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If I'm wrong, tell me why I'm wrong and you're right.. (Reply).
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If there's only one Octonaut you need, that's reasonably easy: ordering n Octonauts and finding that not one of them is the kind you need is equivalent to saying that all of them are chosen from the remaining 7 possibilities, so the chance of that happening is (7/8)^{n}. Hence, getting at least one of the kind you're looking for will happen with probability (1  (7/8)^{n}).
And if you wanted at least one of the five kinds of Octonaut, the same reasoning would apply: you'll get at least one of them with probability (1  (3/8)^{n}), because the only way you can't is if all n of them are one of the 3 kinds you've got already.
(You've gone through that reasoning yourself in your original post, I know. I'm just going through it again myself as a warmup exercise to get my head in gear.)
But if you want all of the five kinds to show up at once, that's more fiddly. I suspect an InclusionExclusion Principle exercise will be required.
Start with probability 1 (all possibilities).
Subtract five lots of (7/8)^{n}, indicating the five subsets of possibilities in which one or other of the Octonauts you want fail to show up.
Now we've doublesubtracted every situation in which two desired Octonauts fail to show up. There are (5 choose 2) = 10 of those and each one has probability (6/8)^{n}, so add 10 × (6/8)^{n} back in.
But now we've got an extra copy of each case in which three fail to show up, so now subtract again, and so on.
So we end up with the probability of getting at least one of each of the five required Octonauts being
P = 1  5 × (7/8)^{n} + 10 × (6/8)^{n}  10 × (5/8)^{n} + 5 × (4/8)^{n}  1 × (3/8)^{n}
And that formula gives me 0.1616 for n=10, and 0.9107 for n=30, so it agrees with your Excel solution.
I suspect there's no nice formula to invert this (i.e. to give you n when you put in P), but it's an analytic solution to half of it at least.