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.

[twittle]

This guy is going through a lot of fancy statitistics jargon to sound impressive, but he is full of it. I agree with your initial analysis.

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%.

He went through all this to tell you the probability of a condom break is 3% when that was a GIVEN in the first place. What a stupid egghead ritual. As you figured (assuming your numbers are valid), the chances in any one given time are 0.006 and the chances over 520 times are as you calculated.

[ex-Texan]

I agree with you. The odds are the same for every throw of the dice or for every lottery ticket you buy. That is why people who buy 20 + lottery tickets a week would be better off buying life more insurance or investing in mutual funds.

With AIDS so rampant in the gay community, and getting worse each day, if a person has stupid, reckless, promiscuous or drug addicted partners they have about a 100% chance of getting AIDS.

[DB]

In truth the AIDs virus is smaller than the latex structure and can eventually pass through it. Latex is permeable. For pregnancy prevention failure you need an actual hole.

[DrC]

"Chance of infection per episode assuming 3% failure, 20% chance of partner infection, and 1/500 infection rate is .000012."

The 3% failure rate seems reasonable (http://www.jr2.ox.ac.uk/bandolier/band64/b64-4.html)

So taking the above one step further, odds of HIV after 10 years = 1 - (1-.000012)^520 = ~1% vs. odds of HIV after 10 years of not using condoms = 1 - (1-.0004)^520 = ~19% so using condoms clearly is smarter than not, but no guarantee against contracting AIDS, i.e., 1 out of every 100 who practices such "safe sex" will end up with HIV.

[DB]

Each isolated event is the same risk. But the overall chances of getting infected increase the more times you take the risk. Each additional event is a 3% chance you wouldn't be exposed to if you didn't do the event again.

[general_re]

Even if you know your partner has AIDS, or assume it for the sake of argument, then the odds of contracting it are whatever the odds of condom failure are, every time, and are cumulative over time because each new sex act is an event independent of the previous event - it's a fresh roll of the dice, so to speak. So if the odds of condom success are .97, then the odds that your condom will be successful 10 times in a row are .97^10, or about .74 - IOW, over ten sex acts, you have about a 26% chance of condom failure, and therefore a 26% chance of getting AIDS, assuming that the odds of getting AIDS from an AIDS infected person are 100% if your condom fails. Which they almost surely aren't, but let's keep it simple ;)

But you don't know that your partner has AIDS, so you have to take that probability into account too. But the probability of condom failure and finding an AIDS infected partner are independent events - the sample spaces are not affected by each other, as far as I can see. So if you choose a partner at random, and the odds that that person will be AIDS infected are .20, then the odds of condom failure and getting an AIDS infected partner for a single sex act is .03*.20, or .006, or six chances in a thousand that A and B will occur, and 994 chances in a thousand that one or both of those events (condom failure or having an AIDS infected parter) will not happen.

But since we're assuming that you only have to lose once to get AIDS, you need to figure up the probability that both of those events will eventually happen given some number of sex acts. Assume you choose a new partner at random every week, for 52 weeks of the year. Then the probability that both events (condom failure and getting an AIDS infected partner) will not occur across the span of a year is .994^52, or about .731. Across ten years, the probability of both events not occurring simultaneously - because you need both events to happen simultaneously to get AIDS - is .994^520, or about .04 - that is, there's about a 96% chance that both of those events will occur simultaneously over the span of ten years.

[CatoRenasci]

From my graduate work in econometrics and statistics some 30 years ago, I believe you are correct. Those coin toss examples always turned on the fact that each event was independent of the prior and subsequent events. I view the HIV/AIDS example as one of those problems where pure statistical probability is meaningless anyway: even if the probability of any event resulting in infection, and even if the events are independent of each other, the downside of infection (probable death) is so great that the expected value of the event happening would have to be very very highly negative.

[<1/1,000,000th%]

Odds of condom failure 3%...

The odds of condom failure are closer to 13%. Somebody posted it on FR a while back.

