“The 13th of the month is more likely to occur on a Friday than on any other day of the week,” or anyway that’s what the bespectacled stranger on the bus insisted to me about ten days ago—”and you can prove it,” he said. I don’t suppose he meant this last as a challenge, and I doubt he gave the conversation much more thought (he no sooner dropped the gauntlet than he disembarked the bus), but there it lay, pulsing, all the way down Halsted, and when I got off the bus it followed me home.

For the next week it took over my life. Listen, I’m no mathematician, and I don’t have time for this sort of nonsense—I’ve got early deadlines to meet and bare cupboards to fill. Listen, I said to myself all week long, as the files of figures invaded my dreams and the sheafs of notes overflowed my desk (by the end of the third day I couldn’t even find the notes for the articles I was supposed to be writing, I think I threw them out on the backsides of a a particular unproductive line of reasoning)—listen, I said: you don’t have time for this.

But a puzzle can get to be like a fever, and it will run its course. Anyway, I think I’ve got it now, finally, the proof, and I’ve even managed to sanitize it of much of they hysteria that accompanied its formulation. So what follows is a message to the stranger on the bus: I did it, OK? Satisfied? I’m sure there must be easier ways to do it, and I’m sure you can think of half a dozen of them, but you know what I think? I think you and your proof should both drop dead.

*The 13th of the month is more likely to occur on a Friday than any other day of the week*, and here’s the proof:

I. Unlike most mathematical puzzles, this one is very much tied, at least at its outset, to the facts as they exist in the actual world, in particular to these:

(a) today, July 13, 1979, falls on a Friday; and

(b) the calendar is organized in such a way that there are two types of year: the “standard year” (Jan/31 days, Feb/28; Mar/31; Apr/30; May/31; June/30; July/31; Aug/31; Sept/30; Oct/31; Nov/30; Dec/31 = 365 days) and the “leap year” (same configuration except that February has 29 days and the total number of days in the year is 366).

The years follow one another in the following way: (1) Every year is a “standard year” except those divisible by four. (2) There is an exception to the above: every centennial year (i.e. 1700, 1800, 1900, etc.), even though divisible by four, is still a “standard year.” (3) There is an exception to the exception: those centennial years divisible by 400 (1600, 2000, etc.) are leap years.

II. Given these ground rules, we can easily perceive that there are 14 possible configurations of a year-long calendar, that is:

—seven possible “standard” calendars (one each with January 1 being a Sunday, Monday, Tuesday, etc.); and

—seven possible “leap year” calendars (derived in parallel fashion).

Furthermore, we can hypothesize that there is a pattern to the sequence of the 14 possible configurations through the centuries, and that at some point this pattern begins to repeat itself.

Now, we could spend endless hours proving the above hypothesis and determining the pattern, but we have the good fortune to know that someone has already done our work for us. The “perpetual calendar,” which can be found in most almanacs, some phone books, and various other popular reference tools, shows an array of the 14 possible calendars (labeled 1 through 14) and an index usually covering a few hundred years (mine goes from 1776 to 2000) telling us which of the 14 calendars applies to any given year.

III. Let us examine that index and see whether we can determine a pattern to the sequence of the 14 possible calendars. Indeed we can.

(a) Using “9, 4, 5, 6” as our opening phrase, we discover that the years 1776-1815 bracket one pattern, which we will call **A**; 1816-1843 bracket a variant pattern, which we will call **B**; 1844-1871 repeat **B**; 1872-1911 bracket another variant, which we will call **C**; but then 1912-1939, 1940-1967, 1968-1994, and 1995-(2022) all repeat pattern **B**.

(b) Let us analyze these patterns:

(1) Pattern **B** consists of a sequence of *28 digits* and seems to occur all the time except when the sequence runs through a centennial year (e.g. 1800 or 1900, years that disrupt the pattern because they are not leap years).

