After a discussion with some colleagues on Friday 13th December 2002, I did the calculations reported here. Of course, there is nothing special about Friday 13th, but this date just happens to be of more interest to some people than any other combination; obviously, the results apply (with minor changes) to any other combination of day and date.

Intuitively, one might expect that Friday 13s are equally spread over all months, and that on average a year has 12/7=1.71428... such days (a year has 12 months, and each month has a one-in-seven chance that its 13th day is a Friday).

However, this turns out not to be true. Friday 13 is least common in August and October, and an average year has 1.72 Friday 13s, slightly more than expected. If you want to know why, read on.

(A common misconception is that Friday 13th should be more likely in months with more days (like January with 31 days) than in shorter months. However, this is incorrect, since every month has precisely one 13th day.)

First of all, a standard calender is rather inconvenient from a
mathematical point of view because of the possible leap day at February 29th.
Therefore, we redefine January and February to be
the months *Undecember* and *Dodecember* of the *previous*
year, so a year runs from March through Dodecember and the leap day is
between the years.

Since there's no leap day in the redefined year anymore, we only need to determine what weekday the first day of the year (March 1st) is; if we know that, then all other weekdays of the year are fixed. We will indicate this first weekday of the year by the number X, with X running from 0 through 6; 0 is Wednesday, 1 is Thursday, 2 is Friday, and so on, up to 6 being Tuesday.

Given this, one can easily construct the following table, which tells for every possible value of X, what months will have a Friday 13th:

X | Months with Friday 13th |
---|---|

0 | October |

1 | April, July |

2 | September, December |

3 | June, Dodecember (= February of next year) |

4 | March, November |

5 | August |

6 | May, Undecember (= January of next year) |

For example, March 1st 2002 is a Friday, so X for 2002 is 2; then September 13th 2002 and December 13th 2002 are Fridays according to this table.

If one compares two normal years, one sees that the days shift by one position: if March 1st is a Friday this year, it will be a Saturday next year. This happens because a year has 365 days, which is 52 entire weeks + 1 day. So our number X increases by 1 every normal year, jumping back to 0 if it would reach 7.

In a leap year however, the days shift by 2 positions rather than 1, because 366 days is 52 entire weeks + 2 days. Thus, the number X of a leap year is 2 more than the X for the previous year (and if this sum is greater than 6, one should subtract 7 to bring X back in the range 0...6).

When do we have a leap year?
In principle, a year is a leap year if it is a multiple of 4.
However, in the Gregorian calender in use nowadays
it is *not* a leap year if
it is a multiple of 100, unless it is also a multiple of 400.
So 1996, 2000, 2004 are all leap years, as are 2096 and 2104,
but 2100 is not.

Let us take a look at one century, say the century starting with the year 2000.
March 1st 2000 was a Wednesday, so the value of X for the year 2000 happens to be 0,
that's easy.
With the above information about leap years, we can easily see that
the value of X for the years 2000...2099 are as follows:

2000...2027: | 0,1,2,3, 5,6,0,1, 3,4,5,6, 1,2,3,4, 6,0,1,2, 4,5,6,0, 2,3,4,5 |

2028...2055: | 0,1,2,3, 5,6,0,1, 3,4,5,6, 1,2,3,4, 6,0,1,2, 4,5,6,0, 2,3,4,5 |

2056...2083: | 0,1,2,3, 5,6,0,1, 3,4,5,6, 1,2,3,4, 6,0,1,2, 4,5,6,0, 2,3,4,5 |

2084...2099: | 0,1,2,3, 5,6,0,1, 3,4,5,6, 1,2,3,4 |

Note the repetition that happens after 28 years; this is a consequence of the fact that 28 is a multiple of both 7 (days in a week) and 4 (leap year interval).

By simple counting the above, we can construct the following table:

X | Number of occurrences in 2000...2099 |
---|---|

0 | 14 |

1 | 15 |

2 | 14 |

3 | 15 |

4 | 14 |

5 | 14 |

6 | 14 |

Now take a look at the century 2100...2199. In principle, we could again write down the entire sequence like we did above, but now starting with 5 (because X for 2099 was 4, and 2100 is not a leap year). Obviously, we would get the same table, except that the entries are shifted 5 places up, or 2 places down.

The same happens for 2200...2299 and 2300...2399, so that we get the following table:

X | Number of occurrences in 2000...2099 | Number of occurrences in 2100...2199 | Number of occurrences in 2200...2299 | Number of occurrences in 2300...2399 | Total in 2000...2399 |
---|---|---|---|---|---|

0 | 14 | 14 | 14 | 14 | 56 |

1 | 15 | 15 | 14 | 14 | 58 |

2 | 14 | 14 | 14 | 15 | 57 |

3 | 15 | 14 | 14 | 14 | 57 |

4 | 14 | 14 | 15 | 15 | 58 |

5 | 14 | 14 | 14 | 14 | 56 |

6 | 14 | 15 | 15 | 14 | 58 |

After 400 years, the pattern of weekdays in a year repeats. So if December 13 of 2002 was a Friday, then so will December 13 of 2402, December 13 of 2802, and so on. (December 13 of 1602 would also, except if you live in an area (e.g., the USA) where the Gregorian calender was not yet in use at that time).

This can be seen by carefully counting the leap years in a block of 400 years, as follows. According to the rule mentioned above, only years dividable by 4 can be leap years: that is 100 out of 400 years. However, 4 of those are also dividable by 100, and one of those 4 is dividable by 400. Therefore, in 400 years, there are 100-4+1=97 leap years, leaving 400-97=303 normal years. Thus in the course of those 400 years, X would increase by a total of 303*1 + 97*2 = 497. However, every time X exceeds 6, we need to subtract 7 in order to bring it back to the range 0...6. Since 497=71*7, the end result is that X after 400 years is back where it started.

So our analysis for 2000...2399 is enough: whatever happens there, repeats itself eternally (unless a new calendar would be introduced).

From the last table we see that in every 400 years, there are 56 years with X=0.
And from the first table on this page, we see that if X=0, then only October
has Friday 13th.
Thus, October 13th is a Friday in 56 out of 400 years.

Similarly, Friday 13th occurs also 56 times in August,
57 times in June, September, December and Dodecember (i.e., February),
and 58 times in each of the other six months.
That is a total of 2*56 + 4*57 + 6*58 = 688 Friday 13s per 400 years,
on average 1.72 per year.

There are (of course...) several other web sites dealing with similar issues. On http://scienceworld.wolfram.com/astronomy/FridaytheThirteenth.html it is shown that the 13th day of the month is slightly more likely to be a Friday than any other day, and http://mathforum.org/dr.math/faq/faq.calendar.html gives a bit of background on calendars.

Comments are welcome at ptdeboer@cs.utwente.nl.

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