$\bullet$ Prove that the $13$th day of the month is more likely to be a Friday than any other day of the week.
$\square$ This problem has several number theoretic and one probabilistic solutions in this website but from none of them I can understand how to calculate the number of Fridays. This website mentions about number of $13$th days of some months out of $146097$ days but the process of calculation is not clear to me.
$\square$ Here is my procedure to prove this problem and I am not sure if this approach is appropriate. If we do not consider the whole Gregorian Calendar and just look at some facts that
$\circ$ If the $13$th day of the month is more likely to be a Friday then the month has to be started at Sunday
$\circ$ If we compare the months and days of consecutive common years and a leap year then we have the required Friday which are the $13$th day of each of the months. So, in this way, if three consecutive common years start on a Sunday, then Monday and then Tuesday and if the leap year starts on a Wednesday then we have total $8$ Fridays, which are the $13$th day of some months. Similarly with three consecutive common years starting on a Monday, then Tuesday then Wednesday and a leap year starting on Thursday we get total $7$ Fridays, which are the $13$th day of some months. So from, $$\mathrm{\{(Y_{Sun}+Y_{Mon}+Y_{Tue}+L_{Wed}) \cup (Y_{Mon}+Y_{Tue}+Y_{Wed}+L_{Thu}) \cup (Y_{Tue}+Y_{Wed}+Y_{Thu}+L_{Fri})\cup (Y_{Wed}+Y_{Thu}+Y_{Fri}+L_{Sat})\cup (Y_{Thu}+Y_{Fri}+Y_{Sat}+L_{Sun})\cup (Y_{Fri}+Y_{Sat}+Y_{Sun}+L_{Mon})\cup (Y_{Sat}+Y_{Sun}+Y_{Mon}+L_{Tue})\}\\ = 8+7+7+6+8+6+6\\=48}$$
Here $Y_{Sun}$ means common year starting on a Sunday and $L_{Sun}$ means leap-year starting on Sunday. Hence, we have total $48$ Fridays out of total $3\times365+366=1461$ days.
$\triangle $ Question :
$\bullet$ If my procedure is not correct, could you please explain the correct way of the calculation kindly?
$\bullet$ If we have to follow the Gregorian Calendar, then would you please explain how to calculate the number of $13$th days presented in this website ?
Any help is valuable and highly appreciated.
Best Answer
I think your conclusion is not entirely correct; we looked at a period of 28 years which has $7 \cdot 1461$ days. Furthermore, the amount of 13th days in this time frame is only $28 \cdot 12 = 336$. So, the probability that Friday the 13th occurs would be $$\frac{48}{336} = \frac 1 7.$$ We can also see this without calculations, but by symmetry of your argument. Let $Y_{\text{Mon}}$ be the amount of Friday the 13ths in a common year that starts on a Monday, and let $L_{\text{Mon}}$ be the amount of Friday the 13ths in a leap year that starts on a Monday (and similarly for the other days). As you noticed, we indeed have $$\text{# Friday the 13th in $28$ years} = 3 (Y_{\text{Mon}} + Y_{\text{Tue}} + Y_{\text{Wed}} + Y_{\text{Thu}} + Y_{\text{Fri}} + Y_{\text{Sat}} + Y_{\text{Sun}}) + L_{\text{Mon}} + L_{\text{Tue}} + L_{\text{Wed}} + L_{\text{Thu}} + L_{\text{Fri}} + L_{\text{Sat}} + L_{\text{Sun}}.$$ However, if we would look at any other day (for example Thursday the 13th) and denote those variables by $Y'$ and $L'$, we would observe that for example $Y'_{\text{Mon}} = Y_{\text{Tue}}$. We only permute the $Y$ values and permute the $L$ values, so the summation stated above does not change. Hence, the probability that the 13th of the month occurs on some day is the same for all days.
The reason why this symmetry argument works, is because our calendar repeats each four years, but we cannot divide the amount of days in four years (1461) by 7. This means that the next period of four days starts at another day of the week. For example because 7 is prime, it means that we will only end up in exactly the same situation after 28 years.
Now let's take the special cases of the Gregorian Calendar into account. It repeats every 400 years (a year that is divisible by 100 is not a leap year, except if it is also divisible by 400). In that time frame, we have 303 common years and 97 leap years, which adds up to 146097 days which actually is divisible by 7! That does not directly imply that the probability of Friday the 13th is larger, but it does imply that our argument by symmetry no longer works. If you work out the summation for a period of 400 years, I would not be surprised if you would indeed observe that the 13th of a month is more likely to be on a Friday.