(2) Patterns **A** and **C** are both patterns of *40 sequential digits*; both happen to run through a centennial year, and both sequences follow pattern **B** up to that centennial, whereupon the **B** pattern is disrupted by a “missed leap” and we note two different strategies (**A** and **C**), such that within the limit of 40 digits the sequence returns to the “opening phrase” (“9, 4, 5, 6 … “), thus beginning a new **B** pattern.

(c) We can thus hypothesize that the calendar follows through in sequences of two types:

(1) **B** type patterns of 28 years, and

(2) **A** and **C** of 40 years.

The pattern will always consist of a succession of **B**types except as it crosses “nonleap” centennials, when an interruption of **A**–**C** type will occur. We note that centennials divisible by 400 (e.g. 1600, 2000) are leap years, and that the sequences running through them will be **B** types.

(d) In any sequence of 400 years, we will therefore have three cases of nonleap centennials, of types **A**, **C**, and **D** (the latter will be explicated presently); all the others will be of **B** type. Furthermore, we can prove that 400 years is the phase necessary for the **B**–**A**–**C**–**D** sequence to begin repeating itself.

We can prove this last contention in the following manner: three phases of 40 years add up to 120 years; 400 minus 120 yields 280; 280 divided by 28 yields 10. Thus in any sequence of 400 years, we will have 3 sequences of 40 years—**A**, **C**, and **D**—interspersed with 10 **B** sequences of 28 years. They will occur in the following order:

** B B D B B A B B C B B B B … **

On the chart 1776-2000, we cut in beginning with an **A** pattern in 1776 thus:

1800 1900 2000

1776 **A** 1816 **B** 1844 **B** 1872 **C** 1912 **B** 1940 **B** 1968 **B** 1996 **B** 2024.

We can, by simple subtraction, extend our chart back to 1624:

1700

1624 **B** 1652 **B** 1680 **D** 1720 **B** 1748 **B** 1776 **A**.

We note, of course, that the first date, 1624, is 400 years earlier than the last date, 2024, and that taking the sequence further back we would only begin to repeat the pattern. Four hundred years is the phase.

And, of course, this stands to reason, since by its very rules (see I.b.2) our calendar goes in phases of 400 years.

IV. (a) Therefore, we have isolated four patterns, **A**, **B**, **C**, and **D**, which constitute all the possible sequences in our calendar. These sequences are:

**B**: 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1.

**A**: 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2 , 3 // 4, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1.

**C**: 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1 // 2, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1.

(Note: two slash marks [//] indicate a centennial.)

We have solved for these with relative ease, because we used the examples in our chart: **B** 1816-1843; **A** 1776-1815; and **C** 1872-1911. But our chart does not provide us with an example of **D** (1680-1719), so we will have to manufacture our own.

(b) How to manufacture a chart for the years 1680-1719 (**D**): Begin in 1680 with the opening phrase (“9, 4, 5, 6 … ) and continue using pattern **B** up to 1699. Then, locate which number calendar applies to 1699, consult that calendar of the standard year that begins the following day, and that’s the number for 1700. Repeat the process for each consecutive year, making sure that on leap years you use leap calendars. The result:

1680 9 1690 1 1700 6 1710 4

1681 4 1 2 1 7 1 5

2 5 2 10 2 1 2 13

3 6, 3, 5, 3, 2, 3, 1

4 14 4 6, 4 10 4 2

5 2 5 7 5 5 5 3

6 3 6 8 6 6 6 11

7 4 7 3 7 7 7 6

8 12 8 4 8 8 8 7

9 7 9 5 9 3 1719 1

Thus sequence **D** will run as follows:

**D** 9, 4, 5, 6, 14, 2, 3, 4, 12, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5 // 6, 7, 1, 2, 10, 5, 6, 7, 8, 3, 4, 5, 13, 1, 2, 3, 11, 6, 7, 1.

