# How to Split Rent with Your Roommates for the Mathematically Inclined

Korok ChatterjeeHere’s another issue many of our users come across – you’re renting a house or apartment with several other people, and each person has their own room. However, the rooms are definitively not of the same quality. How do we split up the rent in a fair way?

The truth is there’s no unique, obvious answer, but we try to outline a fair method in this post. We consider three separate areas of the rental unit – the bedrooms, the bathrooms, and the common areas. First, we must decide what fraction of our total rental experience each area comprises. Then, we assign three variables to them (the reason for this will become clear later). For example:

- Common Area – 45% (A = 0.45)
- Bathroom – 15% (B = 0.15)
- Bedroom – 40% (C = 0.4)

Let’s start with an example scenario: three people are sharing a 3BR/2BA apartment which costs $4500/mo. The master bedroom is 200 sq. ft. and has its own bathroom. The other two bedrooms are 150 sq. ft. and 160 sq. ft. and share the second bathroom. For simplicity, we assume both bathrooms have showers and ignore uncommon cases such as a bedroom having its own half-bath. Now, our job is to determine the fractions of the total rent owed by the three roommates.

Before we dive in, let’s define a few variables (I know, I know, bear with me) so things get easier when we generalize. Let *X1*, *X2*, and *X3*, be the rents owed by roommates *1*, *2*, and *3.* Let’s say roommate 1 is in the master bedroom, roommate 2 is in the 160 sq. ft. room, and roommate 3 is in the 150 sq. ft. room. Let *A1*, *A2*, and *A3* be the portions of the rents that are owed for the common area for roommates *1*, *2*, and *3*, respectively. Similarly, *B1*-*B3* for the bathrooms and *C1*-*C3* for the bedrooms. Finally, let *R* be the total rent for the apartment. Now we can write down a few relationships:

A + B + C = 1

In English, this means that the way we divvied up the three areas must account for the whole unit.

X1 = A1 + B1 + C1

X2 = A2 + B2 + C2

X3 = A3 + B3 + C3

This just says that each roommate’s rent is comprised of their contribution to common area, bedroom, and bathroom.

A1 + A2 + A3 = A*R

B1 + B2 + B3 = B*R

C1 + C2 + C3 = C*R

Again, this says that the total cost of the common areas, bedrooms, and bathrooms, i.e. the full rent, must be accounted for.

Now we just have to figure out what these variables should come out to be. Let’s start with the common areas. In almost all cases, it makes sense to just split this up evenly so that:

A1 = A2 = A3 = (A/3)*R

Let’s move on to the bathrooms. Here, it’s a little bit more complicated. We start with the assertion that roommate 1 who has their own bathroom should pay twice what the other two roommates pay. In other words:

B1 = 2*B2

B1 = 2*B3

Using our statement from earlier that B1 + B2 + B3 = B*R, we arrive at the conclusion that

B1 = (B/2)*R

and

B2 = B3 = (B/4)*R

Finally, let’s do the bedrooms. Following our previous reasoning, it makes sense that each person’s bedroom share should be in proportion to the fraction of the total bedroom space they occupy:

C1/C2 = 200/160

C2/C3 = 160/150

There are many other ways we could have written this, but they are all logically equivalent and result in the same rent distribution. The above two equations plus the constraint that *C1 + C2 + C3 = C*R* yield the following solutions:

C1 = 200/(150+160+200)*C*R = (200/510)*C*R

C2 = (160/510)*C*R

C3 = (150/510)*C*R

Now we can finally calculate *X1*-*X3*.

X1 = R*[A/3 + B/2 + (200/510)*C] = $1718.38

X2 = R*[A/3 + B/4+ (160/510)*C] = $1408.46

X3 = R*[A/3 + B/4 + (150/510)*C] = $1373.16

Hooray! Of course, all roommates have to agree on the allocation of value to the common areas, bedrooms, and bathrooms, but we’ll leave that to you.

In case you’re thirsting for more algebra, let’s generalize to an NBR BR/NBA BA apartment with *N* roommates where the *i-th* roommate has *Ki* sq. ft. of bedroom space and *Li* bathrooms (a bathroom shared between *k* people gives each of those *k* people access to *1/k* bathrooms and a common area bathroom gives *1/N* (or *1/2N* if it’s a half-bath) access to everyone). Let’s also keep all the same variables from above. Then the general solution is:

Xi = R*[A/N + B*(Li/NBR) + C*(Ki/(K1 + K2 + … + KN))]

We’ll also leave it to you to verify that the *Xi*’s add up to *R*. QED and happy renting!