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Assuming Christmas day falls on the 25th of December every year, and using the Gregorian calendar, is it equally likely to fall on any day of the week? (With no prior knowledge about last year's Christmas day, or the year before, or the year before that...)

It would appear from a linked question, and a helpful answer that this is not true for 400 years (since it is not divisible by seven). But it is true for 2800 years (since it is divisible by seven). Is there a flaw in this thought process, or does this question inherently depend on the time period you consider?

flabby99
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    Consider $28$ consecutive years and prove that every day of the week occurs with the same frequency. – Jack D'Aurizio Mar 29 '15 at 15:24
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    I think you actually need $2800$, but yes, same idea. – Ben Millwood Mar 29 '15 at 15:24
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    The sequence of weekdays repeats exactly every $400$ years under Gregorian calendar, because $365\cdot 400+100-3$ is divisible by $7$. So you just need to examine a $400$ year period. – egreg Mar 29 '15 at 15:26
  • @egreg Ah yes, thank you for the link, I did in fact do a search, but I could not find anything in relation to this question – flabby99 Mar 29 '15 at 15:36
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    However, it would seem to me, that all this depends on the time period you take. If we take 400 years, it is not equally likely since 400 is not divisible by 7. If we take 2800 years, it is equally likely since 2800 is divisible by 7. Or is my thought process flawed? – flabby99 Mar 29 '15 at 15:39
  • @egreg, I would argue that it's not a duplicate. The other question asks about the actual distribution. This question only asks if the the days are equally distributed. As my answer shows, there is a very simple way to show that they are not. – Barry Cipra Mar 29 '15 at 15:39
  • "I am not asking for the specific distribution of these Christmas days, only are they equally distributed" - equally distributed is a very specific (type of) distribution, aka uniform distribution. – barak manos Mar 29 '15 at 15:43
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    @barakmanos Fair point, I worded that badly. I have changed the question in light of answers and comments. – flabby99 Mar 29 '15 at 15:50
  • If it's not true for $400$ years, why should it be for $2800$? The sequence (day, weekday) repeats exactly every $400$ years. So if the distribution is not uniform in $400$ years it won't be in any multiple thereof. – egreg Mar 29 '15 at 15:50
  • The argument was that it was untrue for 400 years since it was not divisible by 7. That argument does not hold for 2800 years. – flabby99 Mar 29 '15 at 15:52
  • @flabby99 But $400$ is the minimal period. If the frequency of Monday Christmas is $s$ times every $400$ years, it will be $7s$ times every $2800$ years, so the ratio is the same. – egreg Mar 29 '15 at 15:54
  • @egreg Ah yes, I see the error in my thought process. That makes complete sense. Thank you. – flabby99 Mar 29 '15 at 15:56
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    @flabby99, it doesn't need to. The argument is based on $400$ because that's the minimum repeating period. If, for example, Christmas falls on Sunday $s$ times in $400$ years and on Monday $m$ times, then it will fall on Sunday and Monday $7s$ and $7m$ times, respectively, in $2800$ years. The point is, the ratio remains the same. – Barry Cipra Mar 29 '15 at 15:57
  • @egreg, we agree down to the choice of symbol, except I think of $s$ for Sunday rather than Monday! – Barry Cipra Mar 29 '15 at 15:58
  • @BarryCipra Yes- I completely see what you mean now and I wholly agree with you. I was completely missing that it was the minimum. I have accepted your answer, and am very grateful for the help. – flabby99 Mar 29 '15 at 15:59
  • I will admit, I have used stack exchange for a decent amount of time, but this is my first question. It has been marked as a duplicate, which is fine- but I would say both discussions are useful to someone interested in the problem. As a duplicate, does this question just get deleted, or what happens to it? – flabby99 Mar 29 '15 at 16:15
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    @flabby99, I think it just means that no additional answers can be added. – Barry Cipra Mar 29 '15 at 17:24

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The Gregorian calendar repeats every $400$ years. $400$ is not divisible by $7$. Therefore Christmas cannot fall equally often on each day of the week.

Barry Cipra
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  • Note that the answer to the question I marked as duplicate contains the same argument as yours. – egreg Mar 29 '15 at 16:01
  • I would agree both questions (and answers) are very similar, but would argue that discussion in each one would both be valuable to anyone interested in the problem. – flabby99 Mar 29 '15 at 16:04
  • @egreg, ah yes, I didn't look beyond the question itself. – Barry Cipra Mar 29 '15 at 16:09