(c) We will omit here a discussion of all the interesting patterns within each of these four sequences (those resulting from the parallel 1 through 7 and 8 through 14 sets of possible calendars, and the fact that the first week of January in consecutive standard years pops forward one day of the week). They area readily apparent, but unimportant to our endeavor.

We have, however, in summary, established the sequence within four possible patterns—**A**, **B**, **C**, and **D**—and shown the repeating cycle of these patterns to run through 400-year phases. We have also shown that the given **A**, **B**, **C**, and **D** sequences are the only possible sequences, because the 400-year phase is a closed one

Now let us turn to the vexing problem of the 13th day of the month.

V. We ask ourselves, given the 14 possible variables of “standard” and “leap year” calendars, on what days of the week does the 13th of each month tend to fall? More precisely, within each of the 14 possible year-calendars, how many times does each different day of the week (Monday, Tuesday, Wednesday, etc.) fall on a 13th?

This question can be answered through a simple, if somewhat extended and laborious, process. Let us go through the chart of the 14 possible calendars with pens of seven different colors (Sunday is red, Monday is blue—ain’t it the truth—Tuesday is green, etc.), daubing every 13th of the month in all 14 calendars with its assigned color. This will keep us busy for a while, but when we are finished, let us compile a chart whose horizontal will be the calendars 1 through 14, and whose vertical will be the days of the week upon which the 13th falls (i.e., Sunday through Saturday.) We shall call this chart *Schene 1*.

*Schene 1*: A chart of how many times the 13th of the month falls on each day of the week within each of the 14 different types of calendars.

**1 2 3 4 5 6 7 8 9 10 11 12 13 14**

Sun 1 1 2 2 2 1 3 1 1 3 2 1 2 2

Mon 3 1 1 2 2 2 1 2 1 1 3 2 1 2

Tue 1 3 1 1 2 2 2 2 2 1 1 3 2 1

Wed 2 1 3 1 1 2 2 1 2 2 1 1 3 2

Thu 2 2 1 3 1 1 2 2 1 2 2 1 1 3

Fri 2 2 2 1 3 1 1 3 2 1 2 2 1 1

Sat 1 2 2 2 1 3 1 1 3 2 1 2 2 1

This schene may be checked in the following ways:

(a) Each numbered calendar has 12 13ths of the month (one per month), so that each vertical column should, and does, add up to 12. Indeed, each vertical column includes one 3, three 2s, and three 1s.

(b) Because calendars 1 through 14 include every possible configuration, the total number of times each individual day of the week appears on the 13th will be equal across the expanse of all 14 calendars. This number will be 24 or 12 “standard” times plus 12 “leap” times. Every horizontal sequence of numbers therefore should, and does, add up to 24. Indeed, every horizontal file includes two 3s, six 2s, and six 1s.

VI. Now let us return to a consideration of our four patterns, **A**, **B**, **C**, and **D**. We should now ask how many times each of the 14 constituent integers appears in each sequence of 40 or 28 numbers. This is simply a matter of counting through the four patterns as they are recorded in section IV. Our result is *Schene 2*:

*Schene 2*: The number of times each of the 14 types of calendars appears in each of the four patterns.

**1 2 3 4 5 6 7 8 9 10 11 12 13 14**

**B** 3 3 3 3 3 3 3 1 1 1 1 1 1 1

**A** 4 4 5 5 5 4 4 2 1 1 1 1 1 2 1

**C** 5 5 5 4 4 4 4 1 1 1 2 1 2 1

**D** 4 4 4 4 5 5 5 2 1 2 1 1 1 1

The check here, of course, is that each horizontal column should, and does, add up to the appropriate numbers of integers (that is, 28 for **B** and 40 for **A**, **C**, and **D**).

VII. At last we approach our solution. Our question is this: In the 400-year period that constitutes our cycle, how many times does the 13th fall on each day of the week? If we can show that it falls more often on Friday than on any other individual day, then we will have proved our contention that the 13th of the month is more likely to occur on a Friday than on any other of the days of the week.

