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A Question About Dice

You have a weighted n-sided die. Every side of the die is weighted differently where side n1 has a weight of w1, n2 has a weight of w2, ...

Estimate the average number of rolls needed to see every face at least once.

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marked as duplicate by joriki, Arkamis, Austin Mohr, userNaN, Chris Eagle Oct 30 '12 at 18:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Voting to close as duplicate: although that question is a special case of this one, the answers address the general case. –  Chris Eagle Oct 30 '12 at 18:49

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up vote 1 down vote accepted

This is a variant of the coupon collector's problem

Formula (14a) is what you need.

Updated after comments.

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It's correct that this is the coupon collector's problem, but that expected number can't be right - consider the case where all of the weights are 1 (or so nearly so as to be negligibly different, while still meeting the 'weighted differently' constraint). The formula would suggest that it takes only $n$ rolls to 'collect' all $n$ sides. –  Steven Stadnicki Oct 29 '12 at 18:23
    
Is it still the coupon's collectors problem if the sides are weighted differently? Meaning each side has a different probability of showing up. –  toobsco42 Oct 29 '12 at 18:24
    
@toobsco42 It's still a variant thereof, and I would start with 'weighted coupon collector's problem' as a search term - though a bit of research suggests that the problem may not have any closed-form solution. –  Steven Stadnicki Oct 29 '12 at 18:25
    
But it can't be correct, because the original problem has $w_k=\frac{1}{n}$ and your answer would give us $n^2$ in that case, when it should be $nH_n = O(n\log n)$ –  Thomas Andrews Oct 29 '12 at 18:28
    
Thanks. I was thinking it was a variant. I just thought it had a closed-form solution because I was asked to create a method in Java to help me solve this problem. –  toobsco42 Oct 29 '12 at 18:30

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