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.

[walden]

So, how does this really work? Thanks!

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[drjulie]

Hi there! I'm a psychologist with a graduate minor in stats...so there are likely people here who know more than I do. However, to sum it up, if the probability of getting AIDS is 0.006 then that's what it is each time you have intercourse. Your first model using the multiplication rule, I believe, is incorrect.

[IronJack]

The odds of turning up heads on two consecutive coin tosses is one in four. There are four possibilities: heads-heads; heads-tails; tails-heads; tails-tails.

The probablitity of each coin toss coming up heads is always one in two.

[Mr. Bird]

Because each "gamble" is independent of the other, the odds do not change. Every time you flip a coin, the chances are 50/50. If the condom failure rate is 3%, you have a 3% chance of failure with each use. There is no cumulative risk.

[Spaulding]

There are a number of other problems with your assumptions, having nothing to do with probability theory. First, you are confusing HIV with AIDS. In Africa few nations actually perform HIV tests. When people die of wasting disease today the diagnosis is most often AIDS, because that diagnosis can generate money. Further, the percentage of the US population which is uniformly tested, and which is HIV positive has changed little since testing began. Testing is performed on all US military personel. Take a look at www.virusmyth.org if you are interested in science rather than politics.

[samtheman]

Using your assumptions: condom rate = 3% (which I doubt) and infected-partner-possibility = 10%:

Chances of a single act of sex being "unlucky" =

(the chance of condom failing) * (the chance of the partner being infected) =

.03 * .10 = .003.

Next: the chance of a single act of sex being "lucky" =

1 - "unlucky chance" = .997.

Next: chance of sex remaining "lucky" through 10 encounters =

.997 raised to the 10th power

= .997 ^ 10 = .970402

and the chance of getting infected after 10 encounters =
1 - .970402 = .029598

Next: the chance of remaining lucky through 100 encounters =

.997 raised to the 100th power

= .997 ^ 100 = 0.74048426

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

Next: the chance of remaining lucky through 1000 encounters =

.997 raised to the 1000th power

= .997 ^ 1000 = 0.049563083


and the chance of being infected after 1000 encounters =
1 - 0.049563083 = 0.950436917


So chances are, under those conditions, 1000 encounters means 95% of contagion.

I would argue that your premise is false. I think 3% failure rate of condoms is a very high figure.

[NukeMan]

I can tell this will be a long thread.

I believe you are correct in your methodology. John Allen Paulos used a very similar example in his book Innumeracy. The events are independent, but can still be cumulated to obtain probabilities of long chains of occurrence(as with coin tosses). I am pretty sure you did it right.

[LouisianaLobster]

You should also take into account that unprotected sex with someone who is HIV-positive will not necessarily make you HIV-positive. There's an additional factor that you have to put there, although i have no idea what the actual numbers are.

[Axelsrd]

They are right. In your example, you want to know the probability of getting AIDS. You can only get AIDS if your partner has AIDS and you have a condom failure. Since your partner has AIDS,that condition is satisfied and will always exists. What you now need to do is determine the probability of condom failure, which you state as 3%. The probability of condom failure is the same on the 1st attempt as it is on the 520th attempt. Hope this helps.

[NukeMan]

I think 3% failure rate is actually a decent estimate, and in fact *overestimates* their effectiveness, but I am too lazy to Google the actual stats. The infected-partner possibility is 20% (not 10%) according to the posters estimate.

There is one additional factor to be included. IIRC, the CDC estimated that only 1 in 500 sex acts (with a partner you knew *for sure* was infected) led to you catching aids. Now, this is a general number combining all genders and sex acts. Many sex acts are riskier than others - receptive anal intercourse is the riskiest (much greater than 1 in 500). So, taking into account all that:

Chance of infection per episode assuming 3% failure, 20% chance of partner infection, and 1/500 infection rate is .000012. I would here disagree with the 3% failure rate as being too low, though.

