80085

With Valentine's Day just passed and Ash Wednesday lurking around the corner, I know the topics of sex and pregnancy are on a lot of people's filthy guilt-ridden minds these days. To help people understand their risks, and to show I'm not a prude, I'm hosting a little get-together (an orgy, if you will) of questions all about sex. So turn the lights down low, put on some soft music, and enjoy this special "adults only" post about what we in the math business call "multiplication."*


Dear Dr. Math,
I read in an article that "Normally fertile couples have a 25 percent chance of getting pregnant each cycle, and a cumulative pregnancy rate of 75 to 85 percent over the course of one year." How do you go from 25% to 85? I don't see the connection between those two numbers.
Name Withheld

As is often the case, Name, the way to understand the probability of getting pregnant over some number of time intervals (I almost wrote "periods" there but then reconsidered) is instead to think about the probability of not getting pregnant during any of those intervals. We can use the fact that the chance of something happening is always 1 minus the chance of it not happening. This turns out to be a generally useful technique whenever you're interested in the occurrence of an event over multiple trials. To take my favorite over-simplified example of flipping a coin, if we wanted to find the chance of flipping an H (almost wrote "getting heads"--geez, this is har.., er, difficult) in the first 3 flips, we could go through all of the possible 3-flip sequences and count how many of them had at least one H, or we could just observe that only one sequence doesn't contain an H (namely, TTT). Since the probability of flipping T ("getting tails") is on each flip, the chance of "doing it three times" is . Thus, the probability of at least one H is . Phew.

Similarly here, there are lots of different ways to get pregnant over the course of a year (believe me), but only one way to not get pregnant. If we take the first statistic as correct, that the chance of a normally fertile couple getting pregnant in each cycle is 25%, then we could assume that the chance of not getting pregnant in each cycle was 75%, or 0.75. Assuming a "cycle" is 28 days long, there would be 13 cycles per year, so by the same reasoning as above, we could say that the chance of not getting pregnant in a year is , about 2.4%. So, the chance of "being in the family way" at some point during the year would be , or 97.6%.

Now, that doesn't match up with the observed number you quoted, 85%. In the study, of course, all they do is assemble some group of "normally fertile" couples and count the number of times they get pregnant in a year. We were trying to solve the problem "top down" whereas the data is observed from the "bottom up." What's going on? Well, the problem was our assumption that the different cycles were independent from each other, in the sense that knowing what happened in one cycle doesn't affect our estimation of what will happen in the next. For coin-flipping, this is a reasonable assumption, but for copulation, not so much. It makes sense that there should be some correlation between the different cycles, because the possible causes for infertility one month might continue to be true the next. For example, it could be that either or both partners have some kind of medical condition that makes conception less likely. Or maybe the guy's underwear is too tight, I don't know. But it seems that the assumption of independence probably doesn't hold. Also, it's not entirely clear what's meant by "normally fertile" here, since (as far as I know) it's only really possible to know if a couple is "fertile" if they've succeeded in having a baby. So, it's possible that the data includes some number of couples who were just less fertile and perhaps didn't know it.

The correct way to understand these compound probabilities is to consider the probability of not conceiving in one cycle conditional on the event that you had not conceived the cycles previously. Unfortunately, I don't have access to that information from personal experience, nor a good mental model for what numbers would be reasonable. However, it seems like the probability of not conceiving should be higher than ordinary if you know already that you've gone some number of months without conceiving. As a result, the odds of getting pregnant in a year should be lower than our estimate assuming independence, which does in fact agree with the data.


Dear Dr. Math,
Planned Parenthood's web site says, "Each year, 2 out of 100 women whose partners use condoms will become pregnant if they always use condoms correctly." Is that the same as saying that condoms are 98% effective? If so, does that mean that if you have sex 100 times, you'll likely get somebody pregnant twice? (I mean, if you're a man. If you're a woman I imagine the rate of impregnating your partner will probably slip in the direction of zero.) Yours always,
Name Withheld

Oh, you freaky Name Withheld, you've asked the question backwards! In fact, the statistic you give of 2 women out of 100 becoming pregnant in a year is how the effectiveness of condoms is defined. That is, in the birth control industry, specifically, when someone claims that a particular method is "x% effective," it means that if a group of women use that method, over the course of the year about (100-x)% of them will get pregnant. Now, there are a number of assumptions being made here, not the least of which is that those women (and their partners) used the method correctly. Without actually going into people's bedrooms (or living rooms, or kitchens?) and tallying up on a clipboard whether their condom use was "incorrect", it's impossible to know for sure. Instead, people who do surveys of this kind have to rely almost exclusively on what people say they did. And let me ask you something: If you accidentally impregnated someone/got impregnated by someone while nominally using some birth control method, would you say, when asked, that you had been using it "incorrectly"? Or would you, as all good carpenters do, blame your tools?

