I am wondering if there is a strict argument about the probabilities of Christmas (Dec. 25) on Monday, Tuesday, ..., Sunday. My experiments give:

Sunday 0.145
Monday  0.14
Tuesday 0.145
Wednesday 0.1425
Thursday  0.1425
Saturday 0.14
Friday 0.145

It looks that they are not equal. :)

My question is: 1) How to obtain the probabilities without restricting the counts over certain range of years? 2) why Sunday is more probable than Wednesday, which is more probable than Monday, if it is true?

  • $\begingroup$ Also, the thirteenth of the month is slightly more likely to be on a Friday than on any other day: scienceworld.wolfram.com/astronomy/FridaytheThirteenth.html and math.uiuc.edu/~hildebr/347honors/friday13.pdf . Anyway, what is your question? $\endgroup$ – lhf Mar 15 '11 at 1:28
  • $\begingroup$ @lhf: my question is: is there any "accurate" probabilities for those numbers. By "accurate", I mean without the restriction of certain range of years etc. This is also the problem with the link you posted. $\endgroup$ – Qiang Li Mar 15 '11 at 1:33
  • 4
    $\begingroup$ As the link in lhf's comment tells you, the Gregorian calendar's matching of days of the week to the dates of the year repeats exactly every 400 years (BTW, this is also how the Doomsday Algorithm is possible). So you just need to tally all 146097 days over the 400 years and you'll have an "accurate" answer. Furthermore, since the number of years for the cycle is 400, the numbers should be rational and terminate in finitely many digits in decimal representation. Given the form of the probabilities you list, they are in fact the "accurate" numbers. $\endgroup$ – Willie Wong Mar 15 '11 at 2:04
  • 1
    $\begingroup$ There are no leap hours or leap minutes. Leap seconds happen so we keep midnight aligned with the stars, but do not influence the days of the calendar. $\endgroup$ – Ross Millikan Mar 15 '11 at 2:26
  • 1
    $\begingroup$ possible duplicate of Weekend birthdays I believe the final paragraph answers your question. $\endgroup$ – Arturo Magidin Mar 15 '11 at 3:14

You have all you need to know in the comments to your question.

Assuming you are using the Gregorian calendar, then in its cycle of $400$ years there are $97$ leap years and so $365\times 400+97 = 146097$ days, which is exactly $20871$ weeks, so each cycle repeats the weekdays. All you need to do is count $400$ consecutive Christmases; you could count $2800$ or some other multiple, but it would not change the proportions.

Since $400$ is not divisible by $7$, there is no possibility that each weekday will appear the same number of times. In fact you get the following numbers:

Sunday     58
Monday     56
Tuesday    58
Wednesday  57
Thursday   57
Friday     58
Saturday   56.

Divide each of these by $400$ and you get the proportions in your question.

There is no particular reason why Sunday, Tuesday and Friday are most common; they just are. Some day(s) had to be more common than others since $400$ is not divisible by $7$; in the previously used Julian calendar, each weekday appeared four times for 25 December every $28$ years.

| cite | improve this answer | |
  • 1
    $\begingroup$ doesn't take leap-seconds into account ;) $\endgroup$ – mulllhausen Apr 30 '15 at 8:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.