# Can you figure out why Christmas does not fall on a Friday exactly 1/7 of the time?

Granted it is awfully close, but not exact. There is nothing calendrically unusual about Christmas. I chose it because Friday on Christmas means a 3 day weekend.

If there were no leap years, Christmas would occur on Friday exactly 1/7 of the time, and even if leap years always occurred every four years. But the rule for leap years is slightly more complicated. Every 100 years is not a leap year, except that every 400 years is a leap year. I imagine people would be more aware of this if the new millennium, 2000 had not been divisible by 400.

A simple arithmetic calculation shows why 1/7 can’t possibly be right.

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## 35 Answers

Because of leap year every 4 years, the calendar is on a 28 year cycle, not 7. So if you have a calendar from 28 years ago the days will line up

1/7 can be right, because there are exactly 7 options, and 1 will be correct, so the odds of it landing on any specific day is 1 in 7

So if any given date only lands on a Friday slight less than 1/7th of the time, what day of the week comes up a little more than 1/8th the time?

There is an equal chance of Christmas landing on a Friday as any other day of the week.

You gave the answer in the details.

@marinelife , not quite. There is something special about the 400 year cycle that can only be verified by calculation.

I’m no mathematician, but something about the reasoning behind this question seems intuitively wrong to me.

I don’t see how the answer could be anything but 1/7 for a given day and date because there are only 7 possibilities. It can’t be the case that some other day occurs more often than Fridays. That just doesn’t make any sense.

Isn’t the question of the odds of Christmas falling on Friday *in a given year* a completely different question? That’s when leap years have to figure into the odds.

If you choose too small a sample—say, the present year—you will find that Christmas doesn’t fall on *any* Fridays. But that doesn’t tell you a thing about other years.

Come on, math guys, vindicate my logic here.

Over the course of 2800 years the cycle would be complete and the distribution would be equal.

@Jeruba I agree with you. If Friday is not 1/7th then some day has to be more than 1/7th.

It would take 2800 years to complete the cycle except in one case, which is the case that in fact holds.

Well I give up. Are you going to tell us, or just leave us hanging?

No wait, don’t tell us yet… give people at least a day to think about it.

Still working on this but…

@Jeruba You’re interpretation makes some sense but I don’t think its what the OP is asking. Here is a way to see what the question is:

Pretend we had a leap year every 7 years. Then, we’d have

Su, Mo, Tu, We, Th, Fr…Su, Mo, Tu, We, Th, Fr…etc..

If, instead, it were every 14 years, we’d get half as many Saturday Christmases as others.

So there’s something about our current system that has the same effect.

Edit: In my example, Saturdays are left out. But, I could have also set it up to be Fridays, Sundays, anything. So there is some starting point bias involved.

In math terms, the question is:

Let C(n) be the number of times that Christmas falls on a Saturday from years 1 to n.

Why is it that as n—> infinity, C(n) does not approach exactly 1/7.

@roundsquare, I understand the concept of leap years and the fact that there are only 14 possible calendars (assuming we’re not worrying about the days lost in the transition from the Julian to the Gregorian calendar). Something else is going on here, something that is related to the difference between computing the odds of *any* date being a Friday and the odds of a *particular* date being a Friday. I’m certain that Leonard Mlodinow can explain it, even if I can’t. Somehow we’re casting the question in the wrong terms, and they’re causing us to miss the key to it.

Missed the edit period. One change in my previous post:

Why is it that as n—> infinity, C(n)**/n** does not approach exactly 1/7

@Jeruba How about if we cast it like this?

Take the years 1 to 100 and pick one at random. What are the odds that Christmas is on a Friday?

Take the years 1 to 200 and pick one at random. What are the odds that Christmas is on a Friday?

Now do that for the years 1 to 300.

...

Do that for the years 1 to n.

If we know when Christmas was on year 1, we could calculate the odds.

At first, we’d think that as we did this for bigger and bigger in, the odds would get closer and closer to 1/7.

