Roman women's life expectancy
- Thread starter kev67
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All no, but still many. Death in childbirth was a very, very common occurrence until not so long ago.
I have no idea if those numbers are correct (such estimates must be hard to make), but if they are, childbirth was my immediate guess.
Keep in mind that contraception and abortion weren't as widespread, safe or effective as today. So not only the rate of death in childbirth per pregnancy was much higher than today, but women also got pregnant and carried to term a lot more often than in our modern societies, on average.
Yeah, given that not only the chance of dying as a result of any given pregnancy was pretty high (I don't know exactly how high but possibly 20 or 30%?) but women also tended to get pregnant often, the risk of dying in childbirth at some point was obviously multiplied. The risk also increases as a woman ages and, conversely, it's also greater when she's very young (13, 14 years old, and early marriages were common back then).
It probably wasn't nearly as bad as that. Estimates on the internet (first page of Google results only) are more around 2-2.5%. Still, if you multiply that by, say, 5 pregnancies,* that's about a 10-12.5% chance of dying in childbirth; for 6 pregnancies it would be 12-15%; and so on.
*I don't know if the estimates I found are only about full-term and near full-term pregnancies or if they include miscarriages, which are also dangerous.
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Also, if the risk had been 30% then you'd only have needed four pregnancies to get a 120% chance of dying in childbirth... according to the way I did my calculations above, that is! That's obviously not possible. How do you calculate these things? I've always sucked at math and I've probably overreached myself with my attempts on this thread. I should leave it alone.
This gives the estimate of about 40-50 pregnancies on average to death. Even if getting pregnant twice a year, it's 20-25 years. If you start at 15, its 35 - 40 years of life expectancy. Though of course, there are other causes for death...
That's because this kind of probabilities don't add up this way. Pregnancies, in this context, are what's know as independent events (that is, if having been pregnant previously doesn't increase the likelihood of death in childbirth from ulterior pregancies. Now chances are they actually aren't independent events, but if the only data we have at our disposal is a general risk of death at childbirth, then they should be treated as independent events). It would be like saying that since you have a 1/6 chance to get a 1 if you cast a die, if you cast it 6 times, you'd have 100% chance to have gotten a 1 at some point, which is not the case: you only ever have a 1/6 chance to get a 1.
Now, probabilities is not at all something I'm good at, and I haven't studied that topic much. There's a way to account for the fact that if you cast a cast a die a million time, the overwhelming odds are that you should have gotten at least one 1 at some point, but I can't remember how that works :/ All I can say for sure is that on each specific trial, you only ever have a 1/6 chance to get any number.
And that's if you live to see your 15th birthday. One should not forget that low life expectancy was, for a considerable part, due to infant and child mortality. I heard that in the Middle Ages, roughly 50% of children didn't reach adulthood. I don't know about Roman times but it may not have been much better. Once you did reach adulthood, though, you had a fair chance of living to something like 60 or 70 although the average life expectancy was lower than that because of all the dying children.
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At any given throw. But you've got more than a 1/6 chance to get a 1 at some point if you cast the die several times, and this chance increases with every throw. That's what I was trying, and failing, to calculate regarding the pregnancies: what the overall chance of dying in childbirth at some point is if you get pregnant five or six, etc. times. How do you calculate that? (I won't blame you if you don't know.)
This.
And also, there's a synergy of different death causes. If you catch cold, chances to die are low; but if you do that while pregnant and your immune system is suppressed - chances of complications and subsequent death may be higher (especially given the poor state of ancient medicine).
In such cases you don't add probabilities of getting 1; you multiply probabilities of not getting 1. Probability of not getting 1 is 5/6; probability of not getting 1 two times is (5/6)×(5/6), etc.
Now, there's a simple formula to compute the expected number of attempts to get 1: if the chance of getting 1 in one attempt is X, the expected number of attempts to get 1 is 1/X. In our case X is 1/6 and the expected number of attempts is 1/X = 1/(1/6) = 6.
