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Cumulative probability of HIV infection

  1. May 4, 2010 #1
    I have read that the condom effectiveness in protecting from HIV infection is around 98%.
    Assuming the probability of contracting HIV from a single protected encounter is 2% the probability of getting nfected after 1000 protected encounters is (I took the math from here http://books.google.com/books?id=G4...=probability hiv after n encounters&f=false):
    [tex] 1 - (1 - 0.02)^1000 = 0.99 [/tex].

    This is a pretty high value. Is the math correct here?

    Thanks.
     
  2. jcsd
  3. May 4, 2010 #2

    mathman

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    The math is correct ASSUMING the encounters are all independent.
     
  4. May 4, 2010 #3
    Yes, it is assuming the independence of the events.
    This is quite frightening news! It means that after a thousand protected encounters an individual almost surely has been infected, assuming the effectiveness of condoms is 98%. (Even if you raise it to 99% it doesn't affect the result significantly).
    The frequency of the event overwhelms the probability of a single encounter infection.
    This calculation puts most sex workers at an extremely high risk since they are usually exposed to such high numbers of encounters.

    BTW, I don't understand why the LaTeX code didn't format correctly the exponent 1000.
     
    Last edited: May 4, 2010
  5. May 4, 2010 #4

    CRGreathouse

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    I think the 98% is more questionable than the independence assumption.
     
  6. May 4, 2010 #5
    Yes, the number is questionable, however even assuming 99,9% effectiveness - which I think is unrealistically optimistic, - the probability is at 0.63 which is still high, and this may be surprising for most people.
     
  7. May 4, 2010 #6

    CRGreathouse

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    I think that 99.9% may be reasonable, especially for HIV -- it's very fragile compared to, say, Hep C. But another number to keep in mind is the percentage of HIV-positive individuals. Is your risk going from 100% to 63%, or from 2% to 1%? Obvious translations for varying degrees of promiscuity apply, of course.
     
  8. May 4, 2010 #7
    But I think "independent" here means only that you don't double count "encounters". It's like multiple throws of a die, you're allowed to throw the same or different unbiased dice - the throws will still be independent.

    And before you start spreading panic:

    (a) You're talking about 1000 "encounters" with an infected partner or partners.

    (b) Even if the 98% figure you give is correct, I think it probably means that in a "protected encounter" there is 2% of the risk of infection that there would be in an "unprotected encounter", NOT that there is a 2% risk of infection. So the figure would actually be nonsense.

    Not that I'd like to encourage people to have unprotected "encounters" willy nilly either, of course.
     
  9. May 5, 2010 #8
    I realised after posting the foregoing that my comment on the meaning of "independent" is wrong.

    If the throws are from a set of distinguishable dice, some of which are loaded, the player can affect the outcome of individual throws by monitoring the results of previous throws and selecting the die appropriately.

    In this case, avoiding repeating encounters that previously resulted in infection with a related disease, for example, could possibly affect the chances, even under OP's assumption that the encounters counted are only those with HIV infected partners.
     
    Last edited: May 5, 2010
  10. May 5, 2010 #9

    CRGreathouse

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    Right. You're saying the same thing I was getting at (but more eloquently!) when I discussed the percentage of the population that is HIV-positive.

    I would have expected the chance of infection in an unprotected encounter with a disease-carrying individual to be quite high. Am I wrong?

    Hmm, good point. This really only applies insofar as HIV is less likely than other diseases to be transmitted, though.
     
  11. May 5, 2010 #10

    mathman

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    This exercise seems to make a good case for monogamy.
     
  12. May 5, 2010 #11

    marcusl

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    Who (aside from Magic Johnson) has 1000 independent partners?
     
  13. May 5, 2010 #12

    lurflurf

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    The partners do not need to be independent, onlt the encounters. The results would hold for example if two persons, one infected and one not, had 1000 encounters.
     
  14. May 5, 2010 #13

    Hepth

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    its not 98%. If you heard 98% that meant 98% chance over the course of ONE YEAR or proper use, with average encounters. The more accurate measurement is : 0.9 per 100 person-years infection rate vs. 6.7 without.

    If so, then assume if you had protected sex with say, 100 people in a year, thats 100 people years. 10 years, thats 1000. So your chance of infection after 1000 average-infection-rate-people over 10 years is just about 9%, with it being 67% without.

