Mathematical Marginalia
Classical and Quaint Topics in Mathematics

The Engagement Party

On occasion of its engagement a couple wants to throw a big party. All sisters and brothers of the bride and the bridegroom shall be invited, plus all the partners of these siblings, plus all the sisters and brothers of these partners, plus all siblings of these partners and so on.

How many people will come to the party on average, if we take into account the average size of a family and the percentage of single persons from the available statistical data in Germany (2005-2010). To keep things simple (but less realistic), let's assume that all of the kinsfolk consists of adults, so that we may take the percentage of singles found in Wikipedia (as of 2010), which is 43%. In addition, we have to assume that nobody had passed away (possibly leaving a widower or a widow).

The average family size is also found in Wikipedia on basis of the 2005 micro census.

Children Percentage
1 52
2 36
3 9
4 2
5 0.6
6 0.2
7 0.08
8 0.03
9 0.014
10 0.006
>10 0.005

The percentage is computed relative to the number of families with children. Those without children are not relevant for our considerations, because every partner of one of the siblings necessarily has been born into a family with children.

We want to find out the probabilities, that 0 guests, 1 guest, 2 guests etc. have to be invited to the party. To do this, we represent the particular net of relations by a kind of Feynman diagram: Two persons connected by a partnership will be joined by an undulated line (a photon), siblings will be joined by a line with an arrow (a fermion).

Bride and groom, which are not counted as guests, will be joined by a gluon, to accentuate their connection.

If bride and groom have both one single sibling, the diagram would look as follows:

If one of the couple has two single siblings and the other a sibling in partnership with an only child, we have:

The rules are, that only one  photon or gluon may emerge from a point (no bigamy), and that no additional fermion may be connected to a point at which a fermion arrives(count siblings only once).

As with proper Feynman diagrams, the probability of a configuration can be computed by taking the probabilites of the particular connections and multiplying them. Of course, you must not forget to take the empty connections (representing singles or only childs) into account. The only connection not counted is that of the couple itself, because its probability is 100% by assumption.

The first diagram (both bride and groom are only childs) yields:

and thus a probability of $52\%\times52\% = 27\%$. In $27\%$ of all configurations the couple has nobody to invite!

For the second diagram we compute:

The probability that bride or groom have a sister or brother is $36\%$ for each, the probability that this sibling is single is $43\%$ for each. Hence, he total probability of the configuration (which results in two guests) is \[43\%\times36\%\times36\%\times43\%=2.4\%. \] To solve our problem, we have to draw all possible Feynman diagrams, which result in a certain number of guests, compute the probability of each of these diagrams and sum them up. Thus we get the probability that exactly this number of guests has to be invited. In case of $0$ guests we are already done, and the probability is $27\%,$ because there is only one diagram resulting in $0$ guests. Let's compute the probabilites in case of one guest, two guests and 3 guests: If there is exactly one guest to invite, we have the diagram

and its mirror image

Both configurations yield the probability \[ 36\%\times 43\%\times 52\%=8\% \] so, the probability that exactly one guest is on the list, sums up to $16\%.$ Note, that for every non symmetrical configuration there is always a mirror image relative to the gluon with identical probability. In what follows we won't draw these mirror images for their own, but include their probability with a factor of $2.$ In case of two guests there are more possibilities, of course. At first we have the diagram, already taken as an example, with two single siblings:

This diagram is symmetrical, hence its probability of $2.4\%$ is counted only once. The second diagram may be called only child marries into a family with three children.:

Here the probabilities \[9\%\times 43\% \times 43\% \times 52\% = 0.86\%, \] must be counted twice, because there is a mirror image of this diagram. Note, that the probability for the existence of two siblings ($9\%$) is only counted once, while the fact that each sibling is a single (with probability $43\%$) has to be counted for each sibling separately. The third diagram (sibling has a partner, which is an only child ) looks like that:

Again, we have a symmetrical configuration, so the probabilty of \[ 52\% \times 36\% \times 57\% \times 52\% = 5.5\% \] has to be taken twice. On the whole we get the probability \[ 2.4\% + 2\times 0.86\% + 2\times 5.5\% = 15.1\% \] that exactly two  guests have to be invited. With three guests the method starts to become slightly intricate. We will show only the diagrams without further commentary:

All of the above diagrams have a non identical mirror image, so the probability that exactly three  guests have to be invited computes as \[ 2\times 0,25\% + 2\times 2\% +2\times 0,6\% + 2\times 2\% + 2\times0.08\% = 9,86\% \] We refrain from listing the diagrams, which yield exactly four guests. Instead we print a table, generated by a JavaScript program. The program produces diagrams at random, taking into account the probabilities of the family sizes and the percentage of singles. For every diagram the number of guests is counted and added to the corresponding entry within the table. You may try the program with variant input on ⇒ this page. With 1000000 random samples you may get the following results or the like:

Gäste Anteil
0 27.03
1 15.72
2 15.19
3 9.58
4 7.86
5 5.42
6 4.22
7 3.13
8 2.42
9 1.88
10 1.46
11 1.13
12 0.91

You see that the results are in pretty good agreement with the theoretical values (27%, 16%, 15.1% and 9.86%). Its surprising that the couple will invite only three guests  on average! An engagement party with only siblings and their partners, no matter how far you run into the ramifications, will surely be a boring experience. It would be a good idea, to invite some friends (with their partners) too.