Splitting the Rent

Abstract

The Random Samples item “Splitting the rent, keeping the peace” (7 Jan., p. 37) reminded me of when I and three roommates, all mathematics graduate students, shared an apartment in the late 1950s. We had the same problem as described in the Random Samples item, and we all agreed to a solution I proposed. The idea is that each roommate divides the total rent R among the available rooms, assigning to each room what he or she would consider a fair rent for that room. The person who assigns the highest rent to a room has the rights to that room at that rent, less an equal share of the total rent excess. If anyone has the rights to more than one room, he or she choses one, and the rights to the other is given to the person who assigned the next higher rent. Ties are decided randomly. When everyone has chosen a room, the total T of the assigned rents for all the rooms exceeds the apartment rent R. (It can be proved that this total must be equal to or greater than R. ) Therefore, each person gets a room at less rent than he or she had determined themselves is fair, the reduction equaling ( T - R )/ n , where n is the number of roommates. Whereas the Random Samples item mentions that the iterative solution described there has the flaw that it fails to arrive at a solution if any roommate would not accept an undesirable room even for zero rent, my solution works even for this case by allowing the portion an individual assigns to an undesirable room to be zero or negative. The only constraint is that the total of the rents an individual assigns to all rooms equals R. It would be possible for someone to be paid from the rents of the other roommates to accept an undesirable room, a situation that is perfectly consistent with the problem statement.