I thought I'd ask my eldest son who is a graduate student in mathematics...
So, here's how it works. Let's consider a trait and we are examining collections of n objects to see if any two objects in the collection both have the same version of that trait.
In the Birthday Problem, the trait is "day of birth", of which there are 365. So, letting our trait come in T possible versions, let's write f(T,n) to denote the probability that a collection of n objects possessing various versions of a given trait (which comes in T total possible number of version) contains a pair of objects with the same version of the trait.
For the Birthday Problem, the probability of there being a pair of people among n people possessing the same birth day is given by f(365,n). This is because there are n people in consideration, and because the trait we are measuring (day of birth) comes in 365 possible varieties.
For the situation under consideration here, the probability is applying to following three pieces of information:
A day of the month
A month
A year.
In particular, note that the production date identifier can be thought of as:
One number between 1 and 31 (the day of the month), followed by one name out of twelve (the name of one of the twelve months), followed by a year.
The watches in question were only capable of being produced during the same year for a period of 5 years ('64-'68). Thus, to measure the probability of any two watches out of n watches having the same year among those 5, we use f(5,n).
In those 5 years, all 12 months are present, so, the probability of n watches having the same "month of production" is given by f(12,n)
Finally, we have to deal with the days of the month. Here, the calendar causes us some grief because not all months have the same number of days. However, given that I don't want to have to spend TOO much time on this, let's simplify the problem and assume that all 12 months have the same number of days: 31.
Thus, in examining whether there is a pair of watches in a collection of n watches that have the same day number on their production date, the probability of this is given by f(31,n).
To find the probability of all three of these events happening simultaneously, we multiply them:
P(n) = f(31,n) * f(12,n) * f(5,n)
Thus, P(n) is the probability that a collection of n watches (in an ideal world where every month has 31 days) share the same production date
If you're curious, f(T,n) = T! * T^(-n) / (T-n)!
This then gives
P(7) ≈ 47%
P(8) ≈ 61%
So, if you have 7 watches (all of which were made within a single 5-year period), there's about a 47% that there will be a pair of watches among those 7 with the same production date. The question is how you define the events. If you want to treat each individual combination of day, month, and year as an independent entity, P(n) = f(5*365,n), then you'd need approximately 50 watches to have a 50% chance of finding two watches that share same production date. If you treat the production date as a composite of three deterministic quantities; "day of month", "month", "year'... you will get the first answer, however, if you treat each production date as a single entity you get the second answer which is 50 watches. So, he says...
