Four balls with different colors in a box, how many times do I need to pick to see all four colors?

This is known as the coupon collector's problem.

The average number of tries needed to see each of 4 colors is $$ \frac44+\frac43+\frac42+\frac41 = 8\frac{1}{3} $$

The terms are $4/4$ for the time to take one ball; then $4/3$ for the average time it takes after the first ball until you see another code; $4/2$ for the average time it takes after the first time you see the second color until you see a third, and finally $4/1$ for the time you then have to wait until you see the last color. (These can just be added due to the additivity of expectations).

Denote by $E(n)$ the expected number of additional draws when $n$ of the four balls have not been drawn so far. Then $E(0)=0$, and $$E(n)=1+{n\over4} E(n-1)+{4-n\over 4}E(n)\qquad(n\geq1)\ .$$ This immediately implies $$E(n)=E(n-1)+{4\over n}\qquad(n\geq1)\ ,$$ from which we get $E(4)={25\over3}$ by going through the motions.

Obviously, we stop picking when we take a ball having a color that never occurred before, so we have to count how many strings of length $n$ have only occurrences of $A,B,C$ (no $D$), with at least one occurence of $A$, at least one occurrence of $B$, at least one occurrence of $C$. By the inclusion-exclusion principle, they are: $$ 3^n - 3\cdot 2^n + 3$$ so the wanted expected value is: $$ \sum_{n=4}^{+\infty}\frac{4n}{4^n}\left(3^{n-1}-3\cdot 2^{n-1}+3\right)=\color{red}{\frac{25}{3}}.$$

What is the probability you see all colors after exactly $n$ turns, how many sequences of length $n-1$ contain exactly three colors? first select the three colors in $4$ ways.

When using three colors there are $3^{n-1}$ sequences.

$3\cdot2^{n-1}-3$ use $2$ or $1$ colors.

Therefore $3^{n-1}-3\cdot2^{n-1}+3=3(3^{n-2}-2^{n-1}+1)$ use exactly three colors when those three colors are fixed.

Therefore there are $12(3^{n-2}-2^{n-1}+1)$ sequences.

Then the probability you see three colors for the first time in the $n$'th draw is $\frac{1}{4}\frac{12(3^{n-2}-2^{n-1}+1)}{4^{n-1}}=3\frac{(3^{n-2}-2^{n-1}+1)}{4^{n-1}}$.

Hence you want $3(\frac{1}{4}\sum\limits_{n=2}^\infty (n+2)\frac{3}{4}^n-\sum\limits_{n=3}^\infty (n+1)\frac{1}{2}^n+\sum\limits_{n=3}^\infty\frac{n+1}{4^n})=3(\frac{63}{16}-\frac{5}{4}+\frac{13}{144})=\frac{25}{3}$