Last year, just before my birthday in February, I noticed something unsual. The last two digits of my birth year were the same as my age. How old was I then and was this an unusual occurrence?
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$\begingroup$ When you say last year I assume you mean 2013? $\endgroup$– skvSep 24, 2014 at 2:25
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$\begingroup$ Yes, my age prior to my birthday in 2013 $\endgroup$– JosieASep 24, 2014 at 3:18
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$\begingroup$ this has a name: Beddian birthday (go ahead, search on it) after Bobby Beddian markbialczak.com/2014/03/19/… $\endgroup$– Kate GregorySep 24, 2014 at 15:44
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$\begingroup$ Wow that was an interesting story! And I am pleased to find someone else was fascinated with the age/year quirk. Also that it now has a name. Thanks Bobby Beddian. markbialczak.com/2014/03/19/… $\endgroup$– JosieASep 25, 2014 at 3:54
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$\begingroup$ I'm kind of sad that this won't happen to me until 2086, and that I don't even know if I'm going to live that long. $\endgroup$– user88Sep 25, 2014 at 17:04
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2 Answers
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Generally, people born in $1900 + n$ will have this birthday in the year $1900 + 2n$, when they are $n$ years old. The same can be applied to any century. You'll notice this is an even number; the year in the question is odd because you noticed this just before your birthday. That is, the Beddian birthday (see Kate Gregory's link in the quetion) was in 2012, but you noticed the phenomenon in the next year, 2013.
Working backwards, if this year is $c + m$ (where $c$ is a century such as 1900 or 2000 and $m$ is even and less than 100), those with a Beddian birthday this year were born in $c + \frac{m}{2}$ or $(c - 100) + \frac{m + 100}{2} = (c - 100) + \frac{m}{2} + 50 = c + \frac{m}{2} - 50$.
You can obtain the second option by "going back" a century, making $c - 100$ the new century part and $m + 100$ the new "$2n$" part from the first paragraph. Since $m < 100$, $\frac{m}{2} + 50 < 100$, and so the year is valid. You can't go back another century, or else the "$m$" portion will be more than two digits, which wouldn't make any sense.
Thus, for the year 2012: $$ c = 2000, m = 12\\ \text{birth year is } \boldsymbol{6}=\frac{m}{2} \text{or } \boldsymbol{56}=\frac{m}{2} + 50 $$
As Rob Watts notes, you're more likely to have been 56 than 6. Additionally, this happens every even year for those born turning $(year ~mod~ 100)/2$ and the same age plus or minus 50.
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$\begingroup$ Well that was a very thorough analysis of probabilities. Impressive. $\endgroup$– JosieASep 25, 2014 at 2:23
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You were born in either 1956 or 2006 and were 56 about to turn 57 or 6 about to turn 7. I'm guessing that you're not 8, so you were 56 about to turn 57.
As for how unusual it is for this to happen, it is very common - everyone born in 1956 and 2006 (except for those born on January 1st) were all 56 and 6 respectively on January 1st, 2013. It is a little less common toward the end of a century though. For example, in 1989 people who were 44 or 94 could have noticed the same thing. The only reason it was less common then is because fewer people live to be 94.
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$\begingroup$ Yes. I was born in 1956. I hadn't thought of the other option! $\endgroup$– JosieASep 25, 2014 at 2:22