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).
This can be seen as a state machine, with probabilistic state changes. You start at the state "No numbers seen".
With probability 6/6, you change to the state "One number seen".
At this state, with probability 1/6, you remain there one more throw; with probability 5/6 you change to state "Two numbers seen".
At this state, with probability 2/6, you remain there one more throw; with probability 4/6 you change to "Three numbers seen", and so on.
The expected number of throws is 6/6+ 6/5+ 6/4+ 6/3+ 6/2+ 6/1 = 14.7.
Solution
