[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.

[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…

[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.

[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.

[<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.

[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.

[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.

[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.