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.

[general_re]

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.

There is a "sweet spot", though, where the expected payout becomes large enough relative to the initial outlay that it starts making sense to think about really putting money into significantly increasing your odds. If I have a 1 in 500,000,000 chance at winning $2 billion, and each ticket costs $1, then spending $1 billion on tickets gets me an 86.5% chance of doubling my money. Not that you ever see such situations in real lotteries - the possibility of multiple winners splitting the jackpot complicates things somewhat.

And of course, spending $1,999,999,999 on tickets will give you a 98.2% chance of profiting by a dollar ;)

[general_re]

Note that if there were only 100 numbers in the lottery, then you would win with certainty in case A)

If you make sure that you cover all 100 possible numbers - if your tickets are randomly generated, there's a fair chance that you'll omit at least one number between 1 and 100, and thereby lower your chances to something somewhat less than 100% ;)

[CatoRenasci]

True, I would venture to say that most modern liberals - those who aren't actually Leninists - mean well, in the sense that they would like to see less suffering in the world and better conditions for everyone. The failures of their policies in almost every instance are a wonderful (terrible?) example of the law of unintended consequences. Virtually every policy touted as "liberal" that is intended to "help" actually makes things worse.

The only exceptions (and there are many issues to be considered with them) are the massive public works projects undertaken primarily for other reasons.

[r9etb]

The underlying problem, simply stated, is this:

Suppose I have a coin that has a probability of 0.006 of coming up "heads" on any given toss.

Assume that each toss is independent of any other toss. (I.e., the sexual episodes occur completely randomly, and there are no repeat partners.)

If I toss the coin 520 times, what is the probability that at least one of those tosses comes up "heads?"

Thus, I believe your approach (if not necessarily the probabilities) is sound.

Note, however, that the problem gets a lot hairier if you start assuming repeat partners.

[oh8eleven]

Thanks, but I think you missed the </sarcasm>

[monkey]

I've never played the lottery. Isn't random generation just an option? I always hear about winners "lucky numbers". The point is that you could guarantee a win if you so chose. Of course, in option B), you might win multiple times.

You may have heard about the (Australians?) who bought massive numbers of tickets in lotteries that had accrued big jackpots (and positive expected gains). I'm sure they weren't generating tickets randomly (and I'd hate to get behind them at the 7-11).

[r9etb]

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.

The lottery ticket example is actually counter to what you're saying: If I buy 20+ lottery tickets, my chances of winning the lottery are actually improved -- it's just that the probability of any particular ticket winning is very small, so you'd have to buy a lot of different in order to have a good chance of winning. Note that if you bought half of the possible lottery combinations, your probability of winning would be 50%.

Likewise, when you roll the dice, the chance of getting double sixes on any given roll is the same (1/36). However, the probability that you will eventually get double sixes increases as you continue to roll the dice.

[general_re]

I've never played the lottery. Isn't random generation just an option?

Yes it is, and I just realized that it applies to the example I gave in #81 above. If there's 500,000,000 possible combinations of tickets for a $2,000,000,000 jackpot, you can guarantee a win simply by spending the $500,000,000 on all possible combinations.

Again, though, things are greatly complicated by the fact that there's a good chance of multiple winners who have to split the jackpot. If I spend the $500,000,000 million to guarantee a win, but so do 9 other people, we all end up losing by getting only $200,000,000 in return. We have to keep in mind that, while the odds that a particular person will win the lottery are pretty long, the odds that someone will win the lottery are actually pretty good, and more than one person winning is also a reasonable possibility.

[r9etb]

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.

You're wrong.

Take the coin toss example.

For any given toss, the probability of getting "heads" on that toss is constant, at 0.5.

But we're not talking about individual coin tosses. We're talking about the probability of getting "heads" at least once during a series of coint tosses. Quite obviously the probability of getting at least one "heads" in 10 tosses is much greater than 0.5.

Poke here for a discussion of the topic.

[r9etb]

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.

Got carried away by the beauty of the Central Limit Theorem, no doubt.

[Doctor Stochastic]

The reason to look at cumulative risk is that the consequences of a "success" (in the case of AIDS) is of more importance than the consequence of a "failure."

[r9etb]

And of course, spending $1,999,999,999 on tickets will give you a 98.2% chance of profiting by a dollar ;)

Always neglecting taxes and present value, of course. If you win that $2 billion jackpot and take the lump sum, your net take-home will only be about 40% of the total -- you've just bankrupted yourself. ;-)

[Tauzero]

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

Yes, it would.

But with the lottery it's not that rapid.

With a one in one million chance of winning (odds are typically much worse), 3 million tickets will get you a 95% chance of winning the jackpot. At a buck a ticket, the jackpot better be bigger than $3 million.

Of course it seldom is. The house sets the rules so it makes a net profit, on average. Every once in a while an opportunity comes up that makes buying lots of tickets not such a horrible idea, and people have indeed tried to take advantage of that. But it takes a lot of money, and a lot of hands.

