Odd coincidence: two people with same name living on the same street

[Mark44]

Many years ago, my cousin received a phone call asking him if his name was Bill Perkins (not his real name, for security reasons). He replied that it was. The caller also asked if he had a Saab automobile, which he did. The caller then asked if he had a three-legged dog, to which the answer was also yes - the dog's name was Tripod, for obvious reasons. The caller then went on to ask about the grading of an exam in his Econ course. My cousin replied that he was the wrong Bill Perkins, and that the caller must have intended to reach the other Bill Perkins, who lived a few houses down on the same street. Something to note is that there were only 41 houses on that street.

In a recent conversation with my cousin, he told me this story, and said he thought the caller's confusion was the result of an amazing coincidence. My cousin attempted to work out the odds of such a coincidence and brought this up to me due to my mathematics training.

Although my cousin had worked out some calculations that included estimates for how many Saabs there were in the country at that time, and how many three-legged dogs there might be, all I'm interested in is some guidance on how we might calculate the probability of having two men with the same first and last names living on a street with only 41 homes.

The scenario that I've laid out occurred in 1986, so the figures my cousin used were, I believe, appropriate for that time.
U. S. population in 1985: ~240,000,000
Proportion of people in US with surname "Perkins" : 0.15% (gotten from a table of most-common surnames in the US)
Approx. no. of people in US with surname "Perkins: 240,000,000 X 0.15% = 360,000
Approx. no. of males in US with surname "Perkins": 360,000 X 0.5 = 180,000
First name of Bill or William is or was very common at the time, at about 1% of boys born at this time given this name.
Approx. no of males in US with surname "Perkins" and first name William (Bill): 180,000 X 0.01 = 1,800

My cousin then made the assumption that owners of the homes on that street would be 22 years of age or older, arriving at a figure of 70% of the population in the previous paragraph.
Approx. no. of males in US named William "Perkins" of age >= 22: 1800 X .7 = 1,260

So far, so good. Where I got stumped on this problem was calculating the probability that of all of the houses/homes in the US, two men with essentially the same name happened to live in two of the 41 homes on that street.

Would the calculation involve an assumption that males with that name and age cohort be normally distributed amongst all of the homes/residences of the US? Would the calculation involve some sort of combinatorics?

My cousin did a calculation, but since it didn't involve the matter of the 41 houses on that street, I didn't take things any further. If you have some advice, it would be much appreciated.

 

[DaveE]

I suspect people named Kowalski are more common in the Chicago area than near San Diego, for example.
Of course if it's just a math problem, you can make whatever assumptions you like.

 

[Hornbein]

Would the calculation involve an assumption that males with that name and age cohort be normally distributed amongst all of the homes/residences of the US? Would the calculation involve some sort of combinatorics?

Edited to correct mistake.


What you want is the Poisson distribution. Assume that the likelihood that BP lives in a given house is the same everywhere in the USA. That's not true but is the best we can do. Then divide the houses of the USA into disjoint 41 house blocks. That's somewhat bogus too but simplifies things. AI says there were 60 million home owners in the USA in 1986. AI gave me inconsistent answers but I'm not willing to search out a reliable source. That makes 1.5 million disjoint blocks of 41 homes.

1260/60,000,000 = 0.000021 chance a homeowner is Bill Perkins
0.000021*41 = 0.000861 average number of Bill Perkinses in a block of 41

Find out that the Poisson calculators on the Internet can't do this calculation. Calculate from the formula, which pretty much means we square this number and divide by 2.

Chance of two or more Bill Perkins homeowners in a block of 41 is 3.71 ×10-7

Now we've got a binomial distribution : either a block contains two or more Bill Perkinses or it doesn't. Calculate the probability that this occurs (1-(3.71 ×10-7))^(6*10^7), which is the chance that this coincidence does not occur anywhere in the USA. It's about 97% So chance of this coincidence occurring somewhere in the USA might be 3%.

 

[Mark44]

What you want is the Poisson distribution.

I had the same thought that a Poisson distribution might be a viable approach when I was discussing this with my cousin (but didn't mention it in my post). Thanks for looking into this!

 

[jedishrfu]

Wouldn't this be solved in the same manner as the birthday problem? You know the one where you're at a party of 20 guests, and the question is, what are the odds that two people have the same birthday?

