Generalized birthday problem plus spouse Suppose a group of persons, say of size N (who are all married), are in a room. For any given N, I would like to know the probability of at least two people in the room sharing the same birth date (day/month/year) AND their spouse sharing the same birth date as another person's spouse. Assume that there is an equal chance of a person and their spouse being born on any day of the year and any year between 1930 and 2000 (a 71-year span, or nearly 26,000 possible birthdays, all equally likely).
Thanks. :)
Edit: Sorry, my English was off. I wanted: "...the probability of at least two people in the room sharing the same birth date (day/month/year) AND their spouses both sharing the same birth date (which may or not be the same as their own spouses birth date)." See comments for further clarification.
 A: The probability is not very easy to find. But, I can find the expected number of couples that have matching birthdays.
There are $M=25933$ days between 1/1/1930 and 12/31/2000. We can just assume that the numbers $1$ through $M$ denote these days. So, if I say my birthday is the number 2, then it is understood my birthday is 1/2/1930, etc.
Assume there are $N$  people, all are married couples.  Therefore, there are $N/2$ couples.
Order the couples arbitrarily from $1$ to $N/2$.
Within a married couple, order the birthdays arbitrarily to make ordered pairs. So, the birthdays can be written as
$$(x_1,y_1), (x_2,y_2), ..., (x_{N/2},y_{N/2})$$
where all the $x_i$ and $y_i$ are numbers between $1$ and $M$.
The event the second couple has matching birthdays with the first couple can happen in three ways:
$x_1=y_1=x_2=y_2$
$x_1 \ne y_1 \& x_1=x_2 \& y_1=y_2$
$x_1 \ne y_1 \& x_1=y_2 \& y_1=x_2$
Hence, the probability of matching is the sum of the probabilities of those 3 events:
$$\frac{1}{M^3}+\frac{M-1}{M} \frac{1}{M^2}+\frac{M-1}{M} \frac{1}{M^2}=\frac{1}{M^3}+2\frac{M-1}{M^3}$$
This is also the expected value of the number of matching couples from any two couples.
There are ${N/2}\choose{2}$ different ways to form sets of two couples.  Therefore, the expected number of matching couples is
$${{N/2}\choose{2}} \left( \frac{1}{M^3}+2\frac{M-1}{M^3} \right)$$
If you started with, say, 1 million people (500,000 couples), the expected number of pairs of couples that would have matching birthdays is about 0.06.  Since this is a very rare event, you will almost never see 2, 3, 4, etc. pairs of matching couples.  You can assume it will almost always be either 0 or 1 pair matching. If you repeated this over and over again, you would see 1 pair of couples about 6% of the time and 0 couples 94% of the time.
If you want to consider months and days only, I would ignore February 29 and take $M=365$. With $N=1000$, you would find the expected number of pairs is about 1.87.  So, now you might be able to see 2, 3, 4 etc. such pairs of matching birthday couples. So, you can't use this approach to approximate the probability of seeing 1 or at least 1 pair of couples with matching birthdays.
