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Probability of getting 53 sundays

  1. May 11, 2013 #1
    In a leap year the probability of getting 53 sundays or 53 tuesdays or 53 thursdays is
     
  2. jcsd
  3. May 11, 2013 #2

    HallsofIvy

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    There are 366 days in a leap year so 366/7= 52 complete weeks with two additional days. There will be 53 sundays if and only if sunday is the first or second day of the year- that is if the year starts on a saturday or a sunday. What is the probability of that?
     
    Last edited: May 11, 2013
  4. May 11, 2013 #3

    haruspex

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    There is a subtlety in the Gregorian calendar: there's a whole number of weeks in its 400 year cycle. This makes the odds different for some days of the week.
     
  5. May 12, 2013 #4
    correction in the question

    The actual question says what is the probability of 53 Sundays or 53 Tuesdays or 53 Thursdays in a non leap year
     
  6. May 13, 2013 #5
    If you know one, it should be easy to compute the other since the probability of any day of the week in all years is 1 in 7, and that 97 out of 400 years are leap years.
     
  7. May 17, 2013 #6

    HallsofIvy

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    Your original post said "in a leap year".
     
  8. Mar 2, 2014 #7

    phyzguy

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    I'm not sure what probability means in this context. The calendar is completely deterministic and is known in advance for as far in the future as you wish to calculate it. It's not like you roll the dice to see what happens each year. So it would be pretty easy to just run the next 1000 years (or 10,000 or whatever) and see how many have 53 Sundays or whatever you want to know. It's not clear to me that the probabilistic arguments hold (such as saying 1/7 of the time the first day is a Sunday), since there are deterministic patterns. For example, if you look at the next 200 years and ask how many of the January 1 dates are Sundays, there are 30 Sundays in the next 200 years. This is significantly more than 1/7 of the time. Also, if you look at the number of years between successive Sundays on January 1, you get the following pattern: 6,11,6,5,6,11,6,5,6,11...
     
  9. Mar 3, 2014 #8

    FactChecker

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    You are right that the OP should define the problem better, but many probability problems have deterministic components. If the calendar is determined, but the selection of a year is random, then it is a probability problem. Just like a coin has determined sides, heads and tails, but the result of a coin toss is a probability problem.
     
  10. Mar 3, 2014 #9

    berkeman

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    Thread closed for Moderation

    EDIT -- thread re-opened. Reminder to all posters that we do not post solutions to schoolwork questions here on the PF.
     
    Last edited: Mar 3, 2014
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