[general_re]

The good doctor is much more of an expert than I, so perhaps a ping for help is in order.

That being said, my initial reaction is that condom failure and having an infected partner are independent events - it's not a conditional probability at all, and therefore your simple probability calculation is correct. And even if it was a matter of conditional probability, your opponent muffed his own math - he gives the conditional probability P(A|B) as being (P|A)*(P|B)/(P|B), but it's not. The actual formula is P(A|B) = P(A)*P(B|A)/P(B). The Mathworld page on conditional probability shows how that's derived.

[DB]

If a condom fails 3% of the time AND a condom failure results in getting the other partner infected 20% of the time WHEN one partner is infected AND the other partner is not initially infected AND THEN you repeat the event 520 times using the same uninfected person the result is:

1-((1-(0.03 x 0.20))^520) = 0.956

Resulting in a 95.6% chance the initially uninfected person is infected at the end of the 520 events.

At least that’s how I see it…

[general_re]

You can only get AIDS if your partner has AIDS and you have a condom failure. Since your partner has AIDS,that condition is satisfied and will always exists.

Except that you're not assuming your partner always has AIDS, and therefore you're simply calculating the odds of simultaneous events.

[DB]

If heads is an infection then tossing the coin twice gives you a 75% probability of getting an infection.

[Spaulding]

"You should also take into account that unprotected sex with someone who is HIV-positive will not necessarily make you HIV-positive."

You beat me to it. But I'll go a bit further. There is no direct evidence that AIDS or HIV are communicated through sexual contact - of any sort. The original hypothesis from the Center for Disease Control that AIDS was lifestyle related, and in particular, the result of immune system destruction by drug use, still appears the most plausible explaination.

I've always liked the "Ice cream causes sun stroke" simily. The "fast-track" drug supported lifestyle of a sector of the homosexual community - cocaine, amyl nitrates, tetracycline, night after night after night - clearly includes sex. But drugs are well known to destroy immune systems. Growing up in San Fran, I've know many physically healthy gays in relatively long term relationships which presumably include sex, and no AIDs until after the drug crazed 60s and 70s.

No mechanism by which the tiny amounts of the retrovirus found upon autopsy could cause death has been discovered. The Nobel prize winner for the only mechanism sensitive enough to isolate the virus, Kary Mullis doesn't worry about losing his research grants, and declares flatly that there is no scientific basis for the HIV causal relationship. Where is healthy skepticism?

[Axelsrd]

I think one has to assume your partner has AIDS. The only way you could contact AIDS yourself is if your partner has it and your condom failed. I'm not looking for an argument, but where did I go wrong?

[NukeMan]

Ok, my curiosity has been piqued. I did Google on condom failure rates, and the numbers are fairly distressing. For purposes of preventing pregnancy, about 12-15% fail and for HIV protection purposes about 10% - 30% fail. Yikes. OF course, there are hundreds of online articles touting all sorts of numbers subject to a variety of conditions, which makes a simple summary difficult.

[Agnes Heep]

Even at 3 percent, I wouldn't much care for the odds. Suppose you had a revolver that held 100 rounds, with three live rounds available. Would you care to play Russian Roulette at least once a week with such a weapon? I wouldn't. If you think those are good odds, get 97 white marbles and three black marbles, and mix them in a container. Keep picking one at random, and see how long it takes to get a black one.

People are usually pretty good at instinctively calculating risk, but the problem with AIDS, unlike the Russian Roulette scenario is that the ill effects of the former occur at a distance, while those of the latter are immediate. If AIDS killed instantly, no one would take the risk. And it's the same thing with smoking.

[DB]

If you take a 3% chance of getting an infection twice you have a 5.91% chance of the being infected when done.

You are confusing two different issues.

Each event is 3% regardless of how many consecutive successful chances you took previously BUT if you do the event 100 times you have an overall chance of 95.2% of being infected when you completed the 100 risk events.