What is the expected number of rolls of a fair, normal 6-sided die, one is required to make, so that each of the 6 numbers comes up at least once?

Hint: this is not necessarily an integer answer
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As an aside, it would be interesting to see the computer program simulation of this, but this would not be proof of the solution (merely evidence supporting the proof).

(In reply to re: The other answers by Penny)

(Please excuse the following if the math is inexcusable...)

The probability of NOT having seen a specific number after 14.7 times, if I've understood this correctly, should be

(5/6 ^ 14.7)

Thus the probability of not having seen ANY one number after 14.7 throws should be

(5/6 ^ 14.7) * 6 = 41.13%

So what makes this the expected result? It's not the point where the p of seeing all six number rises above 50%, for instance. Indeed, it hardly seems related at all to the probability of seeing all six.

I understand Federico's answer, I'm just wondering how, if at all, this could be worked out just by the probabilities of seeing (or not seeing) all six.

Posted by Sam on 2004-01-26 21:19:23