Rajib_119's blog

By Rajib_119, history, 4 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.

 
 
 
 
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4 years ago, # |
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. So, answer is .

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    4 years ago, # ^ |
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    Thanks a lot, sir.

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    19 months ago, # ^ |
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    Can you elaborate the function a little bit more,please?

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      19 months ago, # ^ |
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      $$$E(x)$$$ 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.

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        19 months ago, # ^ |
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        Thank you. it helps a lot to understand the function part by part.