Last Name Standing

Submitted for your consideration: In a traditional western society, will there eventually be only one last name left?

Assume the following simplifications:
- 100 married couples with 100 different last names.
- each family on average has 2.0 children (evenly distributed from 0 to 4)
- 50% girls, 50% boys
- there is only heterosexual marriage
- each girl or boy will marry one other boy or girl in the same generation
- the newly created family will always take the husbands last name
- the process repeats for every generation

Question: On average, how many generations will it take until there are only 10 last names left.

Bonus: Find out how long it takes on average for only 1 last name to remain, although humanity surely won't exist that long.

Only three surnames in Korea, I'm told.

Mainly four surnames in Amish country Lancaster, PA.

Iceland would be a mess for calculations.

Its usually sixty percent girls, forty percent guys.

Many women refuse to adopt the husband's name.

Some women create new surnames such as SaraDaughter or Hypenated This That.

Not everyone marries.

Most women wind up preferring the dogs to their husbands who turns out to be the only child they have that never grows up and leaves home.

Well, the first step is to acknowledge that 20% of the names disappear from the first 100 in the first generation, since on an even distribution of 0 to 4, 20 couples won't have kids to carry on the name.

Additionally, another 20 couples will have only 1 child, and 50% of that category will be female. Therefore, only 10 males will be carrying on the name.

We go to 2 kids for the next 20, resulting in 20 names.

This is where it gets tricky - 3 kid couples - each couple will have 1 male, 1 female, and 1 50%. So, 20 names, BUT 10 of those names will be x2.

So, thus far, 10 names from 10 males, + 20 names from 20 males, + 10 names from 20 males AND 10 names from 10 males. So, thus far, 50 names-60 males.

Lastly, 4 kid couples, which gives us 20 names from 40 males.

This is the first generation's spawn results:

40 males - unique names
10 x 2 males with Name (N1)
20 x 2 males with Name (N2)

Here, it gets to complex for my abilities - You would have to assume that each section is equally divided by 5...

So, 8 - 2 - 4 of the categories above are 0,1,2,3,4 kid names - and the math gets really intense after that, because there are going to be some situations that I imagine would be indivisible by 2. If you wind up with 1 unique name having 3 kids, then what?

Quote: 21Flip

Well, the first step is to acknowledge that 20% of the names disappear from the first 100 in the first generation, since on an even distribution of 0 to 4, 20 couples won't have kids to carry on the name.

Additionally, another 20 couples will have only 1 child, and 50% of that category will be female. Therefore, only 10 males will be carrying on the name.

We go to 2 kids for the next 20, resulting in 20 names.

This is where it gets tricky - 3 kid couples - each couple will have 1 male, 1 female, and 1 50%. So, 20 names, BUT 10 of those names will be x2.

So, thus far, 10 names from 10 males, + 20 names from 20 males, + 10 names from 20 males AND 10 names from 10 males. So, thus far, 50 names-60 males.

Lastly, 4 kid couples, which gives us 20 names from 40 males.

This is the first generation's spawn results:

40 males - unique names
10 x 2 males with Name (N1)
20 x 2 males with Name (N2)

Here, it gets to complex for my abilities - You would have to assume that each section is equally divided by 5...

So, 8 - 2 - 4 of the categories above are 0,1,2,3,4 kid names - and the math gets really intense after that, because there are going to be some situations that I imagine would be indivisible by 2. If you wind up with 1 unique name having 3 kids, then what?

Even the two-kid couples might have two girls, thus resulting in extinction, same applies to the 3+ kid families.

Probably the best way to solve this is by simulation. I was hoping somebody here had the tools for it. For a math solution, I changed the premise to a Poisson distribution (for the number of boys) to make use of the Galton-Watson process.

I get a median number of generations until one last name remains of 117 generations (and I am much too lazy to post the math here). However, it may not apply to my original question since I am pretty sure the Galton Watson process assumes surnames within a (much) larger population. I can't find confirmation for this anywhere on the web though.

I seem to remember that there is some threshold below which you run to extinction, and one (the same or another) from which you are (almost) sure that the name maintains. The thereshold depends on population size.
In such a small population, I think only one name will remain.

Countries with few surnames are countries where family names did not exist until recently. In modern times (XIXth century, mostly) the administrative needs called for laws requiring each person to adopt a family name. The first documented example is for Jews in France: they were granted citizenship by Napoleon. Most took names like Levy, Cohen and the such. Then in East Europe, Goldberg, Blumenthal, etc.

