X+2 times, where X is the number of successes you need to achieve, and there are precisely 3 was in which you succeed (where all are successes, where all but one are successes, and where all but three are successes).

So your chance of success overall is:

S^N + S^(N-1)*F*N + S^(N-2)*F^2*N*(N-1)/2

Where:

• N = Required Successes + 2
  S = Chance of Success per Roll
  F = Chance of Failure per roll.

So for example: let's say that you make your skill checks 75% of the time. Apparently because you have a trained skill and a stat modifier of +0. Sure, whatever. This means that for a simple challenge you succeed 83% of the time. And for a complexity 5 challenge, you succeed 28% of the time. Which is probably the kind of thing they were looking for.

But... the basic fact of the matter is that characters will actually have stat modifiers of some kind, and that a base DC of 10 pretty much just means that people who are highly likely to succeed will essentially always succeed. At Complexity 5, you need to succeed each roll on a 4+ or you aren't likely to succeed in the overall challenge.

---

So they addressed the basic flaw in the original writeup where the PCs never ever succeeded. The new one still has a variety of flaws that are quite recognizable from the original bullshit:

• In order to succeed at all, players have to be playing way over in the "nearly always win" bracket for each die roll, meaning that characters who have a snow ball's chance of doing anything they want to do are also living in highly swingy territory where small bonuses and penalties have huge overall effects. A character who has a better than coin flip's chance of victory only needs to get a +3 bonus to go into automatic success territory. Essentially, they aren't taking advantage of the d20 aspect at all. For all the difference it makes, they may as well be using d6s.
• Cooperation is still actively discouraged. Whoever is the best at any task should take all the actions. You are limited by how many actions the party is allowed, not how many actions any character is allowed. So having characters with lesser bonuses contribute in any way spikes the chances for the team.

So yeah, still crap.

-Username17

Well, I'm glad that people with more math experience than me have decided that I don't utterly fail high school statistics.

They just made default skill encounters auto-win instead of auto-lose, as far as I can see. Someone in the party will be able to reliably get +9 to one of the three primary skills (especially since the errata suggests allowing one or two aid another checks), and then you whack that till you win.

It's essentially down to the DM to custom-design decent skill challenges, and the DMG rules are more useful as inspiration than assistance. That's annoying, given the earlier hype about the system.

MartinHarper wrote:They just made default skill encounters auto-win instead of auto-lose, as far as I can see. Someone in the party will be able to reliably get +9 to one of the three primary skills (especially since the errata suggests allowing one or two aid another checks), and then you whack that till you win.

It's essentially down to the DM to custom-design decent skill challenges, and the DMG rules are more useful as inspiration than assistance. That's annoying, given the earlier hype about the system.

In short: Encourage your players to be creative; you don't need a system for it.

Basically, what people have been doing since before 4e.