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Up or Down?A while ago, I had an idea for what I thought would be a nice little calculation. Recently I tried it, and it worked, and here's the result: a mathematical solution to the problem of whether the toilet seat should be left up or down.
The basic idea is to look at some different scenarios and figure out the expected effort involved in each. By effort, here, I don't mean just physical effort, but rather anything that makes an action more undesirable … negative utility, if you're used to thinking that way.
The expected effort will derive from two different units of effort. The first, which I'll call e, is the effort of raising or lowering the seat. It represents a small amount of physical effort, but mostly an expenditure of time, plus the ickiness of having to touch the seat. The second, which I'll call d just to make the letters adjacent, is the effort of bothering to check the state of the seat before sitting down. I think it's negligible, but it doesn't complicate the math to include it, and we can always set it to zero in the end.
Now, on to the scenarios.
If a woman is living by herself (or several women are living together), there's nothing to calculate. The seat stays down, and the expected effort is zero.
If a man is living by himself (or several men are living together), the situation is a bit more complicated. Let p be the probability that, in a single instance, the seat ought to be up, and q = 1-p be the probability that it ought to be down; and assume that the man uses the lazy algorithm of leaving the seat in whatever state it was last in; then the elements of the calculation can be organized as follows.
The letters “s” and “c” in the corner are meant to suggest that the rows represent (initial) states, while the columns represent cases. The row and column headers show the probabilities for the different states and cases, while the main body of the table shows the effort required for each combination. The two cases have probabilities p and q by definition; the two states have the same probabilities due to the lazy algorithm.
Now we can just add up the expected efforts for all the combinations. If, for example, the seat is up (probability p), but ought to be down (probability q), then it needs to be lowered (effort e), and the expected effort for the combination is pqe. Similarly, the expected effort for any other combination is the product of the actual effort with the corresponding row and column probabilities. Here, the only other combination with nonzero effort has the same expected effort pqe, so the total is 2pqe, or rather d + 2pqe, since I didn't include the effort d in the table.
By the way, although here it may look like the columns represent final states, that's just an artifact caused by the lazy algorithm. In later calculations it will be clear that the columns are cases, not states.
Shared Household, Male Algorithm
If men and women are living together, things get much more interesting. As a first step, let's see what happens if everyone uses the lazy algorithm.
Let f be the probability that, in a single instance, the user is female, and m = 1-f be the probability that ve's male. (If there are n people in the household, nf female and nm male, then f = nf/n and m = nm/n, all else being equal. All else may not be equal, but that is something I choose not to discuss here.)
The elements of the calculation look like this.
Just as an example of where the values come from, the seat will be down initially (probability f+mq) if the last user was female (probability f) or the last user was male (probability m) and he wanted the seat down (probability q). Adding up the various combinations, and including d, we obtain the following values for the expected effort.
In particular, the expected effort is not the same for women as for men. Under certain conditions (when f/m < p-q), the expected effort is larger for women; otherwise, smaller. However, I don't think that's the right comparison to make.
Suppose, as often happens, a man and a woman have been living separately, and then decide to move in together. What you want to compare, in that case, is not the expected effort, but rather the change in expected effort due to the new circumstances. From the calculations above, we know the old values of the effort,
then all we need to do is subtract, and we have the changes.
Now the expected effort is always larger for the woman. In fact, even if we allow different numbers of men and women, and ignore the effort d, the expected effort is still larger for women under most conditions (in fact, when m/f > p-q, which by coincidence looks a lot like the previous rule).
That's all pretty abstract, so let's put in some numbers. First, as a rough guess, let's suppose that p and q are about 3/4 and 1/4. In that case, the absolute effort is larger for women if f/m < 1/2, i.e., if there are few women, while the change in effort is larger for women if f/m < 2, i.e., if there aren't a lot of women.
Next, let's suppose that the household consists of one woman and one man, so that f and m are both about 1/2. To keep things simple, let's also set e = 1 and d = 0. Now we can make a nice little table of the various expected efforts.
Thus, moving in together creates more work for everyone, but creates relatively more for the woman, which, I imagine, is part of the reason that women complain about men not putting the seat down, and not vice versa. In other words, there's some justification for the complaint.
Shared Household, Female Algorithm
Since the lazy algorithm is, arguably, unfair to the women, let's look at the other extreme, and see what happens if everyone always leaves the seat down. (For women, that's the same thing as the lazy algorithm, so for “everyone” you can also read “the men”.)
Here are the elements of the calculation. (The “up” row has probability zero, so it could be left out.)
So, the expected effort looks like this,
and the change in expected effort looks like this.
Already we can see how this is going to turn out, but let's put in the numbers, just for completeness.
Clearly this is a far more unjust division of labor!
