I heard someone wanted some MATH. Group check requires at least half the attempts succeed, hence (as already said at least once) odd numbers are bad, but regress to being equivalent. Three is the worst-case scenario. For now, I'll focus only on even-numbered groups, 2, 4, and 6.
2 is simple: it's functionally rolling with advantage, except that each person can also be rolling with advantage, so it's technically that much better. If you can turn a one-person check into a two-person check, you basically always should, regardless of the success rates of the individual participants. E.g. if Alex has 40% chance to succeed and Brooke has a 30% chance to succeed, then on a group check we need both of them to fail for the result to be a failure, which is .6x.7 = .42, meaning their net chance of success is 58%. Nearly doubling Brooke's chance of success alone.
With 4, we just need two successes. Let's say we have Alex (60% success), Brooke (50%), Casey (40%), and Denny (30%). The chance that no one succeeds is very low. .4x.5x.6x.7 = 0.084. The trickier calculation is the chance that exactly one succeeds, but it's not TOO tricky. As each of these events (just A, just B, etc.) is independent, we can just sum their individual probabilities, and each is just (chance N succeeds)(product of the others' failure chances). So we have (.6)(.5x.6x.7)+(.5)(.4x.6x.7)+(.4)(.4x.5x.7)+(.3)(.4x.5x.6)=0.302. But those two give us what we want: if less than 2 succeed, then either none or only one succeeded, and we can sum those chances, giving us a 38.6% chance to fail, and thus a 61.4% chance to succeed.
Surprisingly, even though Denny has only half the percentage chance to succeed, the overall result is about as good as if only Alex had attempted it--and will be decidedly less swingy. However, there certainly could be situations where sufficiently bad third- and fourth-best attempts could actually drag the value down. For example, if we use otherwise identical numbers, but Denny has only a 10% chance to succeed, then the overall chance of "at least half succeed" is actually worse than if Alex had attempted it alone. So that gives us a useful heuristic: Everyone needs to have at least not horrible chance to succeed on the check--which I would define as 30% or better. Anything below 30%, you're playing with fire.
Exactly three participants is definitely the worst-case scenario from a group perspective, strong law of small numbers and all that. Success only occurs if everyone succeeds, or at most one fails. Doing again the 60/50/40 split, success chance = (.6)(.5)(.4)+(.6)(.5)(.6)+(.6)(.5)(.4)+(.4)(.5)(.4) = .5. Which is worse than if just Alex had attempted it (60%), but remarkably, also meaningfully worse than having a fourth person with only 30% chance to succeed!
More precise analysis of 5 or 7 would be pretty tedious, so logical extrapolation is warranted. If the group has to be odd, then a larger group is always better than a smaller group, as that regresses to being equivalent to the even group. If you only have three people that could attempt it, then unless all of them are very good, you may as well just have one person do the attempt, especially if you can get them advantage or other bonuses. If you can ensure the group is even, then unless one of the participants is truly garbage at that check, it is usually worth doing a group check, but it might not be in rare edge cases.
HOWEVER, things are much different if we are doing this iteratively, e.g., Brooke only makes the attempt if Alex already failed, WLOG assuming Alex refers to the person with the best bonus, Brooke the person with the second-best, etc. There, the number of failures is already locked in, and if there aren't any successes at all yet, the attempt will generally give at least a chance to succeed, even if it's slim.
