I need MATH help! (Probability and Statistics) for AIDS discussion

none | 10/15/03 | self

[walden]

I'm having a debate on another board regarding AIDS and the reasons for high infection rates. This was my post:

Simple case, 2 coin tosses, heads=win: What are the odds of winning both times? Odds of winning first time is .5, odds of winning second time is .5, odds of winning both times is .5 x .5 = .5 to the 2nd power = .25, which is 25%.

More complex, add 1 more coin toss for 3 total, heads=win: Odds of winning third time is still .5, odds of winning all 3 times is .25 x .5 = .5 to the 3rd power = .125, which is 12.5%.

Each additional toss lowers the odds of winning every time, at an exponential rate.

This relates quite well to condom usage and sex-- to avoid AIDS, one must win every time. Assume the following:

Odds of condom failure 3% Odds of infected partner 20% (this is quite common in parts of Africa) Assume 1 act of sex per week for 10 years, what are the odds that an individual will "win every time", i.e., remain uninfected at the end of 10 years?

Odds of infection in a single sex act = .03 x .2 = .006 Odds of not getting infected in a single sex act (i.e., winning) = 1 - .006 = .994 Number of sex acts in 10 years= 1 x 52 x 10 = 520

Odds of winning every time (healthy at the end of ten years): .994 to the 520th power = .044 or 4.4 % which means that one has a 96% chance of contracting AIDS in that time period.

This was what someone responded to me:

The math started out well enought, but you're running into problems combining events A and B.

Odds of Failure (P|A) Odds of Infected Partner (P|B) Intersection of (P|A).(P|B) = 3/100*20/100=.006 (which you have right)

Now, you should more accurately treat this as a conditional probability. We'll use (P|B) as the condition, since you have to assume that you can only get infected if B has already occured (infected partner) and we are attempting to find the probability of A (condom failure) happens.

So P(A|B)=(P|A).(P|B)/(P|B)

Sooooo ... .006/.20=.03 Or 3%. You run into a couple problems here. One is that your selecting a sample of 520 different partners over a ten year period. If you know this guy, tell him I could use some tips getting a date for this wknd

What this 3% represents is the Normal Distribution of P(A|B) occuring. And a 97% chance of it not occurring (staying healthy). You can't just raise this by the power times the number of events (in this case, 520). You did this in the coin toss to discover what are the odds of getting a heads on toss 2, given that heads was on toss 1.

Your coint toss experiment is a Joint Occurance (i.e, A has happened, what are the chances B will happen, given A). While correct, we are dealing with Probability of Simultaneous Events (A exists, what is the chance that B will exist at the same time), which is a whole different breed of cat. In the long run, the probability of simultaneous events gets closer to the normal distribution, so you will have close to a 3% chance of P(A|B) occuring today, next week, or ten years down the road, every time. It doesn't compound, just by repeating the experiment. It gets closer to the mean (3%) as N (number of times) increases.

[NukeMan]

"I don't understand how, if I use a condom 33 times without failure, that the next condom I use will statistically be guaranteed to fail. It just doesn't work that way, does it?"

No, it doesn't. It isn't statistically guaranteed to fail, either. It doesn't have to be the *next* one, only *somewhere* in that chain of 33 - an important difference.

[aruanan]

More complex, add 1 more coin toss for 3 total, heads=win: Odds of winning third time is still .5, odds of winning all 3 times is .25 x .5 = .5 to the 3rd power = .125, which is 12.5%.

You let a little error creep in here. Odds for a heads on the third toss may be 0.5, but the odds of "winning third time" is not 0.5 but 0.5 X 0.5 X 0.5 or 0.125 as you correctly stated at the end.

[biblewonk]

Ping for the math.

[DB]

No it doesn't work that way. Read my post 30 and maybe it will make more sense to you.

The 10th coin toss is still 50%. But the overall chances of not tossing heads 10 times in a row is about 1 out of 1000.

