"Why can't the guys put the seat down?" says every cliché woman, ever.
"What's the big deal?" is the standard response. And I agree. But let's be more scientific about it, shall we?
Assumptions: a healthy adult pees about 6-8 times per day (counted as 7 from here on) and poops once. Women perform all activities sitting, requiring the seat to be down. Men always stand to pee and sit to poo. All subsequent results can be adjusted if these assumptions are invalid in your household. For base calculations I'll assume that there are an equal number of men and women in a household. This should generally hold true for couples and families, for generalities' sake. Single people have no one complaining to them, so they are uninteresting for this scenario. We'll also call out adjustments for families with more men than women and vice versa. The math is really household-per-toilet, for example if only mom and dad use the master bath (and they only use the master bath), that toilet's usage is by 1 man and 1 woman regardless of the rest of the family composition.
As a practicality:
Flipping a toilet seat up or down is no one's idea of a good time. Naturally we'd like to do this as few times as possible. The seat will have to be flipped if the previous use required the opposite state. In an evenly mixed household, this means that 7 of 16 (P(up)) times the seat will be up and 9 of 16 (P(down)) times the seat will be down. Assuming the events are totally random (which seems reasonable enough), we don't need to flip the seat P(up)*P(up) + P(down)*P(down) = 130/256 ~= 51% of the time if it's just left in the previous user's required state. This means that ~49% of the time we need to flip it.
Simply leaving the seat alone gets the right next state 51% of the time. Always putting the seat down will get the right answer P(down) = 56% of the time. Not a huge win, really. In terms of number of seat flips, it means we average 0.49 flips per use. Always putting the seat down incurs 1 flip in P(up) and 0 flips in P(down) immediately after. Then there's another 1 flip in P(up) and 0 flips in P(down) for the next user, which comes out to P(up) + P(up) = 0.88 flips per use. This is much higher, nearly doubling the number of expected flips. Always putting the seat down is a worse strategy.
The numbers do change as family composition deviates from half-and-half. Suppose a family with 2 men and 1 woman. P(up) changes to 14/24 and P(down) becomes 10/24. Reworking the equations above for leave-as-is yields a don't flip, still, just barely over 51% of the time. The number grows slowly as men outnumber women. 3-to-1 is at 55% and even 4-1 is only at 58%. Conversely, it means that even in a household with a vast male majority, the percent chance that the woman will need to flip the seat stays safely in the 40s. However, the average number of flips climbs to 1.17 for 2-to-1, 1.31 for 3-to-1 and 1.4 for 4-to-1.
Supposing a family has 2 women and 1 man, P(up) falls to 7/24 and P(down) becomes 17/24. Now leaving the seat as-is will work out 59% of the time. The odds climb to 66% for 3-to-1 and 71% for 4-to-1. In other words, a female-heavy household will tend to work out in women's preference favor anyways. Employing the leave-as-is strategy means the average flips falls to 0.41, 0.34 and 0.29, respectively. Always putting the seat down averages 0.58, 0.44 and 0.35 flips, respectively. As women outnumber men more and more, the two strategies converge. However, it is important to note that leave-as-is will never be worse**, and is therefore the statistically correct strategy.
As a courtesy:
If we consider this altruistically, we should look at how often the next person finds the seat in the right state (which is really P(down)). We've done the math above. P(down) is greater than 50% for even households or those with more women. More men cause it to fall below 50%, though a family of 7 could hit a 50/50.
2 men, 1 woman: 43%
3 men, 2 women: 48%
4 men, 3 women: 50%
The takeaway from this is that family composition might dictate that the women leave the toilet seat up! In cases with more men than women, this would generally be the outcome. However, keep in mind that the misses are going to significantly unfairly affect either the men or the women, externalizing a greater cost onto that subset of the family. Is that really being considerate?
In case someone doesn't check and falls in:
I'm just going to laugh.
Conclusion:
Choosing leave-as-is vs always-down strategies is dictated by the philosophy behind the action. If trying to minimize the number of seat flips, leave-as-is is entirely superior. If being considerate to the next person regardless of cost, the choice is dictated by family composition. Families with more men than women should actually opt for an always-up strategy, while others should opt for always-down. The choice is yours. Now you know what you're fighting for!
** Appendix:
[1] Average flips always leaving the seat down = 2 * P(up)
[2] Average flips always leaving the seat as-is = 1 - (P(up)*P(up) + P(down)*P(down)).
Since P(up) + P(down) = 1 always, we can substitute P(down) = 1-P(up) into [2]:
[3] Average flips always leaving the seat as-is = 1 - (P(up)*P(up) + (1-P(up))*(1-P(up)))
[3.expanded] = 1 - (P(up)*P(up) + 1 - 2*P(up) + P(up)*P(up))
[3.collected] = 1 - (1 + 2*P(up) + P(up)*P(up))
[3.simplified] = 2*P(up) + P(up)*P(up).
