Skill Challenges: Is "harder" really harder?

[Roger]

The thing is, they're calling 8/4 "harder" than 4/2.

Where? They don't do this in the Skill Challenges Excerpt, and I can't find anywhere else they may have discussed it.


Cheers,
Roger

 

[phil500]

also, doesnt a 2/1 take at least 2 rounds to disable, whereas an 8/4 takes at least 8 rounds for a rogue.

 

[Lacyon]

This is not true.

Sigh. It is true. Read all my other posts in here for the math.

Requring more successes even while keeping the ratio of successes to failures the same, does make it more difficulty in that a lucky roles are less likely to succeed (since you need more of them). If you have a simple check, and you need a 10 to succeed, the odds are 55%, while for the 8/4 check you only have a 19% chance. That makes it harder for lesser skilled characters to succeed at more complex tasks. Which strikes me as harder.

Yes. If the DC is such that 55% of rolls succeed, then more complex checks are harder. Contrariwise, if the DC is such that 80% of rolls succeed, more complex checks are easier, because a more unlucky rolls are needed to fail.

It is therefore impossible to tell without knowing a party's skill bonuses whether making a given challenge more complex will make the challenge easier or harder for them.

 

[Kraydak]

...

It is therefore impossible to tell without knowing a party's skill bonuses whether making a given challenge more complex will make the challenge easier or harder for them.

With one sorta-exception. A single skill check is effectively a 1/1 and is always easier than a 2/1.

The additions of variable DCs (easy checks unlocked by moderate ones) and auto-fails makes low complexity skill challenges even wonkier. A 2/1 with an unlockable easy check plays very differently if the players start with the unlocking skill than if they don't. A 2/1 with an auto-fail... doesn't let you recover from mistakes.

If the formula includes both unlockables with easy DCs and auto-fails as default, then at low complexities, running into an auto-fail is catastrophic: if you identify a skill that *can* succeed, chain it, even if the odds aren't great. At high complexities, if your success rate is marginal (near/just below 2/3), you need to hunt for an unlockable or two. Just pray you don't run into the auto-fail.

 

[Lacyon]

With one sorta-exception. A single skill check is effectively a 1/1 and is always easier than a 2/1.

1/1 does not fit anywhere on the complexity scale (to our knowledge). Which means this really isn't an exception.

The additions of variable DCs (easy checks unlocked by moderate ones) and auto-fails makes low complexity skill challenges even wonkier. A 2/1 with an unlockable easy check plays very differently if the players start with the unlocking skill than if they don't. A 2/1 with an auto-fail... doesn't let you recover from mistakes.

It doesn't have to be an unlockable easy check - one PC who is trained versus one who isn't mucks up the math just as much or more.

The point I'm making is that messing with the complexity has unpredictable results. Don't increase the complexity to make the challenge harder (nor decrease it to make the challenge easier). Instead, you should probably set the complexity at a metagame level, based on how much time you want to spend on it. The skill DC is the thing to muck with when you want to make it easier or harder.

If the formula includes both unlockables with easy DCs and auto-fails as default, then at low complexities, running into an auto-fail is catastrophic: if you identify a skill that *can* succeed, chain it, even if the odds aren't great. At high complexities, if your success rate is marginal (near/just below 2/3), you need to hunt for an unlockable or two. Just pray you don't run into the auto-fail.

This is presumably one of the major reasons why the guidelines caution so strongly against campaign-ending skill challenge results.

PS: For those of you using NegBinomDist to calculate probabilities, it's not the right function to use. NegBinomDist calculates the probability of getting exactly y failures before x successes, you want to find the probability of getting at least y failures before x successes.