Conference rusure::math

    Not an answer, but a little experiment:
    
    $ create foo.com
    
$ year = 1964
$ lup: write sys$output "In ''year', Christmas was on a ", -
    f$cvtime ("25-dec-''year'",,"weekday")
$ year = year + 1
$ if year .le. 1984 then goto lup
    
    $ @foo
    
In 1964, Christmas was on a Friday
In 1965, Christmas was on a Saturday
In 1966, Christmas was on a Sunday
In 1967, Christmas was on a Monday
In 1968, Christmas was on a Wednesday
In 1969, Christmas was on a Thursday
In 1970, Christmas was on a Friday
In 1971, Christmas was on a Saturday
In 1972, Christmas was on a Monday
In 1973, Christmas was on a Tuesday
In 1974, Christmas was on a Wednesday
In 1975, Christmas was on a Thursday
In 1976, Christmas was on a Saturday
In 1977, Christmas was on a Sunday
In 1978, Christmas was on a Monday
In 1979, Christmas was on a Tuesday
In 1980, Christmas was on a Thursday
In 1981, Christmas was on a Friday
In 1982, Christmas was on a Saturday
In 1983, Christmas was on a Sunday
In 1984, Christmas was on a Tuesday
          
    /Eric

Spoiler:


In 1991, Christmas will be on a Wednesday.  So, the probability that
Christmas will occur on a Wednesday is 1.

:-)

    Some more experimenting.  Try this .COM file.  Diddle the first
    two lines for examining Christmas during various years.  I tried
    it from 1985-2300, and the days were APPROXIMATELY equal, although
    not as equal as they could have been, so there still might be
    an interesting question here.
    
$ first_year = 1985
$ last_year = 2020
$ ctr = 0
$ i:num'ctr' = 0
$ ctr = ctr + 1
$ if ctr .le. 6 then goto i
$ year = first_year
$ lup:rank = f$locate (f$extract (0,2,f$cvtime ("25-dec-''year'",,"weekday")), -
    "SuMoTuWeThFrSa") / 2
$ num'rank' = num'rank' + 1
$ year = year + 1
$ if year .le. last_year then goto lup
$ ctr = 0
$ write sys$output ("Considering years ''first_year'-''last_year':")
$ write sys$output ""
$ write sys$output "Christmas falls on a Sunday ''num0' times."
$ write sys$output "Christmas falls on a Monday ''num1' times."
$ write sys$output "Christmas falls on a Tuesday ''num2' times."
$ write sys$output "Christmas falls on a Wednesday ''num3' times."
$ write sys$output "Christmas falls on a Thursday ''num4' times."
$ write sys$output "Christmas falls on a Friday ''num5' times."
$ write sys$output "Christmas falls on a Saturday ''num6' times."

    /Eric

Of the fourteen possible years, Christmas falls on a Wednesday in two.

Now if each year were equally likely then Christmas would fall on a Wednesday
2/14 or 1/7 of the time.  However, since half of the years are leap years,
occurring only ~� of the time, Christmas falls on a Wednesday

	�*[1/(14/2)]+3/4*[1/(14/2)]

or 1/7 of the time.  I leave it as exercise for the reader to determine if the
rule about centuries makes a difference.

I believe that the non-mathematical factors, primarily political, override.  A
simple counting of past years would of course provide the most accurate
measure.  Extrapolating in this case is pure speculation, given the nature of
political systems.  I would venture to say that such speculation will be borne
out to a maximum of a century.  Within said time events shall lead to a change
in the calendar and so effect the day of the week on which Christmas falls.

	Jestingly yours,
	David.

Re .2:	there exists an even more interesting answer...

Re .3:	...which can be exhibited using Eric's .com file, perhaps after
	changing the first two lines to

		$ first_year = P1
		$ last_year = P2

Re .4:	sure, but these are math notes, not political ones.  Care to
	extrapolate/speculate into the future?

John

    Answer follows form feed.
                             		John
    
    				.1425

	There are fourteen possible types of year.   Label them Li, Ni
for i = 0,...,6, where L denotes a leap year, and N denotes a normal year.   
The distribution of leap years has a period of 400 years, during which
there are 97 occurences of February 29.   The total number of days during
this period is 400*365 + 97, which (mod 7) is 0.   Thus each cycle of 400
years has the same calendar, and it is possible that some year-types are
more prolific than others with the same number of days.

	Suppose that we start then at year 1, and label it N0.   Then the
structure of the next 28 years will be:

N0 N1 N2 L3
N5 N6 N0 L1
N3 N4 N5 L6
N1 N2 N3 L4
N6 N0 N1 L2
N4 N5 N6 L0
N2 N3 N4 L5

	This means that the first 84 years of every century are evenly
distributed across i.   For the last sixteen years of the century, however
we have:

N0 N1 N2 L3
N5 N6 N0 L1
N3 N4 N5 L6
N1 N2 N3 N4

	And the last four years of the following centuries are:

N5 N6 N0 L1
N3 N4 N5 L6
N1 N2 N3 L4
N6 N0 N1 N2

N3 N4 N5 L6
N1 N2 N3 L4
N6 N0 N1 L2
N4 N5 N6 N0

N1 N2 N3 L4
N6 N0 N1 L2
N4 N5 N6 L0
N2 N3 N4 L5

L  0  1  2  3  4  5  6
# 13 14 14 13 15 13 15

N  0  1  2  3  4  5  6
# 43 44 43 44 43 43 43

Now, it remains only to find what day January 1st 2001 falls on.

1991: Tue =>
2001: Mon

so 0 corresponds to Monday, and so on.

	Now, when does Christmas fall on a Wednesday?   When the year is N1
or L0.   This means (44+13)/400 = 14.25% of the time, as John suggests.

    There is an interesting corollary to this:

    Inspection of which day of the week the thirteenth of the month falls
    upon for each of the years Li,Ni gives the following:

    		    Sat  Sun  Mon  Tue  Wed  Thu  Fri
    		L0   3    1    1    2    2    1    2
    		L1   2    3    1    1    2    2    1
    		L2   1    2    3    1    1    2    2
    		L3   2    1    2    3    1    1    2
    		L4   2    2    1    2    3    1    1
    		L5   1    2    2    1    2    3    1
    		L6   1    1    2    2    1    2    3
    		N0   2    1    1    3    1    2    2
    		N1   2    2    1    1    3    1    2
    		N2   2    2    2    1    1    3    1
    		N3   1    2    2    2    1    1    3
    		N4   3    1    2    2    2    1    1
    		N5   1    3    1    2    2    2    1
    		N6   1    1    3    1    2    2    2

    Multiplying through by the distribution vector for Li,Ni yields the
    number of times the thirteenth falls on each given day of the week
    over the 400 year cycle.

    		    Sat  Sun  Mon  Tue  Wed  Thu  Fri
                    684  687  685  685  687  684  688

    Thus it can be seen that for any month chosen at random, the
    probability that the thirteenth falls on a Friday is higher than that
    of it falling on any other day of the week.

    Regards,

    Pete