But, as the OP says, this is not true. (I’ve verified this on a spreadsheet btw).

Does this way of casting it make it more sensible?

Well, that’s what I said in my previous response, isn’t it?—that the odds of Christmas’s occurring on a Friday *in a given year* is different from the question of how many times a certain date falls on a certain day in a given period of time?

Anyway, I don’t see why we are accepting “as n—> infinity, C(n)/n does not approach exactly 1/7” as a given. Isn’t it a hypothesis that has yet to be proved? If the full cycle is 2800 years, where is the evidence that the distribution is not equal?

@Jeruba Ah, okay. I misread your previous statement then.

As for why we’re accepting it… I did it out in a spreadsheet real quick and found that the OP was right. 3 days of the week get extra Christmases and 4 days get fewer.

Silly, I thought it was common knowledge that it falls on christmas day every single year…who knew?

I will post the answer this evening. Here is an elaboration of the hint I gave. If the rule for leap year was simply once every four years then there would be a 4 year cycle for leap years and a 28 year cycle for the days of the week, which would all be equivalent. In the actual case there is a 400 year cycle for leap years. This would result in a 2800 year cycle with all days of the week equivalent, except in one case, and this is the case that holds. To show which case holds requires a little bit of arithmetic.

You guys are going to be upset with yourselves when you see how simple the answer is.

Okay, one last hint. If January 1 of the first year of a 400 year cycle is on a Monday, what are the possibilities for the day of the week of January 1 of the first year of the next 400 year cycle?

From a strictly practical point of view, I can’t help noting that there was no Christmas 2800 years ago and there probably won’t be one 2800 years from now, so it won’t matter.

The Solution:

From January 1 of the first year of a 400 year cycle to January 1 of the first year of the following 400 year cycle, there is some net change in the day of the week, which will be some number from 0 to 6. If the number is from 1 to 6, it will take 7 cycles of 400 years before January 1 is the same day of the week as the original. However, if the offset is 0, meaning that the day of the week is the same for January 1 of the next 400 year cycle, then we are already there. The total cycle length is not 2800 years, but 400 years. This is in fact the case as I will show below. Since 400 is not divisible by 7, it is not possible for any day of the week to come up exactly the 1/7 of the time for Christmas, or any other day. Christmas will be on a Friday 58 times out of 400. 400/7 = 58 4/7.

To show that the next 400 year cycle starts on the same day of the week as the previous one, we could add the total number of days in the 400 years and check that it is divisible by 7. A much easier calculation can be made. 365 days is 52 weeks plus one day, so that the first of the year will move up one day of the week for the year after a regular year. Similarly, the fist of the year will move up two week days after a leap year of 366 days. Now all we have to do is add the number of regular years to twice the number of leap years, divide by 7, and check the remainder.

There are ¼ of 400 =100 years divisible by 4. Of these, the years 100, 200 and 300 are not leap years. This gives 100 – 3 =97 leap years and 400 – 97 = 303 regular years. We calculate 303 + 2*97 = 303 + 194 = 497 = 7*71, and so is divisible by 7, meaning that each 400 year cycle starts on the same day, specifically Monday. You might find it easier to remember that the start of the year for each year divisible by 400 starts on Sunday.

This also means that the 400th anniversary of any day falls on the same day of the week as the original. July 11, 1611 must have fallen on a Monday. We can’t go too far back, because the Gregorian calendar, which contains the changes for leap years divisible by 100 and 400, was created in the early 16th century.

The first day of the week of the next 400 year cycle would also be a Monday. In any 400 year cycle there can’t be an even distribution of the days of the week for any given date as 400 is not divisible by 7. Christmas will fall on a Friday either 57 or 58 times in that period and more usually 57 times.

As the periods of time get longer and longer I would agree with Jeruba that the distribution has to even out and Christmas will fall on a Friday a seventh of the time.

