As an example of a probability misunderstanding, the book "Mathematics Statistics and Data Analysis" by John A. Rice gives the following quote from the Los Angeles Times:

"Several studies of sexual partners of people infected with the virus show that a single act of unprotected vaginal intercourse has a surprisingly low risk of infecting the uninfected partner--perhaps one in 100 to one in 1,000. For an average, consider the risk to be one in 500. If there are 100 acts of intercourse with an infected partner, the odds of infection increase to one in five.

Statistically, 500 acts of intercourse with one infected partner or 100 acts with five different infected partners lead to a 100% probability of infection (statistically, not necessarily in reality)."

The full article is here.

Rice explains that this is flawed by considering just two acts of intercourse: if we let $A_1$ denote the event that infection occurs on the first act and $A_2$ the event infection occurs on the second, then the event that infection occurs is $B = A_1\cup A_2$ and

$$P(B) = P(A_1) + P(A_2) - P(A_1\cap A_2) \leq P(A_1) + P(A_2) = {2 \over 500}$$

But I'm still confused: I understand that the above is trying to show that the article is overestimating, but surely the probability of $A_1 \cap A_2$ is $0$? You can only be infected with AIDS once... as far as I know. And even if it is not $0$, what's stopping us from exceeding the $P = 1$ threshold eventually if we take enough events (which clearly seems nonsensical since its intuitively its clear that someone might avoid AIDS indefinitely)?

Just to clarify, I'm not disagreeing that the article is flawed, I'm just struggling to locate the exact flaw. Many thanks.

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    $\begingroup$ You need to view the intersection in the absence of knowledge. So $A_1$ doesn't know the outcome of $A_2$ and vice versa. $\endgroup$ Dec 14 '19 at 17:37
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    $\begingroup$ There are different strains of HIV (AIDS is the disease caused by HIV), so it is indeed possible to be infected by HIV multiple times, and having AIDS twice over at the same time is even nastier than having it once. $\endgroup$
    – CJ Dennis
    Dec 15 '19 at 4:11
  • $\begingroup$ In the article author's mind, what is the probability of infection after 501 acts? $\endgroup$
    – svavil
    Dec 15 '19 at 7:21

The terminology is unfortunately misleading. The event of interest is not infection per se, but transmission of the virus. The distinction in language is subtle but important. An individual is considered infected if the virus is present, but a disease may be transmitted on more than one occasion. For a disease with effectively no cure, infection may be regarded as a permanent state--i.e., once infected, one remains so--but transmission is a person-to-person event that may occur any number of times. Therefore, the event that the author speaks of is not the former, but the latter, and $A_1$, $A_2$ should more precisely be regarded as transmission events.

With this clarification, it becomes obvious that if each sexual encounter carries an independent and identically distributed probability $p$ of resulting in transmission of the virus, then $$\Pr[A_1 \cup A_2] = \Pr[A_1] + \Pr[A_2] - \Pr[A_1 \cap A_2] = 2p - \Pr[A_1]\Pr[A_2] = 2p - p^2 < 2p.$$ If $\Pr[A_i] = p$ for each encounter $i$--that is to say, identically distributed but not necessarily independent--then $\Pr[A_1 \cap A_2]$ is not known, but as long as they are not mutually exclusive, the joint probability is strictly positive, thus the inequality stated by Rice holds. And this is where you had difficulty following due to his unfortunate use of terminology, because you correctly infer that if $A_i$ denote infection events, then this probability becomes zero, since $A_1 = 1$ implies $A_2 = 0$.

The natural extension of the iid model assumptions tells us that the outcome of each encounter is an independent and identically distributed Bernoulli random variable, say $$A_i \sim \operatorname{Bernoulli}(p), \quad \Pr[A_i = 1] = p, \quad \Pr[A_i = 0] = 1-p$$ where $A_i = 1$ if the outcome is transmission of the virus, and $A_i = 0$ if the outcome is non-transmission. For an individual with $n$ such encounters, we let $X$ be the number of transmission events experienced, hence $$X \sim \operatorname{Binomial}(n, p), \quad \Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x}, \quad 0, 1, \ldots, n,$$ and this individual is considered infected if $X \ge 1$. Consequently the probability that an individual remains uninfected after $n$ encounters is $$\Pr[X = 0] = (1-p)^n,$$ and the complementary probability--i.e., the individual becomes infected within $n$ encounters, is $$\Pr[X \ge 1] = 1 - \Pr[X = 0] = 1 - (1-p)^n.$$

The original article also makes some sweeping assumptions about transmission probabilities as well as an unwarranted assumption about the independence and person-to-person homogeneity of transmission events, but since we lack source data to model these in a more sophisticated manner, the best we can do is to note that the average of the claimed "number needed to transmit"--what the article calls "risk"--is inappropriately computed; it is not the simple arithmetic mean of $1/100$ and $1/1000$ as claimed. Rather, a geometric mean of these rates is a better estimate of the average rate of transmissions per encounter; i.e., $$p = \sqrt{10^{-2} \cdot 10^{-3}} \approx 0.00316228 \approx \frac{1}{316}.$$ The reason for this is left as an exercise for the reader.

As to the last part of the article quoted, "...100% probability of infection..." this is patently absurd. It is equivalent to saying that you are guaranteed to get at least one head if you flip a fair coin twice, because there are only two distinct outcomes for a coin toss. This illustrates the true nature of the extent of innumeracy of the article's author.


I'm not disagreeing that the article is flawed, I'm just struggling to locate the exact flaw.

The exact flaw is that the article thinks probabilities are additive. In reality they are not (unless the events are mutually exclusive).

For example, if you flip a coin, the probability of heads is 50%. The article would tell you that if you flip the coin twice the probability of getting heads at least once is 100%. In reality there's a 25% chance of getting 2 heads, 50% chance of getting 1 heads and 1 tails, and 25% chance of getting 0 heads. So the probability of at least one heads is 75%.


Let's start with something simpler. A fair, $n$-sided dice (equal probability outcomes in $\{1, \ldots, n\}$, each roll independent of the others) and the event that you see at least one roll with outcome $1$ in $k$ rolls. The chance of seeing a given value in one roll is $\frac{1}{n}$. The change of observing a roll after $k$ trials is not $\frac{k}{n}$, but $$ 1 - \left( \frac{n-1}{n} \right)^k.$$ You can derive this from the relationship $\bigcup_i A_i = \overline{\bigcap_i \bar{A_i}}.$

In the more specific example from the article, the $A_i$ represent that the act would cause an uninfected person to become infected.


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