A few games resolve this situation by dealing with it explicitly in the rules, and building the check system to accomodate how it handles this situation. The most notable one is Burning Wheel and its Let It Ride rule:
The result of one test stands for the duration of the situation.
When in a situation like this, the success or failure of the attempt comes down to one roll. Players may cooperate or not, but once the dice are cast for the first time, the door will/will not be open and there can be no new roll by anyone for that goal until the situation significantly changes (like, they go fetch a battering ram, or they return a month later).
Lots of Burning Wheel's rules are tightly enmeshed in the rest of the rules, but this is one of the few that is easily separated and portable to other games. If your game of choice already has rules for assisting, you don't even need to houserule anything once you let your players know that you'll be following Let It Ride from now on. In the situation you describe, the fiction doesn't even need to change. "We all take turns bashing at the door. Eventually one of us must get through!" After establishing their method and their goal, they decide who's making the roll, everyone else adds bonuses for helping, and then the one roll is made.
This rule was built into Burning Wheel specifically because the author got the same feeling as you, that there's a problem with this common occurrence in games with skill systems. Most games test for the action, which under some circumstances virtually guarantees success (or failure) by just repeating the action. Instead of using task-based resolution like that for skills, Burning Wheel uses an intent-based resolution, where your goal is why you roll (and only once), but what you roll is determined by the method you use. Let It Ride is a key part of its intent-based resolution, in that it reminds everyone that they only get to do this once, so they need to bring all their resources to bear – or not, if someone in the party is opposed to the attempt. Either way, everyone has to commit to either pursuing this goal or not, before the roll happens. There's no hanging back to try after, because there's no second chance.
As an added bonus, it fixes a related problem, because the rule binds the GM too. The GM is forbidden from calling for multiple tests for the same task, so there's none of this sort of thing designed to railroad a failure:
GM: Roll to climb! Hm, success… okay, you get 10 feet up without falling. Roll again! Another success… What's your skill level? Oh, that's pretty good… So you reach a ledge 30' up. You've got 100' to go, so roll again! …Success. You get halfway up. It's a long way down now. Roll climb. Aw, you failed! You plummet to your death. But wait! Giant eagles save you and fly you halfway around the world.
This is how skills are supposed to work!
If you are in a situation where there is only one person doing something, and they are rolling a single skill check, then yes, this is how it's supposed to work. Giving help is a natural thing and should be used in situations like this. There is no reason to prevent it unless the task is clearly something that's not going to benefit from someone else giving you assistance. There are some things you can do to limit it.
It's also worth noting that helping can often save you some table time. As a AceCalhoun points out in the comments, in many cases what happens if you don't help is that everyone in the party tries their hand at the task. This behaves very much like advantage, but with a slightly lower overall modifier (because most likely you'll have one character who is good at a task and the rest that are lower). So Working together only raises the change of success slightly and consumes less table time in these cases.
Be a bit more stringent about what you allow for assistance. Is coaching stealth really all that helpful? do you really want someone looking/talking over your shoulder while you're picking that lock? Evaluate situations where characters attempt to aid more carefully.
Have more than one thing going on at once. If all the characters need to be stealthy, they can't be helping each other. And if you need two arcane characters working on the sigils on opposite sides of the room, maybe they have to choose which one gets help from the third (or don't have anyone to help at all).
Make things take multiple rolls and limit helping on all of them. Maybe the first roll the wizard can be helped, but after that he's on the other side of the trap, or arm deep in the sigil or something to where additional assistance isn't going to help him.
Figure out how to inflict disadvantage for the task. Maybe there are mitigating circumstances.
Create a distraction. A rogue can't help the wizard if he's busy fighting baddies. Make some skill checks happen in an occupied room. Make completing the skill checks the win condition rather than defeating the enemies.
The basic crux of all of this is that helping is supposed to a mundane task that provides advantage. Yes, that's a huge deal, but it also doesn't stack with other things that give you advantage and it can be easily cancelled by disadvantage.