Shared Household, Probabilistic Algorithm
Although the calculations above haven't been done before (as far as I know), I have to admit that the qualitative results were entirely predictable. Now, however, we come to something genuinely new, which I claim as my own modest contribution to the problem of toilet seats. The two algorithms we've looked at have been unfair in different directions, so, clearly, what is needed is something in between. So, let's see what happens if the men, when the seat ought to be up, put it down afterward some fraction x of the time.
Defining y = 1-x, the elements of the calculation look like this.
For consistency, I should perhaps have written 1-mpy as f+mq+mpx, but it doesn't matter, the two are equal. Also, by the way, you can check that the elements are correct by looking at them in the limits x = 0 and x = 1, where they reduce to the elements for the last two scenarios.
Adding up all the combinations, we find the following expected efforts. (Here the second line can be written in many ways; the one below is convenient because it groups all the terms that contain y.)
Then, subtracting, we find the changes.
Now, according to my arguments above, we can make the algorithm fair by requiring the changes in effort for men and women to be equal. But, that's easy enough—we just set the two equal and solve for y.
y = (2p - d/pe) / (1+2mp)
Putting in the standard numbers, we find y = 6/7, so that x = 1/7. Then, as a check on the calculation, we can put the numbers into the same kind of table as before.
Yup, it all adds up.
Now let's step back a second and see what we've got. If the correct measure of fairness is the equality of effort beyond what's required in a purely male or female household, then we know the value of x that makes the algorithm fair. It's pretty clear that that's the correct measure when a man and a woman move in together, but not very clear at all in any other case. If, furthermore, the numbers I've been using are correct, then the value of x for a household of one man and one woman is 1/7, so that the man should put the seat down exactly 1/7 of the time.
Now, here's a funny thing. Before I worked through these calculations for the first time, I imagined that x would be some generic real number, so that you'd need a random number generator next to the toilet in order to apply the algorithm properly. Instead, with the very first set of numbers I chose to use, I got the value 1/7, which, I think you have to admit, immediately suggests a particular way that the algorithm should be applied: the man should put the seat down one day a week. To me it even seems obvious that the day should be Sunday.
Unfortunately, that's not the right way to interpret the result. If, for example, both the man and the woman are away during the day, working, then only evenings and weekends are spent at home, and the fraction of time that Sunday occupies is more than 1/7. (And, we should probably only count waking hours.) Or, if just one person is away during the day, the situation becomes difficult to analyze. Technically, we could just adjust the values of f and m, but that wouldn't take into account the fact that there are now known intervals with only one user, so that the value of putting the seat down depends on the time of day. In both cases, all I can say is that maybe the Sunday rule would be a good starting point for figuring out some workable compromise.
Although the equation above always gives a value for y, the value isn't always a valid solution to the problem. Under certain conditions (approximately m/f < p-q, as explained above), even the lazy algorithm is more work for men than for women, and in such cases the value of y will turn out to be greater than 1. On the other hand, if d is really large, there's no middle ground between always putting the seat down and even the tiniest probability of leaving it up, and in those cases the value of y will be less than 0.
If you believe, with Mr. Spock, the utilitarian sentiment that “the needs of the many outweigh the needs of the few” (for related quotes in Bartlett's, look under “the greatest happiness for the greatest numbers”), then instead of making the expected male and female efforts equal, you might want to minimize the total expected effort … that is, minimize the quantity f times the expected female effort plus m times the expected male effort. (Note that the weights really are f and m, not nf/n and nm/n.) As it turns out, the total effort is linear in x, with positive slope, so that the best solution is always the lazy algorithm … and the same is true if you minimize the total change in effort instead of the total effort! (That last is obvious, because the base efforts don't depend on x.) So, although putting the seat down does have the beneficial effect of redistributing effort, in the end it always creates more effort than it saves.
There's another unit of utility that I haven't talked about so far, namely, the large negative utility of trying to sit when the seat's not down. The reason I haven't talked about it is that it doesn't figure in anywhere. In fact, it's exactly because it's large that it doesn't figure in—if there's any chance that the seat isn't down, it's obviously better to make the small effort of checking the state before sitting.
By the way, it's important to distinguish the small effort of checking the state from the larger and more annoying effort of learning to always check the state. The latter is a one-time cost, so it shouldn't be included in the calculations above, but it should perhaps be accounted for in some other way. For example, when a man and a woman move in together, it's the woman who has to learn to check the state, so probably the man ought to do something nice for her, to compensate.
Finally, I ought to point out that if you're a guest rather than a resident, different rules may apply. My own feeling is that you should always leave things in the state you found them in, even if that means you have to put back down one of those seat covers that have the carpet on top.
@ October (2003)