Another implicit assumption is that the respondents reflect a typical number of sexual encounters in a year. Again, I don't know how they decide what participants to include in this kind of study or how they verify the claims they get, but according to some studies I was able to find, the average "coital frequency", as it's romantically known, for both married and single people in the U.S. is somewhere around 7 encounters per month. Therefore, if we treated the experiences as being independent (with the same caveat as in the previous question), we could estimate the probability of unintended pregnancy in a single sexual encounter:

Let's call the probability p. So the chance of not getting pregnant during a given sex act is (1-p). We'll accept the 7 times/month figure and assume a total of sexual encounters per year, all including correct condom usage. As in the coin example, we've assumed independence, so the probability of not getting pregnant over the course of 84 trials is , which we're assuming is equal to the stated number of 98%. Therefore, we have:

And so , meaning that p is very small, about 0.02%. Therefore, if you had sex 100 times, as you say (and congrats, btw), you could expect to make an average of 0.02 babies.

Some important notes:
1) Our assumption of independence here may be more reasonable than in the previous example, because it's possible that whatever factors contribute to a birth control method failing despite proper use may be due more to chance than any kind of recurring trends.
2) Also, these numbers don't account for the fact that (as we saw above) the chance of getting pregnant in a year even without any protection is something like 85%. So, in a sense, condoms "only" reduce the risk of pregnancy from 85% to 2%.
3) We've only been talking about pregnancy here, not the risks of other things like STDs or panic attacks.
4) Wear a condom, people!


Dear Dr. Math,
Mathematically speaking, what number makes for the best sexual position?
Name Withheld

You seem to be asking a lot of questions, NW.

Personally, I've always enjoyed the ln(2π).

-DrM

*Also acceptable: or "integration by parts".

How Big is That Number? Episode 1

This will be the first installment of a new feature I call "How Big is That Number?" in which I try to explain exactly how big that number is.

Dear Dr. Math,
How much space would a googol grains of sand take up?
xoxo,
Frequent Googol Searcher

Dear FreGooSear,

Let me begin by thanking you for reminding us all of the correct spelling of "googol". A lot of people forget that before it inspired the name of an obscure website, the word "googol", as coined by a man named Milton Sirotta back in 1938, referred to the number . I don't know the whole story here, but I'm guessing that as an infant, he somehow wrote a 1 followed by 100 zeroes (I'm guessing in blue crayon), and when asked by his parents what that was, all he could make were baby sounds. Hence the name. Any other explanation seems highly improbable.

So, how big is this number with the silly name? Well, first off, if we wanted to address it in the nomenclature of billions and trillions, it would be known as ten thousand trillion trillion trillion trillion trillion trillion trillion trillion. That's a ten-thousand followed by 8 trillions. Or, if you prefer to think of it in terms of current financial events, $1 googol would be about 12,706 trillion trillion trillion trillion trillion trillion trillion economic stimulus packages.

Now, to answer your question about grains of sand: let's approximate a grain of sand as a cube 1 millimeter across. That would mean that 1000 grains of sand in a row would be 1 meter long. So, one cubic meter would contain 1000*1000*1000, or , grains. A googol grains, therefore, would take up cubic meters of space. And it would really be "space", because this is way bigger than our tiny little planet could accommodate. For comparison, a sphere the size of the Earth (which has a radius of about 6,371 kilometers, or 6,371,000 meters) has, according to the formula for the volume of a sphere, a volume equal to cubic meters, about grains of sand, so it would take about Earth-sized balls of sand to get a googol grains. Alternatively, if you lumped all the sand together into one giant ball, you'd need cubic meters, and solving for R gives us a radius of approximately meters, which is about 140 trillion light years, or 10,000 times bigger than the size of the observable universe. So maybe space itself wouldn't even be big enough to handle such a massive sand ball.

Carrying the sand idea a little further, imagine there was a huge interstellar alien for whom the Earth appeared to be as small as a grain of sand is to us. Now, suppose this being lived on a giant planet, the planet Gigantica, as big, relative to the alien, as the Earth is to us. This would imply that the volume of Gigantica was in proportion to the Earth as the Earth is to a grain of sand, about times as big; so Gigantica itself would be about cubic meters in volume. If, in turn, there were a still larger alien for whom Gigantica looked like a grain of sand and who lived on an even larger planet, the planet Humonga, say, then Humonga would have to be times as big as Gigantica, for a volume of cubic meters. Now, if yet an even larger alien, from the planet Ginormica, looked down on Humonga as a tiny grain of sand, it would require Humonga-balls to comprise a volume of cubic meters, which you'll recall was the volume of space taken up by a googol grains of our puny Earth-sand. Back here on Earth, grains of sand is 10 cubic meters, about the volume of a nice 10 meter by 10 meter patch of beach (assuming a depth of 10 centimeters). So on Ginormica, this giant giant giant alien could kick back in a beach chair, sip a giant giant giant beer and build a giant giant giant sand castle out of those googol grains with plenty of sand left over to get in its megashoes.

-DrM