How many Friday the 13ths would you expect in a year?

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SUMMARY

The discussion centers on the occurrence of Friday the 13th in a given year, with participants calculating the expected frequency based on mathematical principles. The consensus is that there are typically 1.71 to 1.72 occurrences of Friday the 13th annually, with this year being notable for having three instances. The conversation also delves into the deterministic nature of calendars, contrasting it with chaos theory, and highlights the mathematical certainty behind the distribution of days across months.

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  • Understanding of basic probability and statistics
  • Familiarity with calendar systems and their cycles
  • Knowledge of chaos theory and its mathematical implications
  • Ability to perform simple arithmetic calculations related to dates
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  • Research the mathematical principles behind calendar cycles, specifically the 400-year cycle
  • Explore the relationship between chaos theory and deterministic systems
  • Learn about the distribution of weekdays across different months and years
  • Investigate the historical occurrences of Friday the 13th and their statistical significance
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Mathematicians, statisticians, calendar enthusiasts, and anyone interested in the interplay between probability and number theory.

JakeA
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Today is Friday the 13th. It's the first of 3 this year. Is 3 more than you'd expect in a year?

Here's how I'd calculate it.

12 months, each with a 13th. 7 days in a week, so each month has 1/7 chance of having a Friday 13. 1/7 x 12 = 1.71.

So this year is significantly more unlucky than others, I guess.

OK, if I got this trivial math problem wrong, I'm really embarrassed, but so what.

A more complex question would be whether or not numerically days of the week are evenly distributed, as I have assumed in my calculations. Also, does the occurrence of a Friday 13 effect the probability of another in the year? Interesting number theory questions with hints of chaos theory as well.
 
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Your calculation is spot on and in fact it's possible to determine the maximum number of Friday the 13ths as well as the minimum, but that involves accounting for months lengths and their remaining lengths divisibility 7, whatever days remain can be added successively to determine the amount of Friday the 13ths in any given year leap or otherwise. it is a tad more complicated, in fact I don't want to start working it out as I fear it'll eat up an hour or so and I already know the answer is 3 and 1. :smile:

I wouldn't be surprised though that once you had an answer you could work out the number of friday the 13ths from the first date they fell on accounting for leap years, and in fact I'm sure of it.
 
By the way, JakeA, Chaos Theory despite its name is not even a theory. It is a branch of inquiry.
 
JakeA said:
Today is Friday the 13th. It's the first of 3 this year. Is 3 more than you'd expect in a year?
There are only fourteen possible calendars. January 1 can start on anyone of the days of the week (seven ...) , and February can have 28 or 29 days (... times two: fourteen).

w3390 said:
JakeA said:
Interesting number theory questions with hints of chaos theory as well.
By the way, JakeA, Chaos Theory despite its name is not even a theory. It is a branch of inquiry.
Theory has a rather different meaning in mathematics than in science. Theory in mathematics means "body of knowledge". In addition to chaos theory there's K-theory, knot theory, measure theory, number theory, ...

That said, there is nothing chaotic in this problem because there is no randomness in the problem. There is nothing random about whether April 13, 2525 falls on a Friday (it does). Determining the day of the week a particular date occurs on is a trick commonly learned by autistic savants. (I cheated and looked up a calendar for 2525.)
 
In a 400 year cycle there are 688 F13's
So one expects
E[F13]=688/400=1.72 F13/year
So three is more than you expect in a year
though maybe the 28 year cycle that covers all cases likely to arise in the lives of most now living should be used
E[f13]=48/28=12/7~1.7143
in eith case we may note that about one year every 9 has 3 and the next one is 2015
 
D H said:
That said, there is nothing chaotic in this problem because there is no randomness in the problem.

The irony in this correction is that the main subject of chaos theory is deterministic dynamical systems e.g. the most prominent examples of chaos like the lorenz equations, logistic equation, the 3-body problem, etc have nothing to do randomness. Here is the first paragraph of the wikipedia article on chaos theory:

"In mathematics, chaos theory describes the behaviour of certain dynamical systems – that is, systems whose states evolve with time – that may exhibit dynamics that are highly sensitive to initial conditions (popularly referred to as the butterfly effect). As a result of this sensitivity, which manifests itself as an exponential growth of perturbations in the initial conditions, the behavior of chaotic systems appears to be random. This happens even though these systems are deterministic, meaning that their future dynamics are fully defined by their initial conditions, with no random elements involved. This behavior is known as deterministic chaos, or simply chaos."
 
At the risk of stating the obvious, if Friday the 13th falls in February of a non leapyear, then another must follow in March.

We got two for the price of one this year :smile:
 
confinement said:
The irony in this correction is that the main subject of chaos theory is deterministic dynamical systems e.g. the most prominent examples of chaos like the lorenz equations, logistic equation, the 3-body problem, etc have nothing to do randomness.

There is no uncertainty in the calendar. The calendar repeats every 400 years. That today is Friday the thirteenth means that February 13, 2409, February 13, 2809, and February 13, 20009 all will fall on Friday. Compare this to a deterministically chaotic system, where we don't know exactly the state of the system. If the system is chaotic, a slight change in that state will result in large changes downstream. That we don't know the exact state at any point in time essentially adds a bit of randomness to the system.

There is a key word in that wikipedia article you quoted: "In mathematics, chaos theory describes the behaviour of certain dynamical systems ..."

The calendar is not one of those dynamical systems.
 

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