## Tuesday, February 24, 2009

### 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 $\frac{1}{2}$ on each flip, the chance of "doing it three times" is $\frac{1}{2^3} = \frac{1}{8}$. Thus, the probability of at least one H is $1 - \frac{1}{8} = \frac{7}{8}$. 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 $(0.75)^{13}=0.024$, about 2.4%. So, the chance of "being in the family way" at some point during the year would be $1-0.024 = 0.976$, 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 $7*12=84$ 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 $(1-p)^{84}$, which we're assuming is equal to the stated number of 98%. Therefore, we have:
$(1-p)^{84}=0.98$
And so $(1-p) = 0.98^{1/84} = 0.99975$, 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: $\int e^x$ or "integration by parts".

Anonymous said...

"probability of not conceiving in one cycle conditional on the event that you had not conceived the cycles previously"

I've been wondering about the percentage chance of getting pregnant in a year question.

If you got pregnant in the first cycle that would mean that you couldn't get pregnant in any of the later cycles and lower the 97.6% figure. So shouldn't the probability of conceiving go down if you conceived in the cycles previously, not go down if you didn't conceive?

drmath said...

It's true that getting pregnant in any cycle makes it impossible to conceive in the subsequent cycles (at least for a while), but that's actually irrelevant to the way that we're calculating the probability. As in the coin example in the first part of this post, it's much simpler to consider the probability of not getting pregnant over the course of the year and then just subtract that from 1 to get the probability of conceiving at some point.

Otherwise, you'd be stuck doing something like the following: let the probability of conceiving in any cycle be p. Again, for the time being, we'll assume independence. So the probability of getting pregnant in the different cycles goes like:
--cycle 1: probability = p
--cycle 2: probability = (1-p) * p [we failed in cycle 1 and succeeded in cycle 2.]
--cycle 3: probability = (1-p)^2 * p
--cycle 4: probability = (1-p)^3 * p
...
--cycle 13: probability = (1-p)^12 * p

Now, you could add up all these probabilities, since the events are mutually exclusive, just as you point out. If you know a little statistics, you might recognize this as (a truncated version of) the geometric distribution. However, with a little algebra, you can show that the sum of these probabilities is exactly 1 - (1-p)^13, as we had already calculated.

The important point here is the correlation between not conceiving in any given cycle and not conceiving in any other cycle, due to the various medical possibilities I mentioned. This makes the assumption of independence invalid using either method of calculation.

-DrM

Abigail said...

Very complicated mathematic. I will be having difficulty getting pregnant if I have to understand all these. Great stuff anyway!

hbillions said...

What's the probability of a fertile virile woman getting pregnant over the course of a 40-year active sex life?