I simulated 1 countless completions of the majority of OSRS bosses and graphed the completion kc's. This way you can see the spread of completions. Since conclusion can be somewhat vague I simulated a few distinct options. By default getting 1 of all drops unique to the boss is assumed excluding any rs gold 2007. Calculating average completion is hard to do when two drops have distinct weighting. Additionally, it doesn't show the variance, which I guess you should have the ability to calculate somehow but I'm not certain how.
In this instance it simplifies because the time to collect a new drop having received m-1 of then follows a geometric distribution. But even then its expected value and variance don't have closed types. So the general case (namely this one) is sort of impossible.
On the other hand, what you might have done is take all ordered lists of numbers.,n of a specific length (this is kc) with repeats where every number occurs at least once with the last number occurring exactly once, assigned weights, and summed. Assuming you've got an easy enough tactics to generate the lists, this is going to be an easy enough computation. (An easy means to do this is to take a list of span l-1 with repeats at the numbers without I where every other number appears at least once and then append I to the conclusion ).
Additionally, it is worth noting the complex parts of the coupon collector's problem only arise since the traditional difficulty attempts to calculate the mean, variance, and the pmf directly out of a pmf using analytical procedures, which gets real ugly because there are many nested, iterative calculations. Computational calculation going pmf of only thing - cdf of single item - cdf of all - pmf of all is a lot of dull impossible to perform by hand math, but is computationally very simple. See the section on calculation in that Wikipedia article for the concept. A computer can calculate this without approximations. See my comment above to find the solution that is exact. This solution will not get you the mean or variance, but it will get one of the cdf chart to match the picture above.
That stated this computational method has error except in cases under 1/100 of kc. I mean the points on your curves. Everything you did (I assume) is that you sampled a known distribution. You might have only plotted the same distribution! Ah right, it might have made sense to buy OSRS gold that way. Though I think combining the distrubutions would have given slightly skewed results on more prevalent drops as you can't get 2 different drops in precisely the exact same moment. I wouldn't know how to account for this.