An unsurprising year

I’ve received one those FW:…:FW emails yesterday with the following text:

2011 is an unusual year. Add the last two digits of your birth year to the age you will turn on your birthday this year and you’ll get 111! (…)
This year, July has 5 Fridays, 5 Saturdays and 5 Sundays. This happens once every 623 years. This is called money bags. So, forward this to your friends and money will arrive within 4 days. (…) October 2010 has 5 Fridays, 5 Saturdays and 5 Sundays and that this combination of days only occurs once in 823 years.Based on Chinese Feng Shui the one who does not forward ….. will be without money. I’m not taking any chances!

These ‘facts’ are not terribly surprising. The first one simply boils to down to adding your date of birth to your age to find, surprise!, the current year. Well, meh.

Secondly, having three consecutive days appearing 5 times in a month is a given as soon as the month has 31 days: to use the example given, if the first day of the month is a Friday, the first Sunday will happen on the 3rd and the last and fifth Sunday will be on the 3+4*7=31st. So no surprise there.

But how often do we have this specific instance,i.e. 5 Fridays, 5 Saturdays and 5 Sundays in July? We could calculate that easily given that it happens this year and that there are 365 days/year most years; since 365=1 mod 7, we expect to move forward by one week day every year, and by 2 days on leap years. But we can also have a quick look at the last 100 years using R and not spend too much time on such silliness:

Select All Code:
year<-1911:2011
firstOfJuly<-ISOdate(year,07,01) # all the first of July's of the last 100 years
y<-year[weekdays(firstOfJuly)=='Friday'] # which ones are Friday's?
y

the result:

Select All Code:
[1] 1921 1927 1932 1938 1949 1955 1960 1966 1977 1983 1988 1994 2005 2011

So even the event ‘July with 5 Fridays and 5 Sundays’ is way more frequent than what was claimed (every 623 years). And of course, since there always are the same numbers of days between the 1st of July and the 1st of October, the event ‘October with 5 Saturdays and 5 Mondays’ is true the same years, so as often and with the same frequency.

The event itself follows a regular pattern:

Select All Code:
diff(y)
[1] 6 5 6 11 6 5 6 11 6 5 6 11 6

This is really elementary and one wonders why it’s still passed on so easily. My guess is that it’s an example of availability heuristic: the fact is worded as being surprising and a quick mental check doesn’t provide a counter-example. (although in this case, it’s more the non-availability that induces the bias.)

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2 Responses to An unsurprising year

  1. Don Calpe says:

    >2011 is an unusual year. Add the last two digits of your birth year to the age you will >turn on your birthday this year and you’ll get 111!

    If you have children – born after 1999 , you’ll get 11 ! ;-) . So 111 for all born before 2000 and 11 for the “kids”.

  2. CL says:

    Correct and again very simple arithmetic. If your birth year y starts with 19, the last two digits are equal to y-1900 (e.g.1957-1900=57). Your age in 2011 is 2011-y. So adding the two together gets you: (2011-y)+(y-1900)=111.
    And if your birth year starts with 20, the last two digits are y-2000, so the sum is (2011-y)+(y-2000)=11.
    It has nothing to do with a property of 2011, it’s just a simple sum. Next year, the operation will yield 112 or 12 etc.

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