Assuming 5% (.05) of 5*, and thus 95% (.95) of any other result:
It's much easier to calculate the odds of getting zero 5* (.9511=57%) or at least one 5* (100%-57%=43%). Or eleven 5* (.0511=tinyness).
It's a little trickier to calculate the odds of exactly 1, 2, 3, etc. 5* draws. One pull and you will get either 5* (Five star) or not 5* (Other), so there are two possible results (F or O). Add a second pull, and you have 22=4 possibilities (FF, OO, OF, or FO) and so on until with 11 pulls you have 211=2048 possible chains of results. To get exactly four 5*, you need to figure out how many of those 2048 results have exactly four F and seven O. On top of that, the 2048 combinations are not all equally likely; for example we already know there is a 57% chance of getting the chain of 11 Os, but only a tinytinytiny chance of getting the chain of 11 Fs.
The way we calculate this is as a binomial distribution.
Specifically, the formula you asked for is:
y=[11!/(x!(11-x)!)](.05x)*(.9511-x)
Where X is the number of 5* pulls and y is the chance of that happening. Nicer picture of the formula here.
You can use this calculator to see the results. n=11, p=.05, and x is the number of 5* you want to odds of drawing. You'll see that there's about a 1% chance of getting four 5* and it drops from there (.01% chance for five of them). You can also use the drop-down menu to see the odds of getting greater/less than (or equal to) x amount of five star drops.
I guess while I'm here I may as well fill out the chart:
| Number of 5* pulls | Odds |
|---|---|
| 0 | 57% |
| 1 | 33% |
| 2 | 8.7% |
| 3 | 1.4% |
| 4 | .1%, 1-in-1000 |
| 5 | .01%, 1-in-10,000 |
| 6 | .001%, 1-in-100,000 |
| 7 | 1-in-5,000,000 |
| 8 | 1-in-167,000,000 |
| 9 | 1-in-10,000,000,000 |
| 10 | 1-in-1,000,000,000,000 |
| 11 | 1-in-200,000,000,000,000 |