This happened in Scandinavia (hence all the <Forename>sson or -sen), in Spain (earlier though) and in many colonial countries.

If there are 0 or 3 children, how do you determine the gender mix. If you say it is 50/50 over the whole generation of kids, what if the total is an odd number?

As was noted, extinction is a definite possibility. In fact, I would argue that the probability of extinction is 100% given an infinite amount of time.

Quote: Canyonero

Even the two-kid couples might have two girls, thus resulting in extinction, same applies to the 3+ kid families.

You stated 50% boys, 50% girls in your premise, without stipulating the distribution method - whether over the entire 200 children, or within each family.

Quote: 21Flip

You stated 50% boys, 50% girls in your premise, without stipulating the distribution method - whether over the entire 200 children, or within each family.

For a simulation, it should be random. After offspring number is determined per family, 100 boys and 100 girls would be randomly assigned.

Quote: Wizard

If there are 0 or 3 children, how do you determine the gender mix. If you say it is 50/50 over the whole generation of kids, what if the total is an odd number?

In my original example, there would always be 200 children... - see post above

Quote: Wizard

As was noted, extinction is a definite possibility. In fact, I would argue that the probability of extinction is 100% given an infinite amount of time.

Indeed, the probabilty of extinction tends towards 1. After 197 generations, the extinction probability for each of the 100 original names becomes > 99% (according to the Galton - Watson process).

Quote: Wizard

If you say it is 50/50 over the whole generation of kids, what if the total is an odd number?

To average 2.0 children per couple, would the total number of children per generation not have to be 200? (I don't understand how that could be odd.)

As I understand it, it was stipulated that each generation is 200 people. To get the 50/50 male/female mix, each generation is 100 boys, 100 girls.

20 of the couples in each generation will have 0 children. (0)
20 of the couples in each generation will have 1 child. (20)
20 of the couples in each generation will have 2 children. (40)
20 of the couples in each generation will have 3 children. (60)
20 of the couples in each generation will have 4 children. (80)
(Sums to 200 children per generation.)




Sex is hereditary. If your parents didn't have it, you probably won't either.

Quote: Canyonero

...
Assume the following simplifications:
...
- each family on average has 2.0 children (evenly distributed from 0 to 4)
- 50% girls, 50% boys
...

Using these two assumptions, I find the the following probabilities of one male having a certain number of male children:
0.3875 = 31/80: no boys
0.3250 = 26/80: one boy
0.2000 = 16/80: two boys
0.0750 = 6/80: three boys
0.0125 = 1/80: four boys
1.0000 = Total

So, an average of 38.75 names would be eliminated for the first generation from children.

From the above probabilities, it is relatively easy to calculate the probability of n fathers' (all with the same name) producing m sons. (The method is similar to one used to calculate the probabilities of the sum of n dice.) Below is a table of probabilities of n = 0 to 16 sons for n = 1 to 4 fathers:

Sons column	1	2	3	4
0	0.3875	0.1502	0.0582	0.0225
1	0.3250	0.2519	0.1464	0.0756
2	0.2000	0.2606	0.2129	0.1417
3	0.0750	0.1881	0.2192	0.1878
4	0.0125	0.0984	0.1722	0.1923
5		0.0381	0.1071	0.1591
6		0.0106	0.0536	0.1091
7		0.0019	0.0215	0.0628
8		0.0002	0.0069	0.0306
9			0.0017	0.0126
10			0.0003	0.0043
11			0.0000	0.0012
12			0.0000	0.0003
13				0.0001
14				0.0000
15				0.0000
16				0.0000

By the way, the expected number of sons for n fathers comes out to n as it should, and the variance for the number of sons comes out to be n also. (This brings to mind the Poisson distribution, whose variance is equal to its mean, although these numbers do not match the Poisson distribution.)

To find the number of names eliminated for the second generation, I would multiply the probabilities in cells 1-4 in the n = 1 father column by the corresponding cells in the m = 0 sons row. 0.3250*0.3875 + 0.2000*0.1502 + 0.0750*0.0582 + 0.0125*0.0225 = 0.1606. So a further average of 16.06 names would be eliminated for the second generation. Using a similar procedure, I find that an average of a further 9.11 names would be eliminated for the third generation. However, the method using Excel to exactly calculate more generations is too cumbersome because the number of cells needed grows quickly with each generation.

I think simulation is the best way to solve your problem. Those probabilities 1/80, 6/80, 16/80, 26/80, and 31/80 would be useful in that regard.