# Chances of being ill on Fridays

Background: My son has been ill a few times this academic year and has missed some school. The authorities get involved when attendence is low, and they're basically suggesting he's malingering, which he's not.

So one of the things they're implying is that he's been off a suspicious number of Fridays. I'd like to produce some stats to show it's just a statistical fluke

The problem statement is something like:

Given a student could have attended school on N days, but was absent through illness on M of those days, what's the probability that they were absent for more than P Fridays.

As a follow up, what's the change that they were absent for more than P days for any one of the days of the week, not necessarily Friday.

Pointers to how to work this out for myself also welcomed.

• Is it a theoretical question, making all kind of unrealistic assumptions like uniform distribution, independence, illnesses that can long for any random number of days? Or maybe you want a realistic estimate? If the second, we probably don't have enough data as this will depend on many factors. – Tim Mar 22 '18 at 19:24
• You can't assume indendence and uniformity, unfortunately. In fact, it would have not been in your favor. To me it's likely that on Fri kids have more chance to get sick than on any other day for several reasons. For instance, they're more tired after school week, or they've been exposed to more sick people during the week than during the weekend etc. So, you need to understand the weekly and seasonal patterns in absence rates, then somehow apply them to your son. What fs he has sinus and more likely to catch cold than others, and it's probably going to peak on Fri – Aksakal Mar 22 '18 at 19:28
• Obligatory Dilbert cartoon. – Stephan Kolassa Mar 22 '18 at 19:53
• Perhaps the right way to frame the question is to ask "what is the chance that the authorities in this school, should they choose to look, would notice some anomalous imbalance in absences of some sort--by day of week, month, proximity to holidays, or whatever--among one or more students." The answer is likely almost 100%. It's not obtained through a simple model, simulation, or chi-squared test, but rather by careful consideration of the many processes that lead people to notice coincidences. See David Hand, The Improbability Principle, for a recent account. – whuber Mar 22 '18 at 21:05
• If this were to escalate into an official investigation by the authorities, my advice would be to leave the burden of proof with the accuser. Let them prove that there is "a pattern" and that it is anomalous. – Jim Mar 22 '18 at 21:54

You can evaluate this by running a chi squared or Fisher's exact test on a 2x2 contingency table. Make the columns of the table "Sick" and "Not sick" and the rows "Friday" and "Not Friday". Now fill in the table by counting the number of instances of each case - Sick+Friday, Sick+Not Friday, Not sick+Friday, Not sick+Not Friday. Run a chi squared test on this table, and it will tell you if it's significantly different from the expected distribution (which would be that 20% of sicks days fall on Friday).

• You need data from other students. I think the OP would be well in line to request said data from the school, since they have resources for measuring such things. – AdamO Mar 22 '18 at 21:41

The crudest way I can think of it is this. Certainly not realistic, but pure combinatorial reasoning is sometimes insightful and might even provide a reasonable approximation. Caveat: two consecutive days of sickness count as two separate illnesses.

You are picking $M$ days out of $N$, so you have $N\choose{M}$ possibilities. Out of these, you are interested in those in which $P$ are Fridays, which are $N/5$ (assuming for simplicity the $N$ days span full weeks only).

To count how many cases satisfy your conditions, you can pick the $P$ Fridays in advance, and then pick the rest sick days among the non-Friday days that are left to choose from.

So you have ${N/5}\choose{P}$ Friday configurations, and ${4N/5}\choose{M-P}$ remaining days to pick.

This works out to a probability of $$\frac{{{N/5}\choose{P}} {{4N/5}\choose{M-P}}}{N\choose{M}}$$

This is the chance the poor thing will get sick on $P$ Fridays. You can now just sum over all values at least $P$ to get the desired probability.

$$\sum_{i=P}^M\frac{{{N/5}\choose{i}} {{4N/5}\choose{M-i}}}{N\choose{M}}$$