Rajib_119's blog

Light OJ 1248
By Rajib_119, history, 8 years ago, In English

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.

  • Vote: I like it
  • +1
  • Vote: I do not like it

»
8 years ago, # |
  Vote: I like it +1 Vote: I do not like it

. So, answer is .

→ Reply
  • »
    »
    8 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Thanks a lot, sir.

    → Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Can you elaborate the function a little bit more,please?

    → Reply
    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      $$$E(x)$$$ 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.

      → Reply
      • »
        »
        »
        »
        5 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Thank you. it helps a lot to understand the function part by part.

        → Reply