On a side note:

I calculated that (if the system is truly random) it takes an average of ~21.17 episodes to attain all badges (with a deviation ranging 8 - 69).

So generally at most you have to watch 69 episodes (if you have really really bad luck). Note that it is possible to have an infinite streak of same badges, but this is incredibly unlikely.

At least you have to watch 8 episodes (lucky at every episode, getting a new badge you haven't encountered before).

On average you have to watch ~21.17 episodes.

Only 1 badge to go for me

For anyone wondering how the odds on this game work, I got something slightly different from the previous poster; my break-even point was actually 19.4 episodes. I'll give a few of my numbers, in format "number of turns (chance of winning)": 8 (0.24%), 10 (2.8%), 15 (24.8%), 20 (53.1%), 25 (73.7%), 30 (85.9%), 40 (96.2%), 60 (99.7%), 80 (99.98%).

And for anyone with truly morbid curiosity, here's how the calculation goes:

First, consider the odds of not getting any of one particular piece (say 4, that's the one I'm missing) during a series of N picks. Those odds will be (7/8)^N. There are eight different ways to do this, so to a first approximation, the odds of winning are 1 - 8*(7/8)^N.

However, this isn't quite right (as you can tell by noting that N=8 gives a negative result, when you really should have some chance). The problem is that in some of our sets that lack one piece, there is actually more than one piece missing. In order to correct for this "overcounting," we need to add back on (essentially) the odds of missing two pieces. There are 28 unique pairs of pieces (for those of you who aren't scared off by factorials, this is an N choose n problem, and we get 28 from 8!/(2!*6!)), so our new odds of winning are 1 - 8*(7/8)^N + 28*(6/8)^N.

We can keep playing this game, since in the last bit we neglected the possibility of missing three pieces, and so on. The final odds of winning turn out to be:

1 - 8*(7/8)^N + 28*(6/8)^N - 56*(5/8)^N + 70*(4/8)^N - 56*(3/8)^N + 28*(2/8)^N - 8*(1/8)^N