A chore rota that is even, and how to check it

Most household rotas fail the same way. Somebody writes out week one carefully, week two gets copied, and within a month one person has done the bins every single time. The argument that follows is about fairness but nobody has the counts, so it is unwinnable.

A rota built on one rule fixes both halves: the jobs move, and the counts are checkable.

The rule

Number the people from zero and the chores from zero. In week w, chore j goes to person (j + w) mod n, where n is the number of people.

Each week the whole assignment shifts one person along. With two or more people that guarantees nobody repeats a chore the following week. And over any n consecutive weeks each chore passes through every person exactly once, so a rota run for a whole number of cycles gives everyone an identical total.

Worked example: more chores than people

Three people — Alex, Bea, Chris — and four chores: Dishes = 0, Bins = 1, Hoovering = 2, Bathroom = 3.

1. Week 1, w = 0. Dishes 0 mod 3 = 0 Alex. Bins 1 mod 3 = 1 Bea. Hoovering 2 mod 3 = 2 Chris. Bathroom 3 mod 3 = 0 Alex. Counts: Alex 2, Bea 1, Chris 1.
2. Week 2, w = 1. Dishes Bea, Bins Chris, Hoovering Alex, Bathroom Bea. Counts: Alex 1, Bea 2, Chris 1.
3. Week 3, w = 2. Dishes Chris, Bins Alex, Hoovering Bea, Bathroom Chris. Counts: Alex 1, Bea 1, Chris 2.

Totals over three weeks: Alex 2 + 1 + 1 = 4, Bea 1 + 2 + 1 = 4, Chris 1 + 1 + 2 = 4. Dead even.

Check the other way: four chores every week for three weeks is 4 × 3 = 12 chore-slots, and 4 + 4 + 4 = 12. Run it for six weeks — two full cycles — and each person does 8, totalling 24, which is 4 × 6.

What a four-week rota looks like

Four weeks is not a multiple of three, so week 4 repeats week 1 and the extra load lands on Alex. Totals: Alex 4 + 2 = 6, Bea 4 + 1 = 5, Chris 4 + 1 = 5. Sum 6 + 5 + 5 = 16, which is 4 × 4.

One chore apart, and it evens out at the end of the next cycle. If that still rankles, plan in cycles: with three people, plan three, six or nine weeks rather than four.

Worked example: more people than chores

Four people and three chores means someone is free every week, and the free week has to move as well or it is just a different unfairness.

The same rule handles it. In week 1 chores 0, 1 and 2 go to people 0, 1 and 2, so person 3 is free. In week 2 they go to people 1, 2 and 3, so person 0 is free. Over four weeks each person is free exactly once and does each chore exactly once — three chores each, 3 × 4 = 12 in total, matching 4 × 3 chore-slots. Nobody has to be told the free week rotates; the counts show it.

Publish the totals

The counts are the point. A rota that ends with a total per person converts "I always do the bins" into a number that is either right or wrong, and both outcomes are more useful than the argument. Print it, stick it up, and let anyone check the sum against chores multiplied by weeks.

None of this makes the rota binding. Swaps by agreement are fine and normal — a rota nobody agreed to is a list on a fridge.

The tool

The Chore Rota Builder applies the rule and prints the weeks with a count per person at the end, so the fairness is shown rather than asserted. Type the people, the chores and the number of weeks. It handles the more-people-than-chores case by marking free weeks, and those rotate too.

It runs entirely in your browser, so no names are uploaded anywhere. Use the totals as the starting point for the argument, not the end of it.