Math / probability question re: HIV

November 19, 2014 8:08 AM

I'm trying to calculate the probability of contracting HIV for the husband of an infected woman. Given a risk of 10-50 conversions per 10,000 exposures. And assuming an average frequency of sexual acts of 2-3 X per week. That's 100-150 sexual acts /year. That should give about a 10%-15% yearly risk of contracting HIV, right?
posted by mert to Science & Nature (13 answers total)

Assuming you are only talking about penile-vaginal, ejaculatory sex, you still need to consider the viral load of the infected partner, as that will impact the transmission rate.
posted by stillmoving at 8:11 AM on November 19, 2014 [2 favorites]

It depends on viral levels. Also PrEP is an option where the wife takes HIV medication to stop transmission. It's very effective.

Someone with undetectable viral load has a vanishingly small chance of transmitting the virus.
posted by AlexiaSky at 8:14 AM on November 19, 2014

Oh and PrEP is covered by many many public health departments in the United states for free. Also if they are having sex to get pregnant the likelyhood the baby will have HIV is near 0. There were less than 5 babies born in the US with HIV last year. All of them did not receive neonatal care.
posted by AlexiaSky at 8:17 AM on November 19, 2014 [2 favorites]

In terms of the statistics you've given:

Worst case:
(50 / 10000) * 3 * 52 = 0.78 (78%)

Best case:
(10 / 10000) * 2 * 52 = 0.104 (10.4%)

So there's a bigger variation than what you've calculated. As others have said, there's other issues at play and so while this may give you a very rough ballpark figure it doesn't tell you much more.
posted by leo_r at 8:21 AM on November 19, 2014 [1 favorite]

Are you doing this as an exercise in probability or because you actually want to determine a realistic rate of transmission? Because the latter has been studied extensively and empirically in the literature, and is subject to all of the considerations raised already (and more). The former is much more straightforward, and if it's just a math problem, we should probably remove HIV as a specific part of the question so we can answer without diversions.
posted by telegraph at 8:22 AM on November 19, 2014 [4 favorites]

Assuming you want to treat this strictly as a math question (i.e. independent trials), you would calculate the probability of it this way:

Low estimate (100 trials at 10/10000): 1 - ((10000 - 10) / 10000) ^ 100 = 9.5% chance of infection
High estimate (150 trials at 50/10000): 1 - ((10000 - 50) / 10000) ^ 150 = 52.8% chance of infection

leo_r's math is incorrect (it can yield probabilities greater than 1).
posted by 0xFCAF at 8:23 AM on November 19, 2014 [5 favorites]

In terms of the statistics you've given:

Worst case:
(50 / 10000) * 3 * 52 = 0.78 (78%)

Best case:
(10 / 10000) * 2 * 52 = 0.104 (10.4%)

This gives the expected number of 'transmission events', not the probability that at least one such event occurs. (If there was a 100% chance per sex act, 2 doesn't mean you have a 200% chance.)
posted by PMdixon at 8:25 AM on November 19, 2014 [5 favorites]

Best answer: Assuming a 99.9% chance of not contracting for any given event, then the chance of not contracting across 150 occurrences is (.999)¹⁵⁰, or 86%.
posted by Doofus Magoo at 8:25 AM on November 19, 2014 [1 favorite]

Best answer: I don't know whether medical statistics works like math statistics, but in math, you can't just multiply your per-instance chance by the number of times -- probability doesn't work like that. (Imagine if the probability is 1%...if you add, you're saying there's a 100% chance of infection after 100 go-rounds.)

Mathematically, you calculate it "in reverse:" if there's a 1% chance of getting infected, that means there's a 99% of not getting infected any particular time. So if you want to know the chance of not getting infected after 10 times, the equation is 0.99¹⁰ (99% times 99% times 99% ... ten times) which is 0.9044, or about 90.44% So at 1% chance, after 10 times, you have a 90.44% chance of being non-infected, or a 9.66% chance of having caught AIDS sometime during those 10 times.

[on preview, Doofus Magoo has the actual numbers, if we're talking math.]
posted by spacewrench at 8:27 AM on November 19, 2014 [1 favorite]

Also you need to take into account that women are less likely to transfer HIV to men than a man to a woman. Just like people receiving anal sex are more likely to contract from an HIV positive male than a giver(gah wrong term) to contract HIV from a receiver of an act. This had to do with the likelihood of damage to the penis during the event causing an opening for HIV to be transmitted.

Sorry for all the posts !
posted by AlexiaSky at 8:30 AM on November 19, 2014 [1 favorite]

Response by poster: Thank you all. I am writing an educational session for medical students about confidentiality and partner notification in HIV and wanted to clarify the exact risk of not informing a partner of an infected individual. The numbers I provided are accurate estimates of transmission from female to male in vaginal intercourse. It sounds like the answer is 1-(.999)^150... much appreciated
posted by mert at 8:36 AM on November 19, 2014

It really doesn't make sense to do this using hand wavey math when it has been studied in real life so extensively, especially if your audience are medical students. Here's one study I found in 30 seconds. "Serodiscordant" is a search term that will be of use to you here.
posted by telegraph at 9:19 AM on November 19, 2014 [9 favorites]

I get your impulse here. Math both does and doesn't quite work this way in HIV education and policy though, from my experience. You can't quantify an exact risk. Everyone does sex things differently; everyone is less or more "compliant" with medication; some people have sex five times a week; some have sex five times a year. People have tried to do this math with "condom failure rates" too, but condom failure is about usage, not about defect in the condom. Some people have high condom failure, some people have low.

But it's important to explain this stuff because a lot of people think that "sex with someone with HIV means transmission," which of course it doesn't necessarily.

Looking at actual studies of serodiscordant partners is useful—except that they are tracking people who know their HIV statuses. That's different if you're trying to build a case for partner notification, because at least one person in the couple wouldn't know his or her status. Anyway there's a huge base of literature out there on partner notification too, some of it of course really wrong-headed. (Partner notification + harm reduction might be the winning search terms.)
posted by RJ Reynolds at 10:21 AM on November 19, 2014 [8 favorites]