Skill challenges made more sense in 4e where there weren't any 'non-encounter' abilities for characters.
Other than skills and rituals, you mean?
Obviously, my assumptions are meant to produce a lower baseline, which would then be augmented by high ability scores, expertise, and other character features, but 1 out of 40 times you'd expect a party to succeed facing this sort of skill challenge in 5e?
That's definitely the lower baseline but not the middle ground which is what we should be looking at to be fair.
Like you have mentioned high ability scores, expertise dice and other character features also would play a roll
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May I ask in 4e would you use a skill challenge of 7 success before 3 failures with 1-3 level characters? Just asking, and I don't have the books in front of me and it has been a while.
4e is based around much higher chances of success than 2 in 5 (which is what Quickleaf indicated upthread: +2 bonus vs DC 15). Most of the checks in a skill challenge will be Medium, and a character who is proficient in a skill, or has a good stat, should succeed at a Medium check around 2/3 of the time. If the character is proficient and has a good stat that success rate may be closer to 9 in 10. And that's before factoring in items, power bonuses etc.
Also, once the number of success required gets beyond 6 before 3 failures, 4e (post-Essentials) has the somewhat ad hoc system of "advantages" - basically, ways to mitigate adverse odds by column-shifting DCs or undoing failures.
Here is the "7 before 3" maths for a success chance of 3 in 4, which is not a bad generalisation for the bare bones of 4e:
The chances of 7 successes in a row is 3^7 over 4^7. This = 2187 in 16384.
The chance of 7 successes and 1 failure is 3^7 * 1 [for the failure] * 7 [because there are 7 "slots" into which the failure might fall, before the final success], all over 4^8. Which equals 2187*7 = 15309 in 65536.
The chance of 7 successes and 2 failures is 3^7 * 1^2 [for the 2 failures] * [the number of ways of allocating 2 "slots" out of 8 to failures, = 8!/6!2! = 8*7/2 = 28]. So 2187 * 1 * 28 = 61236 , all over 4^9 = 262144.
Adding all these together over a common denominator:
(2187*16 + 15309*4 + 61236)/262144
= (34922+ 61236 + 61236)/262144
= 157394/262144
which is just a tiny bit more than 60%.
In practice, in 4e, I think the chance tends to be higher because of advantages, the ability to gain bonuses that lift the chance above 3/4, etc.
To get a 3/4 success chance in 5e (ie succeed on a 6 or better on d20), assuming a typical bonus of around +2, is going to require DCs of 8. Even if DCs are 10, the chances of success for characters with a +2 bonus (ie succeed on 8 or better on d20) is going to drop markedly.
If we treat a chance of 13/20 as a 2/3 chance (because I'm not going to do maths that involves 13^7!), then we get the following:
The chances of 7 successes in a row is 2^7 over 3^7. This = 128 in 2187.
The chance of 7 successes and 1 failure is 2^7 * 1 [for the failure] * 7 [because there are 7 "slots" into which the failure might fall, before the final success], all over 3^8. Which equals 128*7 = 896 in 6561.
The chance of 7 successes and 2 failures is 2^7 * 1^2 [for the 2 failures] * [the number of ways of allocating 2 "slots" out of 8 to failures, = 8!/6!2! = 8*7/2 = 28]. So 128 * 1 * 28 = 3584, all over 3^9 = 19683.
Adding all these together over a common denominator:
(128*9 + 896*3 + 3584)/19683
= (1152+ 2688+ 3584)/19683
= 7424/19683
which is between 37% and 38%.
I don't think that's very viable, myself. It also shows that, with these sorts of numbers, the presence or absence of Guidance will make a very big difference to success rates.