    And that's independent sampling I believe. 20 people a year = 1.8% chance vs 13.4% chance without.

    Thats what they mean by the 85%-95% reduction.
     
  15. May 6, 2010 #14

    CRGreathouse

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    Very nice numbers, thanks! Do you have a cite?
     
  16. May 6, 2010 #15
    I read about a study of heterosexual couples one of who had been infected when they met. So far as I remember (I read it rather a long time ago), it took the men about three months on average to become infected and the women one month, which would actually suggest the chances for heterosexuals at least are quite low. However I wouldn't take these figures as authoritative.

    But as far as the original question of whether the maths was correct is concerned, if the quoted 98% should have been interpreted as I suggest then the answer would be no (whatever the actual chances).

    In fact, from the preceding posts it appears that both OP's and my interpretations were incorrect.
     
    Last edited: May 6, 2010
  17. May 6, 2010 #16

    Hepth

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    National Institute of Allergy and Infectious Diseases; National Institutes of Health, Department of Health and Human Services (2001-07-20). "Workshop Summary: Scientific Evidence on Condom Effectiveness for Sexually Transmitted Disease (STD) Prevention" (PDF). Hyatt Dulles Airport, Herndon, Virginia. pp. 13–15.

    http://www3.niaid.nih.gov/about/organization/dmid/PDF/condomReport.pdf [Broken]


    From wikipedia actually.


    EDIT : Seems to have been taken down. I'm trying to find the updated one.

    EDIT: Found. Updated link : http://www.niaid.nih.gov/about/organization/dmid/documents/condomreport.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  18. May 7, 2010 #17
    The exposure risk (ER) for any encounter with 98% protection (2% failure rate) is ER=(P)(F) where P is the prevalence rate in the population "sampled" and F is the failure rate of the protection.

    So the risk of exposure for one protected encounter for the US is (.006)(.02)=.00012.

    The probability of exposure after 1000 independent protected encounters is 1-(1-.00012)^1000)=.1131

    Here's a list of HIV prevalence rates in various countries:

    http://en.wikipedia.org/wiki/List_of_countries_by_HIV/AIDS_adult_prevalence_rate

    The prevalence rate in the sampled population is the critical factor. The case (or exposure) rate per unit time is not the relevant measure here.

    EDIT: Obviously using the prevalence rate for the entire population of a country is unrealistic, but so is this scenario. Who has 1000 random encounters without choosing to go back to the same person even once? Even "professionals" have regular customers. Maybe I'm just old fashioned. In any case, a better prevalence rate would be for non-monogamous sexually active persons.

    EDIT: The probability of exposure for 1000 independent encounters without protection, taking the .006 prevalence rate, is .9976.
     
    Last edited: May 8, 2010
  19. May 7, 2010 #18

    CRGreathouse

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    Rather than use a one-parameter model with number of exposures, how about a two-parameter model with another variable that varies from 'all encounters are with one person' to 'all encounters are with different people'? So as limit (# of encounters) increases without bound, if the monogamy parameter stays at the first bound, the probability of infection rapidly approaches 0.006.
     
  20. May 8, 2010 #19
    By risk of 'exposure', I'm talking about the probability of exposure to an infected person's body fluids by way of sexual intercourse. (Unprotected contact or failure of the protection). If a person is monogamous it's unlikely protection would be used. Strictly speaking however, that person's consort may not be monogamous. For any given encounter with the sole consort, the monogamous person has an unknown probability of infection using this model. It would be wrong to state that repeated unprotected encounters with the same person carries the same risk as with different persons. Otherwise the monogamous person would have a virtual certain chance of being exposed after 1000 encounters with her/his sole consort if we assigned the consort some probability of being infected such as .006.

    If we consider a monogamous uninfected couple, then the probability of exposure is zero. From any point in time, if one member of the couple has other encounters, the risk increases for each first encounter with a new person according to this model. Subsequent encounters with the same person don't change the risk under a static model.

    A dynamic model has to combine prevalence and incidence data and is more complicated. Even with incidence (new cases per unit time) data, it's not particularly relevant to any given person because that person's risk for becoming infected over time depends mostly on his/her behavior.

    EDIT: I'm not equating exposure with infection since infection depends on several factors in addition to exposure. Exposure is a calculated outcome here. Actual infection must be determined by testing.
     
    Last edited: May 8, 2010
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