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We know that every year has at least one Friday the 13th.

What about two Friday the 13ths, in a year? What is the probability of a year having two Friday the 13ths? (interesting subquestion here is if this probability is the same for leap and non-leap years)

Is it possible that a year has three Friday the 13ths? (after posting the question I discovered that it is, as said in comments)


This web page on wolfram MathWorld claims that a year has 1.72 Friday the 13ths on average. It also gives this table:

enter image description here

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    $\begingroup$ 2012 had three of them. $\endgroup$ – Brad Jul 6 '14 at 6:23
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    $\begingroup$ Why was the question downvoted? $\endgroup$ – sayantankhan Jul 6 '14 at 7:28
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Every day of the week (DotW) of any day in the calendar can be determined by two things:

  1. The DotW that December 31 falls on, and

  2. Whether the current year is a leap year or not, for dates in the first two months.

The problem then simply becomes a problem of determining how many Friday the 13ths occur given the DotW that December 31 falls on.

Let the numbers $0$ to $6 \pmod 7$ correspond to the days from $\text{Sunday}$ to $\text{Saturday}$.

There's a well known table of values $T = (0,3,3,6,1,4,6,2,5,0,3,5)$ for determining the DotW offsets of the first day of each month relative to December 31, starting from January. For leap years, the first two values are set to $6$ and $2$ respectively instead.

If we add $d + 12$ to each of these numbers (where $d$ is the DotW of December 31), then we get the DotW of the 13th day of each month. If this is equal to $5$ for any month, then that day is a Friday. Therefore, supposing that December 31 of any year fell on weekday $d$, then any month $m$ that satisfied $T_m + d + 12 \equiv 5 \pmod 7$ (or just $T_m \equiv -d \pmod 7$) would be a month with a Friday the 13th.

That being said, we can now make a table of values for the number of Friday the 13ths a year $F(d)$ will have given the day DotW $d$ that December 31 is on (as well as $F'(d)$ for leap years):

$$\begin{array}{c|c|c} d & F(d) & F'(d) \\ \hline 0 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 1 & 2 \\ 3 & 3 & 2 \\ 4 & 1 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{array}$$

At this point, we can make an estimate, based on the assumption that a year has an equal probability of being on any DotW and a $\frac 14$ probability of being a leap year, of the average number of Friday the 13ths a year could have:

$$p(L) = \frac 34 \left(\frac{12}{7}\right) + \frac 14 \left(\frac{12}{7}\right) = \frac {12}{7}$$

as well as the probability of each number occurring:

$$\begin{array}{c|c} F(d) & p(F(d)) \\ \hline 1 & \frac 37 \\ 2 & \frac 37 \\ 3 & \frac 17 \end{array}$$

The average is equal to $1.714285... \approx 1.72$, as the website you mentioned suggests.

However, because of a quirk in the Gregorian calendar by which there are only 97 leap years in a 400-year cycle, the days of the week don't cycle uniformly every 28 years, but every 400 years. This means that there's actually a frequency chart of the number of years a specific DotW will occur. Consulting a Dominical letter table reveals that each DotW has the following frequency $f(d)$ of December 31s that occur on it every 400 years:

$$\begin{array}{c|c|c} d & f(d) & f'(d) \\ \hline 0 & 43 & 13 \\ 1 & 43 & 15 \\ 2 & 44 & 13 \\ 3 & 43 & 14 \\ 4 & 44 & 14 \\ 5 & 43 & 13 \\ 6 & 43 & 15 \end{array}$$

If we multiply these numbers with the number of Friday the 13ths that occur on that type of year, we get the total number of Friday the 13ths that occur in a 400-year cycle, which is $685$. Therefore, a year has exactly $\frac{685}{400} = 1.7125$ Friday the 13ths on average.

We can also find how many years have two or three Friday the 13ths with this information – simply add up the $f(d)$ values that correspond to an $F(d)$ value of 2 or 3:

$$\begin{array}{c|c} F(d) & f(d) \\ \hline 1 & 173 \\ 2 & 169 \\ 3 & 58 \end{array}$$

So a year has a $\frac{173}{400} = 0.4325$ probability of having just one Friday the 13th, a $\frac{169}{400} = 0.4225$ probability of having two, and a $\frac{58}{400} = 0.145$ probability of having three.

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  • $\begingroup$ The assumption that Dec 31 has equal probability to be any of the seven days of the week is not quite true. I'm sure the calculations of those probabilities exist online somewhere. For a summary table, see here. $\endgroup$ – alex.jordan Jul 6 '14 at 7:50
  • $\begingroup$ I think you address this after a first run-through though, so never mind. $\endgroup$ – alex.jordan Jul 6 '14 at 7:58
  • $\begingroup$ Great anwer, @Joe, but hiw come the site from my question says there are 688 Friday the 13ths in 400 years, ans you say there are 685? $\endgroup$ – VividD Jul 6 '14 at 11:07
  • $\begingroup$ Maybe I have an off-by-1 error somewhere in there. $\endgroup$ – Joe Z. Jul 6 '14 at 20:03

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