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This question already has an answer here:

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?

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marked as duplicate by egreg, George V. Williams, Did probability Mar 29 '15 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Consider $28$ consecutive years and prove that every day of the week occurs with the same frequency. $\endgroup$ – Jack D'Aurizio Mar 29 '15 at 15:24
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    $\begingroup$ I think you actually need $2800$, but yes, same idea. $\endgroup$ – Ben Millwood Mar 29 '15 at 15:24
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    $\begingroup$ 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. $\endgroup$ – egreg Mar 29 '15 at 15:26
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    $\begingroup$ 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? $\endgroup$ – flabby99 Mar 29 '15 at 15:39
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    $\begingroup$ @barakmanos Fair point, I worded that badly. I have changed the question in light of answers and comments. $\endgroup$ – flabby99 Mar 29 '15 at 15:50
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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.

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  • $\begingroup$ Note that the answer to the question I marked as duplicate contains the same argument as yours. $\endgroup$ – egreg Mar 29 '15 at 16:01
  • $\begingroup$ 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. $\endgroup$ – flabby99 Mar 29 '15 at 16:04
  • $\begingroup$ @egreg, ah yes, I didn't look beyond the question itself. $\endgroup$ – Barry Cipra Mar 29 '15 at 16:09

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