Assume the following
- A guy always makes sure the seat is up before peeing.
- Let x = the portion of the toilet usage where it requires the seat up, 0 ≤ x ≤ 1
- For simplicity, assume bathroom going is a memory-less process (i.e. a bathroom going has no impact on the next one).
- Each toilet seat flipping has a cost of 1.
The old rule: keep the seat down all the time
Then when a guy pees, he has to flip up the seat, and then down when he finishes. There is an x probability that happens, so the expected cost of each bathroom going is 2x.
The rule in question: leave the seat as it is
Then when a person go to the bathroom, there is an x probability he/she will find the seat up, and (1 – x) probability down. The seat requires flipping when
- The seat is down, and a guy wants to pee
- The seat is up, and someone needs to sit on it
2x(1 – x) probability that the seat requires flipping. And a flipping only cost 1 in this case, so the expected cost of each bathroom going is 2x(1 – x).
2x ≥ 2x(1 – x) because 0 ≤ x ≤ 1. Therefore, it is more efficient to leave the seat as it is.
Although I have not proved the fairness of this rule, it is sure that men and women both share the burden of seat flipping. And it has to be fairer than men doing all the flipping.
- Proving the lower bound of the expected cost
- Detail on fairness
- A non-independent bathroom-going process
- Empirical simulation