Since P(up) >= 0, [1] <= [2]. If there are any men in the house, [1] < [2] since P(up) > 0
"What's the big deal?" is the standard response. And I agree. But let's be more scientific about it, shall we?
Assumptions: a healthy adult pees about 6-8 times per day (counted as 7 from here on) and poops once. Women perform all activities sitting, requiring the seat to be down. Men always stand to pee and sit to poo. All subsequent results can be adjusted if these assumptions are invalid in your household. For base calculations I'll assume that there are an equal number of men and women in a household. This should generally hold true for couples and families, for generalities' sake. Single people have no one complaining to them, so they are uninteresting for this scenario. We'll also call out adjustments for families with more men than women and vice versa. The math is really household-per-toilet, for example if only mom and dad use the master bath (and they only use the master bath), that toilet's usage is by 1 man and 1 woman regardless of the rest of the family composition.
As a practicality:
Flipping a toilet seat up or down is no one's idea of a good time. Naturally we'd like to do this as few times as possible. The seat will have to be flipped if the previous use required the opposite state. In an evenly mixed household, this means that 7 of 16 (P(up)) times the seat will be up and 9 of 16 (P(down)) times the seat will be down. Assuming the events are totally random (which seems reasonable enough), we don't need to flip the seat P(up)*P(up) + P(down)*P(down) = 130/256 ~= 51% of the time if it's just left in the previous user's required state. This means that ~49% of the time we need to flip it.
Simply leaving the seat alone gets the right next state 51% of the time. Always putting the seat down will get the right answer P(down) = 56% of the time. Not a huge win, really. In terms of number of seat flips, it means we average 0.49 flips per use. Always putting the seat down incurs 1 flip in P(up) and 0 flips in P(down) immediately after. Then there's another 1 flip in P(up) and 0 flips in P(down) for the next user, which comes out to P(up) + P(up) = 0.88 flips per use. This is much higher, nearly doubling the number of expected flips. Always putting the seat down is a worse strategy.
The numbers do change as family composition deviates from half-and-half. Suppose a family with 2 men and 1 woman. P(up) changes to 14/24 and P(down) becomes 10/24. Reworking the equations above for leave-as-is yields a don't flip, still, just barely over 51% of the time. The number grows slowly as men outnumber women. 3-to-1 is at 55% and even 4-1 is only at 58%. Conversely, it means that even in a household with a vast male majority, the percent chance that the woman will need to flip the seat stays safely in the 40s. However, the average number of flips climbs to 1.17 for 2-to-1, 1.31 for 3-to-1 and 1.4 for 4-to-1.
Supposing a family has 2 women and 1 man, P(up) falls to 7/24 and P(down) becomes 17/24. Now leaving the seat as-is will work out 59% of the time. The odds climb to 66% for 3-to-1 and 71% for 4-to-1. In other words, a female-heavy household will tend to work out in women's preference favor anyways. Employing the leave-as-is strategy means the average flips falls to 0.41, 0.34 and 0.29, respectively. Always putting the seat down averages 0.58, 0.44 and 0.35 flips, respectively. As women outnumber men more and more, the two strategies converge. However, it is important to note that leave-as-is will never be worse**, and is therefore the statistically correct strategy.
As a courtesy:
If we consider this altruistically, we should look at how often the next person finds the seat in the right state (which is really P(down)). We've done the math above. P(down) is greater than 50% for even households or those with more women. More men cause it to fall below 50%, though a family of 7 could hit a 50/50.
2 men, 1 woman: 43%
3 men, 2 women: 48%
4 men, 3 women: 50%
The takeaway from this is that family composition might dictate that the women leave the toilet seat up! In cases with more men than women, this would generally be the outcome. However, keep in mind that the misses are going to significantly unfairly affect either the men or the women, externalizing a greater cost onto that subset of the family. Is that really being considerate?
In case someone doesn't check and falls in:
I'm just going to laugh.
Conclusion:
Choosing leave-as-is vs always-down strategies is dictated by the philosophy behind the action. If trying to minimize the number of seat flips, leave-as-is is entirely superior. If being considerate to the next person regardless of cost, the choice is dictated by family composition. Families with more men than women should actually opt for an always-up strategy, while others should opt for always-down. The choice is yours. Now you know what you're fighting for!
** Appendix:
[1] Average flips always leaving the seat down = 2 * P(up)
[2] Average flips always leaving the seat as-is = 1 - (P(up)*P(up) + P(down)*P(down)).
Since P(up) + P(down) = 1 always, we can substitute P(down) = 1-P(up) into [2]:
[3] Average flips always leaving the seat as-is = 1 - (P(up)*P(up) + (1-P(up))*(1-P(up)))
[3.expanded] = 1 - (P(up)*P(up) + 1 - 2*P(up) + P(up)*P(up))
[3.collected] = 1 - (1 + 2*P(up) + P(up)*P(up))
[3.simplified] = 2*P(up) + P(up)*P(up).
Since P(up) >= 0, [1] <= [2]. If there are any men in the house, [1] < [2] since P(up) > 0
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