The solution was found by computing the odds that no one shared a birthday. In your case, you have a few more coincidences to consider, where you compute the probability that no one matches the coincidences, which leads you to the probability that two people did.

It looks like there are 4 coincidences:
- same lastname
- same make of car
- same-legged category of dog
- same street name

Also, did your cousin ever believe he was being pranked? especially considering the caller said it was related to an econ class (lots of stats and these kinds of problems).

What is odd is that if the other guy were a college professor or teacher, then how would the student know about the dog unless the teacher mentioned it in class?

The student would get the last name from the class instructor, the car make by observing what he drove to work in, the dog because the professor mentioned it, and the street name by looking it up in the phonebook, where both Perkins would have been listed under the same street name.

However, I'd discount the street address, figuring the desperate caller would have tried all persons named Perkins listed in the phone book. I figured the student wouldn't know the professor's street or home address.

The coincidence buckets would be:
- 20 common last names
- 20 car makes
- 4 categories

Then the 50% point is at 47 people ie that in a group of 47 people, two of them will match the three buckets above.

 

[Hornbein]

I had the same thought that a Poisson distribution might be a viable approach when I was discussing this with my cousin (but didn't mention it in my post). Thanks for looking into this!

I repaired an error in my calculations. I seldom get it right the first time.

 

[Hornbein]

Wouldn't this be solved in the same manner as the birthday problem? You know the one where you're at a party of 20 guests, and the question is, what are the odds that two people have the same birthday?

That's right. I incorrectly thought the stated problem is whether there are two Bill Perkinses but it's actually whether any two homeowners have the same name. That's about 2%, assuming that there are about 50000 possible names.

 

[jedishrfu]

I based it on the top twenty names in the US in the 1980s. The top name, Smith, was 1%, the top ten names covered 6-7%, and the top 100 names covered 20% of the population. My guess is the cousin's name is in the top 10 names, but I relaxed it to the top 20.

But honestly, I think it's a prank.

There have been other stories of magical coincidences where a hidden connection wasn't shown. One example is the woman who was raised in the Midwest, moved to Paris, and found her old storybook in a secondhand shop. Later, it was discovered that her nanny had also moved to Paris years earlier, and that the nanny's book collection had been donated to charity.

Caveat: This is likely a made-up story, though I remember reading about it and that it was claimed to be real.

Carl Jung used these kinds of stories when he talked about synchronicity ie a coinkydink.

 

[Mark44]

But honestly, I think it's a prank.

To the best of my knowledge, it wasn't a prank. There actually was a person with the same name as my cousin, but different middle name, and living on the same (short) street. According to my cousin, the other namesake also had a Saab as well as a three-legged dog.

 

[jedishrfu]

Interesting, let's hope your cousin wasn't pranking you, knowing your math background.

 

[jedishrfu]

There are a couple of semi-related variations of this problem.

—-

The amazing Randy gave a class on the validity of astrological profiles. He told his class of 30 people that he had submitted their birthdays to a well-respected astrologer, and that the envelopes he's about to hand out contain personalized astrological profiles for each student. He waits until they've read their profiles and then asks, "What did you think? Were they accurate?"

The students said yes; the astrologer was right in more than 50% of cases. He asked what the likelihood was that they’d be that accurate, and they said maybe about 8-9% correct, based on the notion that there are 12 months in the astrological calendar.

Randy then told the students to exchange papers, and they were shocked to find that all the papers were identical. They realized they'd been fooled. The profile attributes were all binary, like you have brown eyes, you have fair hair, … so of course, the astrologer would get it right about 50% of the time.

—-

A person gets an email from a sports prediction company. They claim they can predict who will win a series of 5 games.

They send out two emails: one to the first group saying Team A wins, and one to the second group saying Team B wins. If team A wins, then they focus on that group. They repeat the predictions, keeping the email lists that contained all true predictions. Trust is built.

After a few emails, they offer a paid prediction. Many recipients pay, and the scammer wins. If only they knew it was a scam.

—-

Military officers were polled on what they felt a great general might be. All agreed that they must win 5 battles in a row. When asked how many generals fit that description, the officers responded that it was the top 3%.

Given a pool of generals, one out of 32 would win all 5 battles, which, when converted to a percentage, is 3-4%.

Our notions of probability collide with our intuition.