Odds of something specific happening multiple times in sequence..? More explanation below, struggling to phrase properly I feel really dumb for not figuring this out myself, don't bash me too hard, it's been a while since I've had a stats class. 
I'm playing a mobile game, and simply put, you summon heroes with currency. The odds of summoning an "elite hero" are 4.61%, while the odds of let's say a "common hero" is the opposite, 95.39%. You can also do a 10-summon, which is just what it sounds, a bundle of 10 summons all at one time. Using 1-(.9539^10), you can see that in a 10-summon, you have a ~37.6% chance of pulling one elite. 
My question is... What would be the formula for finding multiple elite heroes in a 10-summon? What would the odds be of 2 elites in a 10-summon, or 3 or 4? 
My first instinct would be something like... (4.61 x 4.61 x (95.39^8)) / 100^10, but this yields a ~.14% chance, and it seems like too drastic of a jump to go from one elite in a 10-summon = ~37.6%, to two elites in a 10-summon is ~0.14%.
Thank you, sorry this probably simple statistics question is blowing my mind!! It's too late for this...
 A: This isn't a dumb question at all, this is a fairly common type of probability question. Assuming that each summon is independent and identically distributed (which basically means that each summon does not depend on other summons and that the odds of each summon are always the same) then this is a Binomial Distribution. I'll try to explain very simply what that is and how we intuitively get the formulas while trying to involve the least amount of math/stat and jargon.
If you were to ask for example, what is the probability of getting 2 elites in a 5-summon given that the probability of getting an elite is p. Then we can list out all possible outcomes of that happening (let's denote E as an elite and C as a common)


*

*E E C C C (this is basically saying, draw an elite, then another elite then 3 commons)

*E C E C C (this is drawing an elite, then a common, then an elite, then 2 commons)

*E C C E C (etc...)

*E C C C E

*C E E C C

*C E C E C

*C E C C E

*C C E E C

*C C E C E

*C C C E E
Now that we listed all possible outcomes of 2 elites in 5 drawings (that's quite a lot), we can now calculate the probability of each outcome happening and then sum those probabilities to get the answer to the question (probability of getting 2 elites in a 5-summon)


*

*$P('EECCC') = P('E')*P('E')*P('C')*P('C')*P('C')=p^2*(1-p)^3$

*$P('ECECC') = P('E')*P('C')*P('E')*P('C')*P('C')=p^2*(1-p)^3$

*$P('ECECC') = ...=p^2*(1-p)^3$
We can see that no matter which of those 10 patterns we calculate the probability for, we end up getting $p^2*(1-p)^3$ 
So the answer is to what is probability of getting 2 elites in a 5-summon = 
$10*p^2*(1-p)^3 = 10*(0.0461)^2*(1-0.0461)^3 = 0.018446348 \approx 1.8 \% $
But now a logical next question is: In this example we could easily list out the all possible outcomes of 2 elites in 5 summons, count them and get 10, but what if we wanted to calculate the probability in 152 elites in 528 summons? Or any other arbitrary number? That's when mathematics comes in and saves us with the Binomial Coefficient. Thus the formula for K elites in N summons is
$\frac{n!}{k!(n-k)!}p^k(1-p)^{(n-k)}$
Where that weird factorial fraction is from the binomial coefficient.
If you want to just plug numbers and get answers, you can use this simple calculator here. If you want to know for example the probability of getting 3 or more elites in a 10 summon, put the numbers 0.0461, 10, and 3 in the first 3 boxes, press calculate and you'll get P(X > x)=0.00920669192 which means in 10 summons there's a 0.92% chance that you'll get 3 or more elites