[monkey]

multiple winners who have to split the jackpot

Consider the following simple game. Three numbers are in the lottery. I pay a dollar each for 1, 2, 3. You pay a dollar for (wlog) 3. No middleman - winner take all, multiple winners split.

Your expected return is 1/3 * 2 dollars = $.67, mine is $1.33. Most people think the expected return for the person betting a buck is $1, i.e., it is a fair game. You can win a little money with games like this, and a lot of money if you can figure out how to apply the game outside of a drinking establishment.

[Mr. Bird]

You're wrong.

Thanks, I figured that out about 50 posts back. :^)

[MainFrame65]

I don't have specific numbers, but I believe that there are some considerations that argue against some of the numbers used here.

1. Gender. Heterosexual, vaginal intercourse presents a much higher risk for male to female transmission than the opposite. Magic Johnson was extremely promiscuous for a long time without infection - until even he finally beat the long odds against catching it from an insfected female partner.

2. Promiscuity. Some condemn it, some practice it, and some just dream about it. Your risk depends not just on the number of your partners, but what their sexual practices and proclivities are. I am sure that there are prostitutes who insist on condoms, forbid risky acts, never use drugs, inspect their clients as carefully as possible under the circumstances, and have themselves tested regularly, but I don't think that is the norm. On the other hand, I'm sure some pretty wild girls (and boys) show up at church picnics. But I'm sure you would agree that infection rates are higher among the kind of partners easily recruited for one-night stands, so I think that statistics about the general population don't adequately define the risks.

3. Circumcision. The rate of HIV infection in Africa is significantly higher in areas where males are NOT circumcised than where they are. The nature of the connection is not known. It could be due to unrelated (but parallel) social norms, or to higher rates of other venereal diseases in uncircumcised males.

4. Risk tolerance and risky behaviors. Visit a casino, and notice that the majority of the gamblers both smoke and drink, although all of these are minority behaviors. These are all minor in terms of immediate hazard, but the fact that they are found clustered together is no accident. Drug use, promiscuity, and "taking a chance", or feeling lucky, tend to cluster as well. In some countries, an overwhelming majority of the prostitutes are infected with one or more venereal diseases, and most large cities have red-light areas where this is the norm. Available women all had their first sexual experience sometime, but you are not likely to be so fortunate.

I have a close relative, a doctor, who was infected with hepatitis C as a result of a broken glove that exposed him to a patient's blood. This happened several years ago, and he still carries the virus, which can be transmitted by intercourse as well as any blood contact. After much soul searching and discussion, his wife decided that they would resume having intercourse, using condoms, and being careful not to engage in anything that might cause any injury or irritation to tissue. He has to be careful around his son, as well, to avoid any blood exposure. So far, so good. But they both know that some virus CAN penetrate the condom, and risk is not absent.

[samtheman]

oops

But the concept is still right.

That was a typo.

[Lonesome in Massachussets]

This is an extremely simple "math" problem. Please don't blame Gauss (inventor of the normal or Gaussian distribution, the central limit theorem and all else which is good in life.) The odds of catching aids are one minus the odds of not catching it. The sequential probability, (independent trials, ect.) would be:

let Po = odds of contracting aids in any one week

Probability of not catching aids in one week = 1 - Po

Probability of not catching aids in N weeks = (1 - Po)^N

Probability of catching aids in N weeks = 1 - (1 - Po)^N

If Po = 0.006 / week and N = 520 weeks,

Probability of catching aids in 10 years = 1 - (1 - 0.006)^520 = 0.95625586


You heard it here first. Your model is overly simplisitic. The trials aren't independent - same partner on a series of occassions, lot failure rates of condoms, etc. Nothing is ever that simple.

In order to make any useful conclusions one would have to know more about the actual mechanisms leading the spread of the disease. I think it is safe to say that limiting the number of ones sex partners is probably a good idea if the object is prevent contagion. Condom giveaways may encourage risky behavior, if that's your point and actually lead to more contagion. Just hard to prove.

[Lonesome in Massachussets]

I get 1 - (1 - 0.006)^100 = 0.4522.

[NathanR]

"So, how does this really work? Thanks!"

Let P be the population of people you are talking about, say all gay males in San Francisco.

Let A be the percent of people with AIDS in this poplulation (expressed as a number between O and 1).

Let C be the condom failure rate.

Let S be the probablility that an uninfected person will get AIDS with unprotected sex with someone with AIDS.

The probability that one will get AIDS from one sex act with anyone in the population using a condom is A*C*S and the probability that one willn not get AIDS is 1 - A*C*S. Therefore, the probability that one will not get AIDS after n sex acts is (1 - A*C*S)**n (where * means multiplication and ** means to the power of).

That was probability and was the easy part. The hard part is determining what A and C and S *are*. That is statistics.

But, let us say that A is .1 or 10%, a reasonable guess in the above population, that C is .3, and that S is .1. Then A*C*S is .003 and 1 - A*C*S is .997. After 100 random sex acts, the probablity of not getting AIDS is .741, and gets worse with more sex.