A random walk through mathematics -- mostly through the random part.
A problem in the model: it assumes John must do operation #1 with the seat up. In fact, John has an option.What's needed is another parameter to measure the preference John has for having the seat up during #1. If John has no strong preference (or, indeed, prefers the seat down) the optimal solution is clearly to leave the seat down at all times and pay no costs whatsoever.
Your solution is provably optimal.
Obviously, this is a trick question: everybody knows the only acceptable state to leave a toilet in is seat down and lid down.
As a grad student in math at a small school, I shared the "graduate dorm" with about 12 other grad students. There was a bathroom with 5 or 6 toilets. One of them ended up reserved for a certain Robert, who is fortunate that I can't remember how to spell his last name, assuming he's still among the living.His reservation technique was to do operation #1 while standing up without bothering to lift the seat. In addition, he drank heavily each night and was in a certain hurry to use the facilities each morning. After that, he started the day with a Coke and an asprin (which he called a prophylactic for the headache he would otherwise develop).
Post a Comment