We can do this in the following way:

We recall that 400 years disperse themselves through the following formula:

10 **B** + **A** + **C** + **D** = 400 years.

Let us first determine the distribution of 13ths of the month within **B**, **A**, **C**, and **D** respectively, and then run the results through our formula.

This can be accomplished through an intersection of our two schenes. Schene 1 tells us how many times any given day (for example, Friday) occurs on the 13th during each of the 14 possible calendars. Schene 2 tells us how often the 14 possible calendars occur respectively in our four possible patterns **A**, **B**, **C**, and **D**. To determine how many times the 13th falls on Day X in Pattern **A**, therefore, align the **A** values for **1** through **14** in Schene 2 above the X values for **1** through **14** in Schene 1, and then multiply each of the 14 (vertical) pairs of number and add up the 14 resultant totals (horizontally). For example, let us take the example of Friday/**A**:

Calendars: **1 2 3 4 5 6 7 8 9 10 11 12 13 14**

**A** values: 4 4 5 5 5 4 4 2 1 1 1 1 2 1

Friday values: 2 2 2 1 3 1 1 3 2 1 2 2 1 1

Totals: 8 8 10 5 15 4 4 5 2 1 2 2 2 1

Adding the bottom line of totals across, we get a grand total of 70.

What this chart shows us, for example, in its first vertical column, is that the **1** calendar appears four times in Sequence **A**, and that the 13th of the month is a Friday twice in that **1** calendar, and that therefore there are eight Friday the 13th in **1**-type calendars in the **A** sequence.

The grand total is the number of times that any given day of the week appears as the 13th of the month during that particular sequence. In this particular example, we have determined that there are 70 Friday the 13ths in sequence **A** (there would thus be 70 Friday the 13ths between 1776 and 1815, for example).

Now, by repeating the process through multiple small additions and multiplications, we can arrive at the figures in *Schene 3*.

*Schene 3*: The number of times the 13th of the month occurs on each day of the week in each of the four possible sequences.

**B A C D**

Sun 48 69 68 70

Mon 48 68 69 69

Tue 48 68 68 69

Wed 48 69 70 68

Thu 48 68 68 68

Fri 48 70 69 69

Sat 48 68 68 68

To check Schene 3, consider that **B** consists of 28 years, while**A**, **C**, and **D** consist of 40 years each. The totals of the vertical columns should equal 12 times the number of years in each pattern, since each year has 12 months and each month has one 13th. They do. (The **B** column consists of 7×48=336, and 336 in turn equals 28 x 12. The **A**, **C**, and **D** columns all consist of 70+69+69+68+68+68+68=480, which in turn equals 40 x 12.)

Now, recalling the formula

10**B** + **A** + **C** + **D** = 400,

we can produce our final schene, *Schene 4* by multiplying **B** figures in Schene 3 by ten and then adding the figures in the three other columns horizontally. Our result will be how many times in 400 years the 13th of the month occurs on each given day of the week.

*Schene 4*: The number of times the 13th of the month occurs on each day of the week over a phase of 400 years.

10 **B** + **A** + **C** + **D** = Grand total

Sun 480 69 68 70 687

Mon 480 68 69 68 685

Tue 480 68 68 69 685

Wed 480 69 70 68 687

Thu 480 68 68 68 684

Fri 480 70 69 69 688 … da winnah!

Sat 480 68 68 68 684

Thus, over a period of 400 years (which we have determined to be the period of our repeating cycle), the 13th of the month will fall on Friday more times than on any other single day (once more, that is, than on either Wednesday or Sunday).

*Hence, it is true: the 13th of the month is more likely to occur on a Friday than on any other day of the week.*

Just barely.

And of course, this is a complete fluke, owing entirely to the fact that today, July 13, 1979, is a Friday, rather than, say, a Wednesday, in which case the day with the one-point margin of 688 would have been some other day altogether. (*Quiz*: which one?)