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Recently, we looked at the extent to which incomplete infection figures in the Corona pandemic can be compared with warranty data for vehicle components. Today, we use statistical quality control to reject some conspiracy theories surrounding Covid-19.
A tweet by Florian Aigner sums up the current findings about the true background of the Corona virus very well:
Translation: »China accidentally released a virus in an experiment, but the virus doesn't exist, but Bill Gates invented the virus, but in fact it's 5G networks, but politicians worldwide are under one roof, but they can agree never.«
These conspiracy theories can be approached in terms of content or psychology. And, new: you can approach them statistically. This requires some tools that we usually use in statistical quality control at ITWM.
We pose ourselves next to an assembly line where breathing masks are produced. A certain amount of rejected items is tolerated. For example, as long as 99 percent of the masks are in order, we consider the production to be good. In quality control, these 99 percent must be checked regularly, for which some masks are taken randomly to be tested. The question is: how many masks do you need to test to verify the claimed 99 percent?
To do this, you need a basic course in probability to know the binomial distribution. Let us shift our problem to the lottery booth. There, »Every tenth lot wins!« is advertised. How likely is a win if we draw exactly ten tickets? Think about it briefly; the answer will follow.
Before that, two comments: On the one hand, the usual result of the individual mask test is a success (99 out of 100) in quality control, while the usual result for the lottery booth is a failure (nine out of ten). In probability theory, one always speaks of probability of success, for example, the failure probability of 90 percent or the probability of winning of 100 percent. »Success« is not meant literally here, but means »the result on which we focus« has occurred.
On the other hand, quality control is often calculated backwards. While our lottery question was »If the probability of success is ten percent, how likely is a win among ten tickets?« quality control asks rather »We have not found a defective mask among 150 tested. What is the probability of success then at least?« For our application to the conspiracy theories, the forward thinking is sufficient.
So what is the probability of winning from ten tickets? Let's start with the marginal cases: not a single win and not a single failure.
Each ticket is independent of the previous one (at least if there is large number of tickets, otherwise one must strive for the so-called hypergeometric distribution). The chance of winning at e.g. ticket number ten is thus exactly the same as for tickets one to nine. The probabilities simply multiply:
What is the chance of (exactly) one winning ticket? To answer this, we need ten factors again. One of them is a 10 percent probability of winning, the remaining nine are the failure probability of 90 percent:
10 percent x 90 percent x ... x 90 percent = 3.87 percent
This is not yet the probability you are looking for. It is now necessary to take into account that there are ten positions on which the winning ticket appears. Therefore:
P (exactly one winning ticket) = 10 x 3.87 percent = 38.7 percent
In order to calculate, for example, »exactly two winning tickets« some combinatorics is necessary. You need to know how many ways you can position two winning tickets in ten tickets. This is then done by the binomial distribution for one. Fortunately, we can take a shortcut here. If we are not interested in exactly one, but at least one profit, this is exactly the opposite of »no profit«:
P (at least one profit) = 100 percent - P (no profit) = 100 percent - 34.86 percent = 65.14 percent
Let us sum up briefly. Only profits almost never happen. In about 35 percent of cases we only have failures, and in 39 percent of cases exactly one winning ticket. In the remaining 26 percent of cases, there are two, three or more wins. At least one win is made in 65.14 percent of cases.
There is a well-known article by David Grimes, published in 2016 in the journal »Plos One«. Using the Poisson distribution, he calculates how likely it is to plunder a conspiracy over time, or what the average life time of a secret would be before it flies. For the sake of simplicity, we leave the temporal component behind and ask: How likely is it that the secret will remain one for, say, five years?
In order to do this, one has to consider how likely it is that the individual accomplice will not talk anything out. Suppose each individual manages this with a 99.99 percent probability (i.e. one person of ten thousand chats). The following considerations may be made by the inclined reader for other probabilities.
Let us start with the WHO and large pharmaceutical companies wanting to do us some kind of evil with vaccinations. Grimes estimates that about 730,000 people must be involved in the conspiracy. So we have to multiply 99.99 percent 730,000 times with ourselves. Let's make it short:
The table shows that it is impossible to keep the secret for five years as early as 100,000 people (with the hint that the five years here were chosen arbitrarily). But at least if only 100 people knew about it, it would remain secret.
Even if we better protect the secret in the individual and assume 99.999 percent security (one accomplice out of one hundred thousand leaks) it usually does not remain secret with 100,000 accomplices, with one million accomplices in no way:
Bottom line: You can immediately discard any conspiracy theories that are said to have kept a secret from more than 100,000 accomplices who have been with knowledge over a few years. These include, for example, moon landing, climate change lie and vaccine fraud.
Smaller conspiracy theories (among 100,000 people) you would have to discard on the content level.
Fraunhofer Institute for Industrial Mathematics ITWM