@flutherother , The distribution for each 400 year cycle is identical, because they all start on the same day.There are 58 Christmas days that fall on a Friday and that will be true in each cycle, meaning that over time the number will be slightly less than 1/7 of the total.

@Jeruba , There must have been some delay in adopting the Gregorian calendar. This is the calendar for 2411, which is identical to this year’s calendar.

@LostInParadise That us true, I can’t argue with the maths though the result seems counter intuitive.

@LostInParadise I don’t like to be the Grinch that stole Christmas but aren’t there only 57 Christmas days that fall on a Friday rather than 58?

One other question, @LostInParadise (and all I’m doing here is raising questions, because when it comes to mathematics I don’t feel qualified to make assertions about anything but my own doubts):

If we grant that in the 400-year cycle that starts today, and hence each 400-year cycle that elapses thereafter, we come up short on Fridays, won’t the opposite be true of the 400-year cycle that starts tomorrow or the next day? so that if it’s a *rolling* 400-year interval, and we compare results among the intervals, we *will* get an even distribution?

In other words, it’s still an issue of sample size, as I suggested in my first response above. Isn’t it?

@LostInParadise, you said: *It would take 2800 years to complete the cycle except in one case, which is the case that in fact holds.* What is that one case in which it does not take 2800 years to complete the cycle?

@Jeruba, One of the reforms of the Gregorian calendar was not just to keep future years in better sync with the Earth’s revolution around the sun, but to correct for the accumulated error. As I recall, this was done by yanking a few days out of October. This would mean that there was a difference between the Julia and Gregorian calendars in when the new year started.

As to the second question, suppose we labeled the years from 1 to 400. If we take 400 years starting from some point other than 1, there will still be exactly one copy of the years labeled from 1 to 400 and they will each have the associated number of days.

@fluterother, I have to confess that I did not think up this problem, though I wish I did. I got it from the book Mathematics For Pleasure In the answer to the problem, Jacoby said that there are 58. Before reading the answer, I wrote a computer program to do the calculation and I also got 58. I could send you the code. Did you make sure to start with Monday?

I hadn’t counted them but 400/7 is 57 and a seventh.

Thanks for correcting my division, So it turns out that Christmas falls on a Friday slightly more often than 1/7 of the time.

@LostInParadise, yes, thanks, I mentioned that I didn’t need help with the Julian-to-Gregorian anomaly.

Suppose we label the years from 1 to 400 and start them on a Monday. Let’s call that Interval A. Suppose we also label the following day, Tuesday, as day 1 of a 400-year period called Interval B. Will Interval A and Interval B have the same number of Mondays or not?

And…why is it 400/7? Why isn’t it ([400×365] + 99) / 7?

And what’s the answer to this part?

@LostInParadise, you said: *It would take 2800 years to complete the cycle except in one case, which is the case that in fact holds.* What is that one case in which it does not take 2800 years to complete the cycle?

@Jeruba , The other cases were the 6 other days of the week that the next 400 years could have started on, each of which would have resulted in a 2800 year cycle. It is an interesting coincidence that the next 400 year interval starts on the same day of the week as the previous one.

Interval B has one copy of each of the years labeled from 1 to 400, in the same way that any consecutive 7 days has each day of the week and any consecutive 12 months has each month of the year. Interval B starts from year 2, goes up to 400 and ends with year 1. Each labeled year has a specific day associated with it. Changing the order in which the 400 years are visited does not change anything. The total for each day of the week will still be the same for the two intervals and any other consecutive group of 400 years.

Interval B as I defined it does not begin a year later. It begins a day later.

I asked questions because I took an interest in your puzzle. I guess the answers just aren’t forthcoming.

Sorry, The same logic applies, The 400 years cycle through the exact same days. The cycle starts at day 2 of year 1 and will finish at day 1 of year 1. All the days in all of the years are identical. If you want to be able to picture it, imagine the days arranged in a large circle. If you break into the circle at any point, 400 years will bring you back to where you started.

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