Intuition for two combinatorial situations Sitution 1: There is a box that, when opened, can contain nothing, a red ball, a blue ball, and a yellow ball. Each ball has a $1/100$ chance of being in the box when opened, otherwise there is nothing in box. What is the expected amount of boxes that need to be opened to obtain 1 of each ball?
Situation 2: There is a box that when opened, can contain nothing or a red ball. The red ball has a $1/100$ chance of being in the box.  What is the expected amount of boxes that need to be opened to obtain $3$ red balls?
My guess is that, for situation 1, the three balls' probabilities are independent of each other so you would expect to get 1 of each by box 100. But for situation 2, it would be 300 boxes. Is this intuition right?
 A: HINT
Your intuition for situation 2 is correct.
However, in situation 1 the expected number $> 100$.  Let say $N_r =$ no. of boxes to get a red ball, and similarly for $N_b, N_y$ for blue, yellow balls.  It is true that each of them is geometrically distributed and $E[N_r] = E[N_b] = E[N_y] =100$, but you need to get one of each color, which requires $N = \max(N_r, N_b, N_y)$ boxes, and it is intuitively clear that $E[N] > 100$, because you're looking at a sort of worst case (the $\max$).
Both situations can be solved using a similar technique: let $N_1 = $ no. of boxes to get the first useful ball, and $N_2 = $ no. of additional boxes to get the second useful ball, and $N_3 = $ no. of additional boxes to get the third useful ball.  Then the total no. of boxes is $N = N_1 + N_2 + N_3$ and you want $E[N] = E[N_1] + E[N_2] + E[N_3]$.
Situation 1: How is $N_1$ distributed?  What's its mean?  Then how is $N_2$ distributed and what's its mean?  Finally how is $N_3$ distributed and what's its mean?  
Situation 2: Exact same questions as above (but different answers of course).
Lemme know if you need more help / want me to verify your work.