[js1138]

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.

So all those dead gay guys were drug users? I think not.

[DB]

If heads is an infection and you toss the coin 10 times even though each time the chances of getting an infection is 50% the overall chances of being infected after the 10 tosses is 99.9%.

[DB]

Ditto

[general_re]

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.

Right - that's the odds for a single occurence. But he wants to know the cumulative odds for multiple sex acts over some period of time, so you have to take that probability and raise it to the power of the number of discrete acts in order to calculate those odds. IOW, the odds that you will not flip heads on any given coin toss are 50%. The odds that you will not flip any heads across ten coin tosses are pretty small - about 1 in 1000.

[Mind-numbed Robot]

You are leaving out important factors. i.e., how the sex is performed. If it is doggy-style then the it is 100% safe because dogs don't get AIDS and almost the same with missionary style. Now if you are doing super kinky stuff, like guy on guy with both feet off the ground, the chances of infection increase. If there is a monkey involved the odds go up, also.

I hope this helps.

[FLAUSA]

I think you are quite right in calculating the probability of not being infected in 520 occurrences. While the probability of each individual occurrance remains the same (ie., .03 x .1) the probability of multiple occurrances resulting in a specific outcome are indeed exponential (assuming replacement, which means that each potential sex partner is available for all 520 cycles).

The answer that you received is the probability for each individual occurrance, which does stay the same, assuming that all potential sex partners have an equal chance of being selected.

All potential sex partners do not have an equal chance of being selected. In fact, the rate of HIV infection is greater in those potential sex partners who are more likely to have sex with you. Do you think that the probability of a 65 year old woman, who has only had sex with her husband of 45 years is the same as a 20 year old homosexual, who has had multiple partners? Of course not. Therefor, if the rate of HIV infection in the general population is 10%, then the rate of HIV infection in the population willing to have casual sex is much, much higher.

The condom failure rate is usually determined by virginal hertosexual encounters. Anal penetration, either homosexual or hertosexual would have a much higher failure rate.

The probability is also affected by the transmission of body fluids or as some say whether you're a pitcher or a catcher.

I'm sure you could calculate the odds if you knew all the probabilities of the individual pieces, but common sense would tell you that having sex with certain sub-groups or certain sexual practices would increase your chances of being infected with the HIV virus. But alas, common sense is not a common commodity anymore.

[general_re]

If heads is an infection and you toss the coin 10 times even though each time the chances of getting an infection is 50% the overall chances of being infected after the 10 tosses is 99.9%.

Hey, stop saying what I'm about to say. Or at least stop being faster than me ;)

[Mr. Bird]

I only took baby stats in college, so I acknowledge I could be wrong. But it seems to me that the fallacy in the cumulative risk equation is that the risk too rapidly approaches 100%. Let's stipulate to the 3% failure rate. 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?

[DB]

Ya, I saw your example was virtually identical...

Good thing I did it first…

People would say I was lifting your work ;-)

[NukeMan]

You might be right. The numbers vary wildly from my Google search. 3% failure was mentioned, as was 10%, 12%, 15% and even higher numbers. For pregnancy prevention purposes, the failure rate was lower than for HIV-infection purposes. It defies a simple summary.

I'll also note that the numbers seemed to vary by socio-economic class and by country. Anyone using these numbers could make that argument, but had better be prepared to withstand the fury that would surely follow!

[NukeMan]

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.

[general_re]

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 really work that way - your odds on the 34'th time are still 3% for that event. What the cumulative risk calculation does show is that eventually that 3% risk will catch up with you - eventually, you will lose. Might be the first time out of the gate, or it might be the 100'th, or it might be the 1000'th - but the odds that you will win every single time start getting vanishingly small the more times you try it. You may have only a 1 in 6 chance of shooting yourself if you play Russian Roulette, but if you play a hundred times, I practically guarantee you're going to shoot yourself at least once - and like AIDS, once is all it takes to lose big ;)