P.S. I only graduated high school, and I took the GED to do that early... I'm an engineer now...

[js1138]

The 1 in 500 rate is probably not applicable to the folks most at risk, for reasons already stated.

[NukeMan]

I totally agree. I believe that the numbers were from North America anyway, and the applicability to Africa is dubious.

[Doctor Stochastic]

The original math is correct under the stated assumptions.

First that condom failure is "random" in that one doesn't get a bad batch of condoms or a good batch.

Second, that the partners are chosen "randomly" (independently and identically distributed); a succession of one-night stands would be an example. Things would be different with a single partner who was known to be HIV positive.

The math is correct. The actual probabilities would have to be obtained through observation.

I didn't really understand the objection; it seemed to be poorly written.

[CatoRenasci]

I don't think we disagree. Since the probability of heads is .50 for any fair toss, assuming a perfectly fair coin, over a random number of tosses (greater than, say, 100), you would expect 50% of the tosses to be heads and 50% to be tails. You are interested in how likely it is that at least 1 out the sample universe will be tails (condom failure). That probablity would, as you say, approach unity.

However, it seems to me that this problem is not amenable to sound statistical analysis on the facts given. The second probalistic variable, the HIV infection rate at 20%, does not adequately capture the decision tree facing individuals: after all, who sleeps randomly with 500 different partners over 10 years from a population which mirrors the overall probablity of having aids. All of us move in circles that can best be characterized as subpopulations at greater or lesser risk of aids: i.e. the monogomous spouse who only has relations with his or her spouse to whom he or she has been married since before 1980 faces a very different risk probability (assuming exactly the same facts about condom use and failure rates) than does the promiscuous homosexual who frequents the bath houses for a bit of rough anal trade every week.

While statistic analyssi can be highly enlightening in many circumstances, you can't do meaningful work when you can't control for the variables in an analytically usefeul way.

[walden]

"See post 11 for numbers on catching AIDS with a known-infective partner (1/500 per sex act). It's an important part of the methodology that was left out. "

That seems awfully low to me-- do you have a source on it?
Thanks!

[MosesKnows]

There is no cumulative risk.

So many people encounter extreme difficulty in grasping the significance of this absolute.

[samtheman]

The purpose of my post was to show the mechanics of the math in this problem, which is, in fact, a two-part problem.

Part One: coming up with a risk figure based on the chance a condom will fail and the chance a partner will be infected.

Part Two: coming up with a risk figure based on the concept of "at least one hit" in a series of trials.

I think my post is correct in how to solve the problem. You can plug in any numbers you want. Use an excel spreadsheet for best results and you can easily play around with the variables.

As far as condoms go, I think there are three kinds of failure:

1. Actual structural failure of the condom itself. A break or a split or an opening of some kind.

2. A barrier failure. In other words, the condom worked but the pathogen "got around the barrier" in some way, such as cut or scrapes on the skin of sexual partners.

3. Operator failure. Inappropriate condom use.

I would bet that 99% of all condom failures are operator failures. You have to use the things correctly to prevent pregnancy and disease.

In the case of AIDS, I would bet that the incidents of barrier failure are extremely few in number. Not worth talking about.

As to the actual structural failure of the condoms themselves, I believe it's larger than barrier failure but, as I said, something like 1% of the total failure rate.

If you buy condoms in America and use them properly, you can sleep with an AIDS-infected person and not get AIDS. If that's what you want to do. I personally wouldn't want to do such a thing, but not because I'd fear catching the disease, but rather just because the whole idea would gross me out.

Plus, I'd feel sorry for the poor girl and that would change the mood.

