2014-03-18: Added calculations for the expected value and standard deviation and corrected results. Big thanks to ASSASSIN for calling attention to the previous mistake.

Alright, I guess the question is pretty clear from the description.

Every boss in the game drops one random warframe component blueprint for one specific warframe when he is defeated. There are three different warframe components: Helmet, Chassis and Systems. You need one of each to craft the warframe. So every time you kill a boss, you have equal chances to obtain either of the three component blueprints. But how often on average do you have to kill one to obtain all three components at least once?

That's a rather simple question, and when I asked it to myself, I expected to answer it quickly with some basic probability theory. But as I tried to solve the problem, it turned out to be much trickier than I imagined. I took the challenge to solve it anyways and it took me about six hours, four different approaches and a lot of transformations, but I did it. So I thought, why not share my results?

Alright let's get to it. ;)

I drew all the schemes and wrote all the formulae on paper, and I don't think many people are interested in the specifics, so I won't copy it as of now, maybe I will add it later. You can skip the following abstract if you're not interested and scroll down to the final formulae, or even further to the diagrams in the next chapter, I won't be mad at you. ;)

In short, the approach that allows to create a universal formula is to only distinguish between "new part" and "duplicate". The first run always gives a new part, and the last run does (since you won't continue once you have all three). Between them, there is one more time where you get a new part, all the others get you duplicates. For the first few steps (the probabilities to get the 3rd part in the 3rd, 4th and 5th run), you can do the individual calculations to get your first probability values to recognize a pattern. This is comprised of two parts:
1) Which possibilities exist?
For getting them from the third run:
For getting them in the fourth run:
For getting them in the fith run:
For getting them in the sixth run:
That's enough. As you can see, there are always n-2 possible ways to get the last missing part in the "n"th run. However, these ways aren't all equally probable. In the next step, we have to calculate the probability values of these possibilities.
2) How probable are these possibilities?
For n=3, this is rather easy: You have a 100% chance to get a new part on the first run, then a 2/3 chance to get a new again on the second, then a 1/3 chance to get the last one in the third. For all the following, duplicate drops are added. The chance for a duplicate before the 2nd new one is 1/3, and for after the 2nd new one it's 2/3. The three new ones always keep their probabilities, no matter at which position they occur. For the product, the order doesn't matter either, it's just to distinguish how many of the duplicate drops are one-thirds and how many are two-thirds. For n=4, the one duplicate there can be in either of the two states, thus wielding us the two possibilities, and its probability is:
For n=5, there are three possibilities for us, since we don't distinguish between duplicates and between different orders, there can only be either two two-thirds or one two-third and one one-third or two one-thirds. For our equation, this means:
We can keep extending this sequence, the pattern should be clear by now:
and so on.
Note that for each summand within the bracket, the divisor is always 3^(n-3), and the dividends are 1, 2, 4, 8, ... depending on how many summands there are. Since the divisor is always the same, we can add the sum together. The divisor keeps the same, the dividends add: 1 + 2 + 4 + 8 + ..., and there are always n-3 summands. If we include the 2/9 outside the bracket now, the dividend changes to 2+4+8+16+..., and the divisor 3^(n-3) is multiplied by 9, or 3^2, which can be put together to 3^(n-3+2) = 3^(n-1). Furthermore, the doubling sum is more elegantly described as sum of 2^i with i as the running index starting at 1 and ending at n-2 (our initial start and end values were 0 and n-3, the number of possibilities, but were pushed up by one each by the multiplication with 2 from outside the bracket). So finally, writing that down, we have our shiny fresh pmf, the probability mass function:
The right side is equivalent, it has been transformed as a geometric series, getting rid of the annoying sum.

Now, that's fine and all, but it only gives us the probability to acquire the last missing part in exactly the n-th try. What we actually want to know is the probability to have acquired the last missing part by the n-th try. That is the sum of the probability masses of all the "n"s up to the current one, not just the current one itself, Pr(n=<nk) rather than Pr(n=nk). You can get the values by simply summing up the previous values and the new value for each step, or you transform the summation again to get this pretty neat formula on the right side:
Which is our cumulative distribution function, which is more revealing to our initial problem.


Alright, so we have our probability distribution well defined now. But what is the actual, definite answer to our initial question? How often will I have to try now, on average?
What we are asking for is the expected value of this probability distribution. The expected value of a distribution is the weighted average of its values weighted by each value's probability. Each value's probability is its probability mass function value. Unfortunately, we have an infinite number of possibilities, because you're never 100% guaranteed to get the last missing part in the next try. Thus:
We can't keep adding up for an infinite amount of time, but fortunately, we don't have to so: As we start adding, we soon see that the sum, though infinite, converges towards one specific value, and this value is 5.5, the expected value.
Once again, instead of mindlessly adding up the partial sums to vaguely see where it's going, one could also employ highly sophisticated maths to analyze the convergence behavior of this distribution. I tried to dig into it, but ultimately gave up, it's way too difficult for me to comprehend it. As I refuse to use formulae I don't understand for the sake of this guide, we have to be content with the method used above, which I'm okay with since we got such a nice value.

One last additional question you may ask is: By how much, on average, will my actually required number of tries deviate from the expected number of tries?
The answer to this question is the standard deviation of this probability distribution. The standard deviation is the square root of the variance. The variance is the weighted average of the squared differences between each value and the expected value, weighted by each value's probability.
As you see, the way it's calculated is actually similar to the expected value, including the fact that it gives us an infinite sum to mess with. Once again, we may just observe the development of the partial sums and see whether it converges, which it seems to do (though less quickly so): towards 2.598076, or rounded with still fully sufficient accuracy to 2.6.

We're done, let's look at the results! :)

Here are the actual values of the pmf and cdf as shiny, easy-to-understand diagrams:


and


The expected value is 5.5 tries - that's the answer! :)

The standard deviation is 2.6 tries.

Don't forget though, it's a random variable, so don't yell at me if you are unlucky! :D

Anyways, good luck and happy hunting! :)

Thanks for reading my guide! Feedback would be greatly appreciated, as well as suggestions for other math problems related to the game you would be interested in seeing me take a shot at! :)