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: solution by Sam)

Two corrections on the difficulties presented:

Regarding:
"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% "

Aside from the fact that you can't actually have 14.7 throws (a minor point here), you can't add up (which is what multipliction by 6 is here) all six probabilities to come up with the total probability that any one of them would happen. Just think: after one throw, the probability that you haven't gotten a particular side to show is 5/6. The probability is not (5/6 ^ 1) * 6 of not having seen ANY one number after 1 throw. The events are not mutually exclusive, so you can't just add the probabilities. I can't give you an actual value for 14.7 throws, as such a number of throws does not exist. However, from the table I posted as The other answers, the probability of having completed all 6 at 14 throws is 0.582845348851636399901396, and at 15 throws is 0.644212738595100396845553. The probability that one or more did not yet come up is the complement of each of these, or 41.7% and 35.6% respectively. Interpolating to 14.7 would give 39.23%.

Which brings us to the second item:
"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. "

The expected value is the average, or mean, value. What you are describing is the median value, which in this instance in the above referenced comment I note is just below 13. Suppose there were four boxes with money inside: $1, $2, $3 and $100. You pick one at random. Your median winnings would be 2.50, but the mean (the expected value) is 26.50, even though you only have 1 chance in 4 of picking a larger amount.

Posted by Charlie on 2004-01-26 23:04:56