Using Statistics to Better Understand Reality

About Lottery Booths, Donald Duck and Mutated Corona Viruses

Streuspanne – Unser Blog zu Statistik und Jurojin / January 14, 2021

Just in time at the end of 2020, we were surprised to learn that a mutation of the Corona virus is circulating in England. It is not said to be more deadly, but more contagious. Directly, neighboring countries have expressed concern that the mutation could also reach them. So let's ask ourselves today: can we tell from the reported infection numbers if something like this has happened?

A blogpost by Jochen Fiedler and Sascha Feth.

We start with a harmless everyday example. Everyday life as it was before the pandemic: We visit a county fair. There are two lottery booths. The first, run by Gustav, is known for its higher chance of winning. The second, run by Donald, lets you win much less often. We now send 100 people to the fair and ask them at the end how many winnings they drew. Can we tell from this how many people were on Gustav?

Using School Mathematics for One Opinion

Suppose that at Donald's lottery booth five percent of all tickets are a win, while at Gustav's it is 20 percent. Now the 100 people come back and bring us 16 winnings. What do you guess: How many were at Donald's, how many at Gustav's?

Let's try some school math first: If everyone had been at Donald´s, we would have expected five wins. If they had all been at Gustav´s, we would have expected 20 wins. Now our result is in between with 16 wins. A little bit lazy we guess: There were 50 persons at Donald´s and 50 persons at Gustav. That would lead to the following expectation of wins:

50*5% + 50*20% = 12.5

This is still a little smaller than our observed 16. Presumably, therefore, a little more than half of the people were at Gustav´s. With some school mathematics we can save the trying and calculate explicitly: If x persons were at Gustav´s, we have the following expectation about the number of wins:

(100 – x)*5% + x*20% = 5 + x*15%

Setting this equal to the observed 16 wins, we get for x:

5 + x*15% = 16

x = 11/0,15 = 73.33

This is also called the maximum likelihood estimator. The result says that the most plausible explanation for our 16 observed wins is when about 73 people were at Gustav´s and 27 people were at Donald´s. So far, so plausible. So let's prepare the press release »Most people buy lottery tickets at Gustav's«.

With Confidences to the Other Opinion

How reliable is this result? Do the numbers really prove that 73 people were at Gustav's? For this we have to work with confidences or with hypotheses. We have encountered both concepts several times on the blog. Unfortunately, they are not concepts that fall into your lap, so we want to approach them again slowly.

Could it be that no one was actually at Gustav's? Or to put it another way: How likely is it that 100 people at Donald's will ensure 16 wins?

This is still relatively easy to answer if you know the binomial distribution. Here you only need some combinatorics and patience. For example, how likely are ten wins out of 100 tries at Donald? Answer:

(100 over 10) * 5% ^10 * 95%^90 ~ 0.0167

To understand this, it is best to imagine a tree diagram. There are 100 steps and on each step you turn left to a loss or right to a win. So, to get past ten wins, we have to turn right ten times (each with five percent probability) and right 90 times (each with 95 percent probability). Now we have to figure out how to distribute ten such turns to the right over 100 turn decisions. This is what the binomial coefficients have done: (100 over 10) = 17.310.309.456.440, or about 17 trillion possibilities.

If we calculate in this way how probable 16 or more wins are at Donald´s, we get about 0.0037 percent as an answer. So it is almost impossible that nobody was at Gustav´s. If we now calculate a confidence interval of how sure we are about the 73 persons, we get with 95 percent confidence the statement that there were at least 39, with 99 percent confidence even only 27 persons.

No, we won't even try to explain this methodology in a box. Instead, here's our post about confidences again: When do you say confidence, when do you say significance?[only available in German]

Let's draw an interim conclusion: Probably the majority of the people (73 out of 100) were at Gustav's lottery booth. However, we should not vouch for it. After all, taking into account the uncertainty due to the sample size, we can only say that at least 27 or 39 people were at Gustav's (depending on whether you prefer to work with 95 percent or 99 percent confidence). The majority ratios are tipped here! Now our example was deliberately generous in that we put a factor of four between the winning probabilities. Now imagine if there had only been a factor of 1.7 between them. Now Donald would still have a 5 percent probability of winning, Gustav 8.5 percent, and we would probably have observed about 6-7 wins among 100 people. You guess right: Now it is hardly possible to say anything with confidence. It could have been everyone at Gustav´s as well as everyone at Donald´s. The sample is simply too small to determine this reliably.

Back to Corona

Transferred to Corona, pulling at the lottery booth corresponds to a social contact. There is also a certain probability of infection. This probability is 1.7 times higher in the mutated variant. The number of reported infections of a new day then corresponds to the number of winnings. Since this is currently unfortunately always several thousand, we have here also the chance to resolve smaller differences. But wait: There are two dark figures!

On the one hand we do not know all infections. This is just the dark figure problem from our first blogposts about Corona. Here we can make approximately estimates to know roughly the number of gains or infections. Second, we have always assumed that we know the exact number of people who bought lottery tickets at the fair. Transferred to Corona, these ticket buyers correspond to those who had contact with infected people. Unfortunately, we do not know the exact number and its evolution, so changing infection numbers can be explained purely by this dynamic. Both dark figures together result in an uncertainty that again makes it impossible for us to resolve to a factor of 1.7. This problem is made more difficult when events such as Christmas or New Year's Eve fall within the observation period.

Conclusion

Even under very well controlled conditions, as in the example of the fair, we can tell little about spreads of various mutations from reported infection numbers alone. The reason is that we do not know the number of contacts with infected persons well enough. This can only be done relatively well when the new variant has already become very widespread and the corresponding effects are immediately apparent – which seems to have been the case in the UK.

In order to detect a much more infectious mutation as early as possible, good monitoring of all mutations is needed across the board.

Blog Entry

Using Statistics to Better Understand Reality

About Lottery Booths, Donald Duck and Mutated Corona Viruses

Using School Mathematics for One Opinion

Further Information

With Confidences to the Other Opinion

Back to Corona

Conclusion

Contact Press / Media

Dr. Sascha Feth

Contact Press / Media

Dr. Jochen Fiedler