Finding probability of all outcomes occurring in n repetitions Suppose you have numbers 1, 2, 3, 4. You can pick each of these numbers with probability 1/4. After n iid selections, what is the probability that you have seen all 4 numbers? Express this as a function of n.
For example, if n=5, and you draw the numbers 1,1,3,2,1 then this is unsuccessful, as 4 was never picked. If you draw 3,4,2,3,4, this is also unsuccessful as 1 was never picked. However, 3,2,2,1,4 is successful.
I am aware you can use a discrete Markov chain and find the n-step transition probability matrix on Wolfram, but is there a way to find this without a calculator and with reasoning alone?

You may just count systematically as follows:
Number of sequences of length $n$ containing only $1$ digit: $$\color{blue}{\Rightarrow 4}$$
Number of sequences of length $n$ containing exactly $2$ digits:

*

*Choose two of the digits: $\color{blue}{\binom 42}$

*Number of sequences of length $n$ containing at least one of these digits: $\color{blue}{2^n}$

*Number of sequences of length $n$ containing only one of the $2$ selected digits: $\color{blue}{2}$
$$\color{blue}{\Rightarrow \binom 42\left(2^n-2\right)}$$
Number of sequences of length $n$ containing exactly $3$ digits:

*

*Choose three of the digits: $\color{blue}{\binom 43}$

*Number of sequences of length $n$ containing at least one of these digits: $\color{blue}{3^n}$

*Number of sequences of length $n$ containing at least one of only two of the $3$ selected digits: $\color{blue}{\binom{3}{2}2^n}$

*Number of sequences of length $n$ containing only one of the $3$ selected digits: $\color{blue}{3}$
$$\color{blue}{\Rightarrow \binom 43\left(3^n- \binom{3}{2}2^n+3\right)}$$
All together, number of sequences where at least one digit is missing:
$$\color{blue}{4 + \binom 42\left(2^n-2\right) + \binom 43\left(3^n- \binom{3}{2}2^n+3\right)=4-6\cdot 2^n+4\cdot 3^n}$$
Now, the probability you are looking for is
$$\frac{4^n - \left(4-6\cdot 2^n+4\cdot 3^n\right)}{4^n}=\boxed{1-\frac 1{4^{n-1}}+6\frac 1{2^n}-4\left(\frac 34\right)^n}$$
