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Our Daily Bleg: The Old Roommate/Rent Dilemma

Conor Hunt, an I.T. consultant in Chicago, writes with a dilemma that, while common, seems to be always unsatisfactorily solved.

Two friends — a merchandising analyst and a law student — and I are attempting to split up rent of a three-bedroom apartment with two common bathrooms. All rooms have their pros and cons, with the major differentiators being closet space and sheer square footage:
Room No. 1: 15 ft. x 15 ft.
Room No. 2: 12 ft. x 12 ft.
Room No. 3: 20 ft. x 8 ft.
Rent is $2,200 per month and the apartment is approximately 2,200 square feet.
Simple math would show that one would pay per square foot, but that goes out the window with the ranking intangibles, and the fact that no one necessarily wants the big room.
The roommates threw out these prices:
No. 1: $800/month
No. 2: $710/month
No. 3: $690/month
So given their prices over the course of a year, Room No. 1 would have to yield $1,080 and $1,320 more in value than room Nos. 2 and 3, respectively. That’s an insanely high premium for a little more square footage and a closet! It is still just a bedroom, after all.
How do you recommend solving this situation?

I am sure Conor and his friends will welcome any suggestions you have. I am not sure why Room No. 3 is considered worse than Room No. 2 even though it is larger, but I’m sure there’s a reason. In advising Conor, feel free to consider a few of our own suggestions:
1. Just settle it on a coin flip or, better, Rock Paper Scissors.
2. Rotate rooms every three months.
3. Price all rooms equally but tax Room No. 1’s occupant higher for household goods, or cooking/cleanup chores.
4. Give the smallest room to the guy least likely to have sleepover guests.
5. All three roommates hold hands over open flame; whoever lasts longest gets room of his choice.