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Light OJ 1248

Revision en1, by Rajib_119, 2017-05-02 22:39:10

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.

Tags probabililty


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en1 English Rajib_119 2017-05-02 22:39:10 461 Initial revision (published)