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.
Thanks in advance.
Thanks a lot, sir.
Can you elaborate the function a little bit more,please?
$$$E(x)$$$ 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.
Thank you. it helps a lot to understand the function part by part.