I know how to calculate such things in the case of coin tosses, but not sure how to apply it to the issue at hand... Specifically, I don't know how the fact that if you die, you can't be pregnant anymore, changes how you should calculate the probabilities. (and yes I know it sounds dumb, but it drastically reduces the number of possible outcomes, which is crucial in how you calculate probabilities, for reasons I'll get back to later) And of course, there's the fact that the probabilities are unequal.
But in the case of a fair coin (independent events, only two possible outcomes with equal odds), the odds of getting only tails over 3 throws is 0.5^3, so 12.5%.
The probability of getting at least one heads over 3 throws is 1 minus the possibility of getting only tails, so 1 - 0.125, or 87.5% of getting one heads (or one tails, for that matter) over 3 throws.
That's a result you can get to without using a formula, but simply by listing all the different outcomes. In the case of 3 coin tosses; the only possible outcomes would be:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
You can't possibly get any other combination of tails and heads from 3 coin flips in a row. So if you want to know what are the odds of getting at least one heads, you just count the number of outcomes out of those 8 where you have at least one heads, namely, 7/8, or a 87.5% chance, as per the formula.
Now suppose you want to do the same with the problem at hand. You first have to list all the possible outcomes, and that's where the problem I was speaking about earlier comes up: if you use "S" for survives childbirth, and "D" for dies at childbirth, DS wouldn't make any sense, since you can't survive after having died. So for 5 pregnancies, the only possible outcomes are:
SSSSS
SSSSD
SSSD
SSD
SD
D
So, in 5/6 outcomes the mother dies... but obviously, contrary to the fair coin experiment, not all those outcomes are equally likely.
And I'm failing to find a way to take that into account, even though it seems to be like probabilities 101.
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Thinking about it, it looks very possible to me that I overthought it. Plainly put, the odds of dying should be: 1 (because all odds should add up to 1, or 100%, in any experiment) minus the odds of surviving 5 deliveries. If we take the death rate to be 2.25%, and so the surviving rate to be 97.75%, we get 1 - 0.9775^5 = roughly 10.76% chance of death at childbirth for 5 deliveries, assuming they are independent events.
That's huge considering that 1) several estimates I've seen give a birthrate of 6 to 9 children per woman in ancient Rome, which by definition doesn't take into account stillbirths, which were probably very common, and 2) deliveries are probably not completely independent (and so the death rate could increase with each delivery for instance, not to mention that it would indeed increase anyway if only because of the age of the mother).
That's huge considering that 1) several estimates I've seen give a birthrate of 6 to 9 children per woman in ancient Rome, which by definition doesn't take into account stillbirths, which were probably very common, and 2) deliveries are probably not completely independent (and so the death rate could increase with each delivery for instance, not to mention that it would indeed increase anyway if only because of the age of the mother).
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If we're dealing with life expectancy at birth, that number for men is almost certainly too high. All estimates of premodern life-expectancy are super problematic, but the modern research that I'm aware of (essentially Walter Scheidel) puts Roman life expectancy somewhere in the 20s to mid-30s and doesn't find nearly so significant a difference between men and women. For example, based on Egyptian census returns, Scheidel finds a female life expectacy around 22 and a male life expetancy around 25+ [1]. (Indeed, for Roman Elite, some evidence suggests that women had a longer life expectancy at birth (27.5) than men (25.3) [2].)
By contrast, 45 is going to be closer to the adult life expectancy (i.e. from around 20 y.o.) in Rome, although this is probably closer to 50.
1: Summarised in Walter Scheidel, "Physical Wellbeing in the Roman World" Princeton/Stanford Working Papers in Classics (2010), 2. (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1505815)
2: Walter Scheidel, "Emperors, Aristocrats, and the Grim Reaper: Towards a Demographic Profile of the
Roman Élite", The Classical Quarterly 49, No. 1 (1999), 266.
Read Lysistrata
This is the gamblers fallacy https://en.m.wikipedia.org/wiki/Gambler's_fallacy Your odds of rolling a one are always one in six no matter how many times you throw the dice, I think....