Suppose we have $3$ different types of coupons $C_1,C_2,C_3$, and the probability we draw each coupon is $p_1,p_2$ and $p_3$ for $C_1,C_2$ and $C_3$ accordingly. How many times in average will we have to draw until we get a coupon of each type?
I said the following: Observe the last coupon drawn. With probability $p_1$ it is $C_1$. Now observe the first coupon drawn, it must be different than $C_1$ since otherwise the trial wouldn't have ended. With probability $p_2$ it was $C_2$, and in the middle we would've needed to draw $C_3$ once, and for that to happen we would need $\frac{1}{p_3}$ draws. Doing the same logic for all possibilities of last coupon drawn, we get that the average number of draws is:
$p_1 \cdot (1 + p_2 \cdot (1 + \frac{1}{p_3})) + p_1 \cdot (1 + p_3 \cdot (1 + \frac{1}{p_2})) + p_2 \cdot (1 + p_3 \cdot (1 + \frac{1}{p_1})) + p_2 \cdot (1 + p_1 \cdot (1 + \frac{1}{p_3})) + p_3 \cdot (1 + p_1 \cdot (1 + \frac{1}{p_2})) + p_3 \cdot (1 + p_2 \cdot (1 + \frac{1}{p_1}))$
However, I feel that something is off with this. Any guidance towards whether this is correct or where a mistake has been made?
This solution steps through the process from beginning to end. As noted in the comments, you can also approach this recursively. The recursive approach is laid out here.
Definitions:
S[0] = starting point
S[1,2,3] = ending point
S[x,y] = state of having Cx and Cy, order you got there doesn't matter
Level 0 = S[0]
$L_1$ = S[1], S[2], and S[3]
$L_2$ = S[1,2], S[1,3], and S[2,3]
$L_3$ = S[1,2,3]
$E_x(L_y)$ = Expected number of steps to get from $L_x$ to $L_y$
Solution:
$E_0(L_1) = 1$
$E_1(L_2) = \frac{p1}{1-p1} + \frac{p2}{1-p2} + \frac{p3}{1-p3}$ (each term is probability times expected wait time)
If exiting S[1] the possible expected wait times at next level are:
$\frac1{p2}$ with probability $\frac{p3}{p2+p3} = \frac{p3}{1-p1}$ if you step to S[1,3], and
$\frac{1}{p3}$ with probability $\frac{p2}{p2+p3} = \frac{p2}{1-p1}$ if you step to S[1,2]
Similarly for exiting S[2] and S[3].
Weighting the three possible $\frac1{p_x}$ $L_2$ to $L_3$ waiting times by the ways of getting to the various S[x,y]
$$\begin{align} E_2(L_3) &= \frac1{p_1} \cdot ({p_2}\cdot \frac{p_3}{1-p_2} + {p_3}\cdot \frac{p_2}{1-p_3}) \\&+ \frac1{p_2} \cdot ({p_1}\cdot \frac{p_3}{1-p1} + {p_3}\cdot \frac{p_1}{1-p_3}) \\&+ \frac1{p_3} \cdot ({p_1}\cdot \frac{p_2}{1-p_1} + {p_2}\cdot \frac{p_1}{1-p_2}) \end{align} $$
Combining all $E_x(L_y)$ above gives
$$\begin{align}E_0(L_3) &= E_0(L_1) + E_1(L_2) + E_2(L_3) \\&=1 + \frac{p_1}{1-p_1} + \frac{p_2}{1-p_2} + \frac{p_3}{1-p_3} \\&+ \frac1{p_1} \cdot ({p_2}\cdot \frac{p_3}{1-p_2} + {p_3}\cdot \frac{p_2}{1-p_3}) \\&+ \frac1{p_2} \cdot ({p_1}\cdot \frac{p_3}{1-p_1} + {p_3}\cdot \frac{p_1}{1-p_3}) \\&+ \frac1{p_3} \cdot ({p_1}\cdot \frac{p_2}{1-p_1} + {p_2}\cdot \frac{p_1}{1-p_2}) \\&= 1+\frac{p_1}{1-p_1}\cdot(1 +\frac{p_2}{p_3} + \frac{p_3}{p_2}) + \frac{p_2}{1-p_2}\cdot(1 +\frac{p_1}{p_3} + \frac{p_3}{p_1}) + \frac{p_3}{1-p_3}\cdot(1 +\frac{p1}{p_2} +\frac{p_2}{p1}) \end{align}$$
For $p_1=p_2=p_3=0.33$ the result is 5.5