Bathroom Mathematics

In the interest of improving the overall quality of human life, and also of filling up an hour of free time, I have conclusively proved that men should leave the seat up after using the toilet.

First, let’s construct a simple model. In this model, which resembles real life closely but not exactly, there are two types of people, male and female. To make calculations easier, let’s say that the populations of these people are identical – there is the same number of women in our model as men – and they both use the bathroom at the same frequency. Furthermore, there are two activities that these people engage in: we’ll call them number 1 and number 2, both of which are conducted an equal amount of times. Both before and after conducting said activities, the people have the option of lifting the seat, dropping the seat, or leaving the seat in place. For the sake of convenience, let’s say that there is one woman and one man, and that they both do number 1 and number 2 each once a day. We’ll call them Ellen and John.

We will state that when Ellen conducts either activity 1 or 2, the seat must be down. When John conducts activity 2, the seat must be down, but when he conducts activity 1, the seat must be up. The goal is accomplish these conditions with the lowest expenditure of energy. In our model, let us say that the act of moving the seat up or down costs exactly 1 unit of energy.
As you can see, when a person enters a bathroom in our model, there are four possibilities.

1. Ellen wants to do activity #1.
2. Ellen wants to do activity #2.
3. John wants to do activity #1.
4. John wants to do activity #2.

Each of these possibilities feature a starting condition: whether the seat is already up or down.

Condition 1, seat up:
1) Ellen/Activity #1: Ellen must put the seat down. Cost: 1eg.
2) Ellen/Activity #2: Ellen must put the seat down. Cost: 1eg.
3) John/Activity #1: John leaves the seat up. Cost: 0eg.
4) John/Activity #2: John must put the seat down. Cost: 1eg.
The total cost for condition 1: ¾ (3 units of energy out of a possible maximum of 4).

Condition 2, seat down:
1) Ellen/Activity #1: Ellen leaves the seat down. Cost: 0eg.
2) Ellen/Activity #2: Ellen leaves the seat down. Cost: 0eg.
3) John/Activity #1: John must raise the seat. Cost: 1eg.
4) John/Activity #2: John leaves the seat down. Cost: 0eg.
The total cost for condition 2: ¼ (1 unit of energy out of a possible maximum of 4).

But we’re not done yet! There is another part to our simulation: what Ellen and John do after they finish their business. In the first scenario, John puts the toilet seat down when he finishes activity #1, so the end state is always TS-down. In the second scenario, everyone leaves the seat exactly where it is. It is these two scenarios that we will be comparing.

Scenario 1, TS-down:
1) Ellen/Activity #1: Ellen leaves the seat down. Cost: 0eg.
2) Ellen/Activity #2: Ellen leaves the seat down. Cost: 0eg.
3) John/Activity #1: John must lower the seat. Cost: 1eg.
4) John/Activity #2: John leaves the seat down. Cost: 0eg.
Total cost for scenario 1: ¼ (1 unit of energy out of a possible maximum of 4).

Scenario 2, TS-neutral:
No one does anything. Total cost, 0 (0 units of energy out of 4).

Okay, now let’s do some very basic math. Here’s our energy calculation formula: ProbCond1(EnergyCostCond1) + ProbCond2(EnergyCostCond2) + EnergyCostNxtM = TotalEnergyCost.

That is, the probability of encountering condition 1 multiplied by its energy cost, plus the probability of condition 2 multiplied by its energy cost, plus the energy cost of the next move (i.e. putting the seat down) equals the total energy cost of the given situation.

Given scenario #2 (TS-Neutral), when someone enters a bathroom they will encounter condition 1 (TS-up) exactly 1/4th of the time. Why? Because out of four possible combinations of participant and activity, only one results in the seat being left up. So our equation looks like this:

¼(3/4) + ¾(1/4) + 0 = 3/16 + 3/16 = 6/16 = 3/8 = average energy usage in scenario 2.

Now, let’s take scenario 1, where John must put the seat down after he concludes activity #1. In this case, no one ever encounters condition 1, because the seat is always down.

0(3/4) + 1(1/4) + ¼ = ¼ + ¼ = 2/4 = ½ (or 4/8) = average energy usage in scenario 1.

Clearly, the average energy usage in scenario 1 (1/2eg) is greater then that in scenario 2 (3/8eg). The reason why is easy to understand once you think about it. In scenario 1, a unit of energy will have to be expended exactly 1 out of 4 times. The benefit you get from this is that you never have to cope with condition 1. But in scenario 2, condition 1 only costs 3/16! So in order to save 3/16, you expend 1/4. That’s not good business.

Or, you may think of it this way. In scenario 1, Ellen never has to expend energy (her seat is always down when she enters, and she never changes it). But John has to expend 2 units of energy exactly half the time he goes. If he needs to do #1, he will ALWAYS have to lift the seat up and then he will always have to put it back down.

In scenario 2, Ellen has to expend 1 unit of energy exactly 1 quarter of the time. She always needs the seat down, but there is a 1 in 4 chance that John was in before her doing a number 1. John, on the other hand, expends 0 energy 1 of the time, and 1 energy the other half. That’s because he has a ½ chance of doing a number 1, at which point he has a ¼ chance of encountering the seat already up. There’s 1/8. He also has a ½ chance of doing a number 2, in which case he has a ¾ chance of encountering the seat down. That’s 3/8. 1/8+3/8 = ½.

Scenario 2:
1/4(1) [ellen] + ½(1) [john] = 1/4 + ½ = ¾
Scenario 1:
0 + ½(2) = 1

In case you didn’t notice, the ratio 1:3/4 is the same as ½:3/8. Expressed in more conventional format, 4:3. For every 4 units of energy expended in the world of scenario 1, only three are expended in scenario 2!

So, let’s examine the advantages to scenario 2 over scenario 1. Less energy is used, of course, but what does that mean? Well, the less energy Ellen and John expend, the less food they have to consume, saving them money. Since each energy-using action also takes time, you’ll see that scenario 2 also saves time. In each case, we’re talking about only a tiny fraction….but over the course of a lifetime, this adds up! Say we considered each movement of the seat to take 1 second. Well, we already know the ratio, 3:4. If every day John and Ellen together use up 4 seconds moving the toilet lid under scenario 1, they would only use 3 if they switched to scenario 2! That’s a one second saving per day, 365 seconds over the course of a year, or just over 6 minutes. Ten years give us 3650 seconds or just slightly over an hour. Now if there are ten million couples like John and Ellen, that’s one million hours of life lost every decade for no good reason! And one million extra units of energy as well.

Furthermore, scenario 2 is clearly the more equal of the two in terms of distributing labor. In scenario 1, John does all the work while Ellen does none. Scenario 2 redistributes the labor so that John does 2/3rds of the labor and Ellen 1/3. While this is not truly equal, it is certainly better than the other option.

I find this evidence to be incontestable. Men of the world unite, you have nothing to lose but the extra energy you expend by lowering the toilet seat!