### Rajib_119's blog

By Rajib_119, history, 4 years ago,

Help in this problem

My Think: Say E(x) is the expected value when we see x sides of dice.

E(x) = (x / n) (1 + E(x)) + ((n — x) / n) (1 + E(x + 1))

where at this time, the probability of getting the old side of dice which is already done is (x / n) and renaming are (n — x) which need to see at least once, probability of that (n — x) / n.

•  » » » 19 months ago, # ^ |   0 $E(x)$ is the expected number of moves if you have seen x faces already. so with $\frac{x}{n}$ probability you roll a side which you have already seen. And with $\frac{n - x}{n}$ probability you roll an unseen face.