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.

[Saundra Duffy]

This is sick. Math gives me a headache.

[monkey]

For grins, assume the subject sleeps with one partner the whole time (let's say he doesn't know if the partner has AIDS - there is a 20% chance, and all your other assumptions are in place).

Then, if the partner has AIDS, the chances of contracting it would be 0.99999986777857955389708680154366, zero otherwise.

So the overall chance would be .2*1 (rounding) + .8 (0) = .2

The "moral" of the story (subject to the assumptions): you're better off sleeping with one person, even if you don't know if they have AIDS, than lots of people, some of whom have AIDS and some of whom don't.

[PMCarey]

It seems to me that your friend knows more statistics than he or she can handle. I'm not sure why the Normal distribution gets tossed into this discussion when we're really looking at the Binomial distribution or more accurately the Negative Binomial. As I understand you're looking for the probability of "being lucky" throughout a sequence of events and your friend has confused that with the idea of "being lucky" in the next event which is a totally different thing.

So IF all of your assumptions and probabilities are correct, your cumulative probability value is also correct.

[CatoRenasci]

No, it was a failure of nerve in the public health officials in California (San Francisco and LA) and New York, especially in San Francisico, who didn't want to alienate gays. It should have been emphasized from the beginning that this disease was spread by voluntary behavior and could be stopped if the behavior stopped and the infected were removed from the population as soon as they were discovered.

[Tauzero]

"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's not, of course. That's not what the cumulative risk equation says.

Wouldn't you agree the odds of living through 100 rounds of Russian roulette are lower than the odds of living through 1 round? The odds of the 100th round, though, are not affected by the result of the first 99. If they were, then you'd get a different equation for the cumulative risk!

[Tauzero]

You are right. He's a doofus who knows just enough math to hurt himself.

[Mr. Bird]

I intuitively understand, but it's still a little shaky. With such a rapid approach to 100%, how many times should I play the lottery? The same principle would apply, would it not?

[general_re]

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.

Indeed - the problem being, of course, that people don't tend to think of it that way.

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.

I agree again, and I say that as someone who tends not to be on the strict social conservative end of the political axis - I have no personal objection to homosexuality per se, as I think many people also do not, but the fact that this infection hits a particular subpopulation harder than others cannot be used to justify the risk to the population as a whole. Unfortunately, the fact that this subpopulation is a fairly vocal minority made quarantines politically impossible, to the detriment of society at large. I think people were also further lulled into complacency by the exceptionally long period of time between infection and the onset of symptoms for HIV - if HIV caused death within 72 hours of infection, I think we would have seen quarantines tout suite, homosexuals or no.

[Doctor Stochastic]

Corrcect. You have 1/70,000,000 (roughly) chance to win $10,000,000. Do the numbers. Don't forget to add the $1 cost of a ticket.

Remember Hollywood Henderson.

[Mr. Bird]

Do the numbers

I was hoping someone else would do that for me! LOL.

[CatoRenasci]

Agree as to all points. Thinking in expected value terms, for me however, is one of the by-products of a couple of years doing graduate work in mathematical economics.

[Doctor Stochastic]

There were political problems. Congressperson Henry Waxman worked hard to keep the bath houses open even though the epidemiology was known. Unfortunately, the politics led to far more suffering and deaths than was necessary. Most of the deaths hit the community that Waxman was supposedly helping.

[CatoRenasci]

Bingo!

[general_re]

With such a rapid approach to 100%, how many times should I play the lottery? The same principle would apply, would it not?

Yes, but how rapidly the odds of winning or losing approach 100% depends on the initial probability and how many times you try. If your odds of winning on a single lottery ticket are 1 in 500,000,000 (i.e., you have a 99.9999998% chance that your ticket is a loser), then buying 100 tickets doesn't improve your odds very much - the odds jump all the way from the 99.9999998% odds for one ticket being a loser up to a 99.999980000002% chance that all 100 are losers. Not much of an improvement. If, on the other hand, you bought 20,000,000 tickets, your odds that they would all be losers is about 96%, giving you about a 4% chance of winning ;)

[monkey]

When the probability of success is very low, like in the lottery, your chance of success improves by a multiplicative factor, e.g., 3 tickets are 3 times as likely to win as 1 ticket. That is not perfectly accurate, but it is so close that calculating it exactly isn't worth it (unless you are planning on buying a very large number of tickets).

[general_re]

Most of the deaths hit the community that Waxman was supposedly helping.

With any luck, a generalized form of that statement will serve as the epitaph of modern liberalism, given that it seems to fit regardless of which favored minority we're discussing - "They killed the people they wanted to help".

[Porterville]

There are different strains on the different continents and sub continents. Some strains are more insidious then the others. Some are transferred more easily through mucus membranes while others are generally transferred by the blood (the strain we have in the USA)

Certain ethnicities are predisposed to transmission because of epidemiological history of race....i.e. the black plague in Europe had an immunity effect that built in a level of resistance.

[Mr. Bird]

I was thinking more in terms of playing successive lotteries. For example, are you more, less, or equally likely to win the lottery if you A) buy 100 tickets on one lottery, or B) buy one ticket in 100 successive lotteries? Just curious....

[Triple]

Your model would be better if you added rate of infection given exposure (maybe 5%) and controlled for exposure potential - (it should include number of different partners, not just number of vile acts.)

PS - Health professionals record diligently 'sticks', when they prick themselves with a needle. The infection rate following an AIDS stick is thankfully low, @5%-10%.

[monkey]

You are more likely to win in case A), but the difference is trivial (unless you buy a lot of tickets, relative to the size of the lottery).

Note that if there were only 100 numbers in the lottery, then you would win with certainty in case A), but not case B).