When taking elevator one is possible to calculate the probability of being the first one to leave the lift.
In my situation, assume that:
- An elevator can just stop from floor to floor .
- Infinite capacity.
- We start from ground floor.
Given that I will leave at floor , where , then the average probability of taking the first leave at floor is
Reason. Denote , , the event that floor is the first one to stop with buttons on the elevator control panel being pressed.
We discard the case as is too big that pollutes our average value.
We have (for )
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Note that we are just considering choosing buttons because floor being the first to stop has conditioned our sample space to combintations of selected buttons that contain .
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Here the numerator represents the combinations of buttons that is bigger than .
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The demoninator represents the available combinations of buttons apart from .
Note that the edge case of is and . Our average probability will be
Here follows from the following lemma:
Lemma. Let be positive integers and , then
Proof. [In case you want to try, don't unfold it]
Denote and . We note that
We rearrange to conclude
as desired.
I am at floor , therefore on average I just have
chance of being the first one to leave if there are at least 2 buttons pressed in control panel.