Expected Value

thepillow said:

I was hoping someone could tell me if my work on the question below is correct.

---

Q: If we roll a normal die \(\displaystyle n\) times, what is the expected number of different sides of the die that we observe?

---

And here's my attempt at a solution:

Let \(\displaystyle S_i = 1\) if side \(\displaystyle i\) is observed at least 1 time, and \(\displaystyle S_i = 0 \) if side \(\displaystyle i\) is not observed, where \(\displaystyle i = 1, 2, \dots, 6 \). Then for all six possible values of \(\displaystyle i\) we have

\(\displaystyle Pr(S_i = 1) = 1 - Pr(S_i = 0) = 1 - \left(\frac{5}{6}\right)^n \),​

since the probability that a side is not observed on each roll is 5/6 and we roll \(\displaystyle n\) times.

And the expected value of each of the six \(\displaystyle S_i\)'s is therefore

\(\displaystyle E(S_i) = 0\cdot Pr(S_i = 0) + 1 \cdot Pr(S_i = 1) = 1 - \left(\frac{5}{6}\right)^n\).​

Now, let \(\displaystyle N \) be the number of different sides observed. Then we can write \(\displaystyle N \) as

\(\displaystyle N = S_1 + S_2 + \dots + S_6 \).​

So the expected number of different sides observed is just the expected value of \(\displaystyle N\), which is

​

\(\displaystyle E(N) = E(S_1 + \dots + S_6) = E(S_1) + \dots E(S_6) = 6\cdot E(S_1) = 6\left(1 -\left(\frac{5}{6}\right)^n\right) \).

​

Thanks for your comments!​

Click to expand...