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Thread: Friday the 13th

  1. #1
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    Friday the 13th

    Just noticed my sister's birthday will be a Friday the 13th this year. Here is a question:

    Under our present calendar, what day of the week is the 13th most likely to fall upon?

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    Re: Friday the 13th

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Friday!</span hide> Here are the frequencies over a 400 year period (enough to take the exceptions to leap years into account)

    <table border=1><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Day</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Frequency</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Sun</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">687</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Mon</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">685</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Tue</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">685</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Wed</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">687</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Thu</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">684</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Fri</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">688</span hide></td><td><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Sat</span hide></td><td align=right><span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">684</span hide></td></table>

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    Re: Friday the 13th

    <img src=/S/hmmn.gif border=0 alt=hmmn width=15 height=15>Good job - I don't think I even want to know how you achieved this data (not manually I Hope )!
    Scott

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    Re: Friday the 13th

    I used Excel.

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    Re: Friday the 13th

    <img src=/S/bravo.gif border=0 alt=bravo width=16 height=30> <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16> <img src=/S/clapping.gif border=0 alt=clapping width=19 height=23>
    <img src=/S/bash.gif border=0 alt=bash width=35 height=39>

    <img src=/S/stupidme.gif border=0 alt=stupidme width=30 height=30> <img src=/S/exclamation.gif border=0 alt=exclamation width=15 height=15> <img src=/S/exclamation.gif border=0 alt=exclamation width=15 height=15> <img src=/S/exclamation.gif border=0 alt=exclamation width=15 height=15>
    Scott

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    Re: Friday the 13th

    Statistically I would say that the day should be <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">it makes no difference</span hide>.

    But based on your data are you trying to imply that it is <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Friday</span hide> since in the past 400 years that this is the most likely day to have the 13th? <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

    Steve

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    Re: Friday the 13th

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Yes, I'm saying that the 13th falls on a Friday more often (in the long run) than on any other day of the week.

    The 13th day of a month isn't distributed randomly, since there is a fixed progression: if January 13 is a Sunday, February 13 is a Wednesday etc. Complicating factors are that a year is not exactly 52 weeks, and the occurrence of leap years:
    Years divisible by 4 are leap years, except
    Years divisible by 100 aren't leap years, except
    Years divisible by 400 aren't leap years.
    So in 400 years, there are 100-4+1 = 97 leap years.
    The total number of days in 400 years is 400*365+97 = 146097. This is divisible by 7. So in a 400 year cycle, you take the occurrence of leap years and their exceptions into account, and since it is a whole number of weeks, the distribution of the 13th over days of the week will repeat after 400 years.</span hide>

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    Re: Friday the 13th

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">It appears to me that your sample was just not large enough and you are only seeing sample variation.

    There are only 14 possible calendars for a year (a non-leap year starting on a Sun-Sat, and a leap year starting on Sun-Sat). If you look at each of these possible calendars, then yes for any year there may be more of one day than another (eg if the year starts on Sunday in non-leap year, there is only 1 Sun, Tue, Sat 13th, there is 2 Wed, Thu, Fri 13th, and 3 Mon 13th), but if you look at all 14 of them, then each day of the week has 12 occurences in the 7 calendars without leap days and 12 occurences in the 7 calendars with leap days.

    I can't imagine that over time the distribution of the starting day of the year would not be random and that there should be over a long stretch of time an equal number of each non-leap years starting on Sun-Sat, as well as an equal number of leap years starting on each of the days of the week. If there are an equal number for each of the 7 calendars in each of the subtypes, the 13th day should be equally distributed.

    To get Friday 13 to show up more often would require would indicate that non-leap years are more likely to start on a Thursday, and/or that leap years more likely start on Sunday over time instead of being equally distributed among all possbily starting days

    It is possible for this to occur in a small sample of only 400 years especially since it is not evenly divisible by 14 (or even 7)</span hide>

    Steve

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    Re: Friday the 13th

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">I was wrong, the distributions of starting days is not random, as I originally thought.

    The way the calendar is setup now, Tues and Thurs are most likely non-leap year starting days and Sunday and Friday are most likely days that a leap year starts on. Which works out that Friday is the most likely day to fall on the 13th...</span hide>

    Steve

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    Re: Friday the 13th

    Try it out in the attached workbook.

    Added: modified attachment to be more flexible.

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    Re: Friday the 13th

    Thanks, I used a different approach (and was writing my correction when you were working on the spreadsheet).

    My "problem" was the paradigm that the starting days of the year were not randomly distributed. I had to convince myself that of the 7 possible non-leap year calendars and 7 possible leap year calendars, that one did not get equal numbers of each one in the type....

    Steve

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    Re: Friday the 13th

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Actually, the sample size does not matter. The key to the whole thing is that the calendar repeats every 400 years and that the number of days in 400 years is divisible by 7. If we were on the Julian calendar instead of the Gregorian calendar, there would be no preference for Fridays. If the number of days in 400 years were 146096 or 146098, there would be no preference for Fridays.</span hide>

    Oh the things that Y2K work brought to light.

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    Re: Friday the 13th

    <P ID="edit" class=small>(Edited by mbarron on 21-Aug-06 23:11. Forgot the Hide tags)</P>Some interesting things I "discovered":


    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">With a start date of 1/1/2001:
    Friday will hold (or share) the title most years for having accumulated the most 13th, but it's not unit 3392 that Friday will hold the exclusive title of most Friday's accumulated.

    As late as 2991, a day other than Friday will hold the title.

    From 2183 to 2198 Friday is not the leader. After 2199 the longest non lead span is 1 year.



    Marco I used (I took it out of the workbook, as not to raise any flags):

    Option Explicit

    Sub fridays()
    Application.ScreenUpdating = False
    Dim i As Long, chDate As Date, mMonth As Integer

    For i = 2001 To 4000
    Cells(i - 1999, 1) = i
    For mMonth = 1 To 12
    chDate = DateSerial(i, mMonth, 13)
    Cells(i - 1999, Weekday(chDate) + 1).Value = Cells(i - 1999, Weekday(chDate) + 1) + 1
    Next mMonth
    Next i
    Application.ScreenUpdating = True
    End Sub </span hide>

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    Re: Friday the 13th

    I love it! I hadn't thought of actually listing out the years like that. <img src=/S/clapping.gif border=0 alt=clapping width=19 height=23>

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