Expected number of different outcomes

Question

Let's say I roll a fair die $4$ times. What is the expected number of different outcomes to be seen?

What I mean by "number of different outcomes to be seen" is as follows: If the result of the outcomes is $1233$ then it is $3$.

My try:

I tried enumerating like in the previous question I asked. It seems tedious. Any other method to solve this?

Help appreciated!

Answer 1

One method:

• the probability a particular outcome is not seen is $\left(\frac{5}{6}\right)^4$
• the expected number of outcomes not seen is $6\,\left(\frac{5}{6}\right)^4$
• the expected number of outcomes seen is $6-6\,\left(\frac{5}{6}\right)^4 = \dfrac{6^4-5^4}{6^3} \approx 3.10648$

Answer 2

For $i=1,\dots,6$ let $X_i$ take value $1$ if outcome $i$ shows up and let it take value $0$ otherwise.

Then $X:=X_1+\cdots+X_6$ is the number of different outcomes.

With linearity of expectation and symmetry we find:$$\mathsf EX=6\mathsf EX_1=6\mathsf P(X_1=1)=6\left(1-\left(\frac56\right)^4\right)$$