Since we don't repeat, this is equivalent to just having a random order. Thus, one is equally likely to have positions 1-N, so the expected random position is (N+1)/2
After how many laps could be interpreted as one less than this, or(N-1)/2
Leo S.
asked 12/04/20N ski slopes meet at the bottom of a valley, the lifts to the slopes are here in one place, there are waiting lines for the lifts.
After how many laps are we expected to ski on the most difficult piste (one particular slope),
a) if we follow the tactic of always standing in the shortest queue? (The queues have random lenght.)
b) And if we don't slide down on a track until we did one slide on all the others?
Since we don't repeat, this is equivalent to just having a random order. Thus, one is equally likely to have positions 1-N, so the expected random position is (N+1)/2
After how many laps could be interpreted as one less than this, or(N-1)/2
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