So get creative! Build some situations into your adventures that prevent your heroes from helping each other (or make the opportunity cost higher). But don't do it all the time, that might get tiresome. Adventurers like to help each other out, let them, but don't make it easy all the time.
Best Answer
You want a table that models "stacked" probabilities.
For example, if the chance of any individual attempt's success is 40%, your table might look like this:
\begin{array}{cc} n\text{ to success} & \text{d}100\text{ roll}\\ \hline 1 & 01-40\\ 2 & 41-64\\ 3 & 65-78\\ 4 & 79-87\\ 5 & 88-92\\ 6 & 93-95\\ 7 & 96-97\\ 8 & 98\\ 9 & 99\\ 10 & 100\\ \end{array}
But how do you generate this table?
Given the probability \$p\$ of succeeding on any given attempt, then the probability of succeeding on the \$n^\text{th}\$ attempt is given by $$P(n)=(1-p)^{n-1}\cdot p$$
This is because in order to succeed on the \$n^\text{th}\$ attempt we must first fail, with a probability \$(1-p)\$, \$n-1\$ times, then succeed (with probability \$p\$).
Thus in the example above we see that \begin{align*} P(1)&=p=0.4\\ P(2)&=(1-p)\cdot p =0.24\\ P(3)&=(1-p)^2\cdot p \approx0.14\\ P(4)&=(1-p)^3\cdot p \approx 0.09\\ \vdots\quad& \hspace{2cm}\vdots \end{align*}
But in order to stack the probabilities we recognize that the "break points"--the highest number in each of the percentile ranges--are given by the sum of all probabilities up to the \$n^\text{th}\$. Luckily, this is just a geometric series: $$\sum_{i=0}^{n-1}{(1-p)^i} = \frac{1-(1-p)^n}{p}$$
With this in hand it's easy to generate a table of "break points" for any given \$p\$:
\begin{array}{c|ccccccc} & \text{d}100\text{ roll}\\ n\text{ to success} & p=0.1 & p=0.2 & p=0.3 & p=0.4 & p=0.5 &p=0.6 &p=0.7\\ \hline 1 & 01-10 & 01-20 & 01-30 & 01-40 & 01-50 & 01-60 & 01-70\\ 2 & 11-19 & 21-36 & 31-51 & 41-64 & 51-75 & 61-84 & 71-91\\ 3 & 20-27 & 37-49 & 52-66 & 65-78 & 76-88 & 85-94 & 92-97\\ 4 & 28-34 & 50-59 & 67-76 & 79-87 & 89-94 & 95-97 & 98-99\\ \hline 5 & 35-41 & 60-67 & 77-83 & 88-92 & 95-97 & 98-99 & 100\\ 6 & 42-47 & 68-74 & 84-88 & 93-95 & 98 & 100\\ 7 & 48-52 & 75-79 & 89-92 & 97-97 & 99\\ 8 & 53-57 & 80-83 & 93-94 & 98 & 100\\\hline 9 & 58-61 & 84-87 & 95-96 & 99 & \\ 10 & 62-65 & 88-89 & 97 & 99 & \\ 11 & 66-69 & 90-91 & 98 & 100 & \\ 12 & 70-72 & 92-93 & 99 & & \\\hline 13 & 73-75 & 94-95 & 99 & & \\ 14 & 76-77 & 96 & 99 & & \\ 15 & 78-79 & 96 & 100 & & \\ 16 & 80-81 & 97 & & & \\ \vdots & \vdots & \vdots\\ \end{array}
Notes:
1. In a few places the same single number appears twice. (p=0.2, n=14,15, for instance.) In these cases multiple break-points round to the same percent-value. You could randomly choose from among n values if this is rolled, simply take the lower, or devise some other scheme.
2. Obviously, there is a non-zero probability that success would take longer than I've indicated, ending my tables at the first appearance of a rounded-to-100 percent-value. However, by construction there is less than a 1/2% chance that any n greater than the last presented could occur. I felt fine leaving the long tail off the table.