Re: Triplets
- From: msb@xxxxxxx (Mark Brader)
- Date: Fri, 04 May 2007 17:32:31 -0000
Peter Moylan:
I've known twins who were born in the more typical way - within a few
minutes of each other - but who still had birthdays on different days.
Sara Lorimer:
Ooh -- imagine if one was born on February 28, and the other on February
29.
Zamarrafack, a friend once said to me in email something like "Imagine,
both twins are the same age! What's the probability of *that*?" He was
joking, but I realized that this was actually an interesting
question, and answered as follows.
It depends on the probability distribution of the time interval
between the two births, and I have no idea what that is. However,
I do know that it's very rare for the births to occur more than a
fraction of an hour apart (though, particularly in these days of
modern medicine, it's actually possible for them to be weeks apart),
so I'm going to make the simplifying assumption that they'll always
be less than a day apart -- also that all dates of birth are
equally probable, which definitely is not really true, and that
the births do not occur on a vehicle crossing the International
Date Line. [Added in posting: since the usual convention is that
your age changes when your birthday starts, it also matters whether
the twins are in the same time zone then, and I assumed they were.]
Let's suppose that the average time between the births is 5 minutes;
the actual probability of them NOT being the same ages will vary in
proportion to the correct number.
If it is 5 minutes, then the chance is 287/288 that they were born on
the same day, and hence have the same age always.
The remaining 1/288 probability now splits into four cases: the days
could be February 28 and 29 (probability 97/146097, in the Gregorian
calendar), February 29 and March 1 (same probability), February 28
and March 1 (303/146097), and none of these (145600/146097).
For the two February 29 cases, though, there is a psychological element:
it depends on how that twin celebrates birthdays in non-leap years.
For instance, I think there would be a certain appeal to having separate
birthdays only in leap years. But I understand that the common practice
is to treat a February 29 birthday as falling on February 28 in non-leap
years. In that case, the case of February 29 and March 1 can be
combined with the common case, which increases its probability to
145697/146097.
In the common case, they have different ages if it is now the older
twin's birthday (400/146097), and hence the same age with probability
145697/146097.
In the case of February 28 and March 1, they have different ages if
it is now either February 28 or 29 (497/146097), and hence the same
age with probability 145600/146097.
And in the case of February 28 and 29, with the latter treated as
a February 28 birthday in non-leap years, then they have different
ages only if this is February 28 of a leap year, probability 97/146097,
and hence the same age with probability 146000/146097.
So the answer (subject to the qualifications already stated) is
(287 + (145697^2 + 303*145600 + 97*146000)/146097^2) / 288
= (287 + 21285894609/21344333409) / 288
= 6147109582992/6147168021792
[but there is a common factor of 2337552(!) there...]
= 2629721/2629746
Hope this helps.
(The probability that they have *different* ages under these assumptions
is therefore 25/2,629,746, or about 1/105,200.)
--
Mark Brader, Toronto cat>/dev/null got your tongue?
msb@xxxxxxx -- Jutta Degener
My text in this article is in the public domain.
.
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