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On Average (Posted on 2004-01-26) Difficulty: 4 of 5
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).

See The Solution Submitted by SilverKnight    
Rating: 3.7500 (4 votes)

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re(2): solution | Comment 9 of 11 |
(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

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