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
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.