[general_re]

Oh, I agree - this example is, shall we say, highly idealized. The odds of getting HIV if you have unprotected sex with an HIV infected person, while surely high, are undoubtedly not 100%. And as you point out, I don't know of anyone who really chooses sex partners completely at random - you would at least like to know the rates of infection among the pool of people that make up your likely partners. If you prefer your sex partners to be 70 year old Caucasian widows from the United States, the odds of encountering a partner with HIV are very likely to be somewhat lower than if your preferred partners are gay intravenous drug-using Haitians ;)

And in the final analysis, people tend to do this sort of odds calculation without really weighing the costs and benefits of winning and losing. You may have a pretty good chance of not contracting AIDS if you use a condom when you have sex with the girl you just met in a bar twenty minutes ago, but if you do lose, you lose everything.

[ko_kyi]

I thought it was common knowledge that tails leads to infection more than heads.

yuk yuk

[monkey]

I didn't really understand the objection; it seemed to be poorly written.

That is being charitable. It's sad to think that the person must have taken some math/stats.

[general_re]

The original math is correct under the stated assumptions.

I suspected as much, but it's always nice to have some backup ;)

[Ronaldus Magnus]

and the chance of being infected after 100 encounters = 1 - 0.74048426 = 0.74048426

1 - .740 = .260

[general_re]

And in the final analysis, people tend to do this sort of odds calculation without really weighing the costs and benefits of winning and losing.

And in reviewing your initial post, I see you anticipated exactly that, so I think we're in pretty much perfect agreement after all ;)

[CatoRenasci]

And in the final analysis, people tend to do this sort of odds calculation without really weighing the costs and benefits of winning and losing. You may have a pretty good chance of not contracting AIDS if you use a condom when you have sex with the girl you just met in a bar twenty minutes ago, but if you do lose, you lose everything.

Exactly! Any really serious statistical analysis of this, even correcting for all the factors we've both cited, that included even a small portion of potential sex partners who were potentially infected at a 20% "average" rate would have to be an expected "value" or expected "cost" analysis. And, even subjectively I doubt many people, if really put to it, would value a number of random sex acts sufficiently highly to offset the cost (even at lowish probabilities) of contracting HIV/AIDS.

Truly, if AIDS had been treated as a serious public health problem from the beginning, like the plague, anthrax or SARS, we would have immediately quarantined all known infected persons. The cost would have been far less than all we spend now. Infections would not have skyrockedted the way they did had we started quarantine in the mid-1980s or even earlier. Add to that making knowing transmission a felony punishable by life in prison w/o parole.

[oh8eleven]

Truly, if AIDS had been treated as a serious public health problem from the beginning, like the plague, anthrax or SARS...

Gotta' be Ronald Reagan's fault. </sarcasm>

[Cboldt]

Odds of condom failure 3% Odds of infected partner 20% (this is quite common in parts of Africa) Assume 1 act of sex per week for 10 years, what are the odds that an individual will "win every time", i.e., remain uninfected at the end of 10 years?

Your math is okay. What is being challenged (mostly) are your assumptions, and with valid reasons. One poster has pointed out that you assume "use of a failed condom with an infected partner" results in infection. Another has alluded to the fact that "odds of infected partner" is in fact highly variable -- some individuals will have less risky behavior in this regard than others.

But putting the assumptions (of condom failure rate, of rate of contact with infected partner, and of 100% probability of infection upon failed-condom contact with infected partner) aside -- that is, for talking purposes assuming the assumptions are accurate -- your conclusion is mathematically correct.

Even if the odds of a outcome are low on a single attempt, the odds of that outcome occuring at least once increases as the number of trials increases.

An event with a 1 in a thousand chance (.999 chance of failure) has a 41% chance of occuring at least once in 520 tries (.999**520 = 0.59 chance of failure).
An event with a 1 in ten-thousand chance (.9999 chance of failure) has about a 5% chance of occuring in 520 tries (.9999**520 = 0.95 chance of failure).

Your model was that each of 520 incidents of contact had (the same) .03*.20 probability of resulting in infection. The chance of infection being 6 in one thousand, or about 1 in 167, for each incident of contact. An event with this probability has about a 96% chance of occuring in 520 tries (.994**520 = 0.04 chance of missing the event).