7 ms·

If my friends have more friends than me, do their friends have more friends than them? And if I'm my friend's friend, am I not included in their friends that ha

If my friends have more friends than me, do their friends have more friends than them? And if I'm my friend's friend, am I not included in their friends that have more friends?

lupire 4y ago
No, because the statement is about averages (normal people), not the outliers (celebrities) that average people are friends with.
- tshaddox 4y ago
  Celebrities are a much less counterintuitive example. The effect is quite obvious if you look at asymmetric relations like being aware of a famous person (or, say, Twitter follows). It's pretty obvious that a lot more people know Justin Bieber or Barack Obama by name than the pop star or U.S. President know by name. What's slightly less intuitive about the friendship paradox is that this also tends to be true with symmetric relations.
vikingerik 4y ago
The real title and the principle are "your friends more likely have more friends than you". It's not true in every case, and obviously not on both sides of a reciprocal pair.
- taneq 4y ago
  Exactly. The heavy lifting in this case is being done by the phrase “on average.” They could have just said “a small number of people have WAY more friends than usual.”
  - tshaddox 4y ago
    > They could have just said “a small number of people have WAY more friends than usual.” They do say that, but that's a way to explain why the friendship paradox is true. It's not some trivially equivalent statement. If the statements were trivially equivalent, then people wouldn't be complaining that one is counterintuitive while the other is obvious. The whole point is that one fact is a counterintuitive result of another more intuitive fact.
micah94 4y ago
Whenever you start talking about averages, you have to be careful. The average nuclear family having 2.5 kids, for example.
- Oxidation 4y ago
  More like 2.499 once you consider the mass defect when you fission the third child.
aidenn0 4y ago
If you select a random person, and then select a random friend of that person, most of the time, the latter has more friends. This is because friendship graphs tend to have a small number of highly-connected nodes and a large number of less-connected nodes. Therefore most of the connections are between a less-connected node and a more connected node. So, if you just select a random node, you are highly likely to select a less-connected node, which also means that if you follow a random connection from that node, you are likely to reach a highly connected node. So to answer your original question: No.
- taeric 4y ago
  That summary is still confusing to me. Rather, "If you select a random person, and then average how many friends all of their friends have, most of the time the latter will be higher." I don't think it holds nearly as strongly in the case of selecting a single neighbor. (I can't say it doesn't hold.)
  - aidenn0 4y ago
    Let's simplify this a bit. Divide people into "lots of friends" (L) and "few friends" (F). Most nodes are F, most edges are F<->L. Therefore if you randomly select a person, you probably end up with an F. Since most edges are F<->L, following an edge from an F is likely to get you to an L.
    - taeric 4y ago
      I could see a plurality of edges are actually F<->F, though? Such that most people you connect with are connected with everyone else you also connect with. You will just have a few people in any large group that are hyper connected outside of that group. Such that, as likely, you have the same number of friends as any given friend. Similarly, though, most likely you will have fewer friends than the average of all of your friends.
      - aidenn0 4y ago
        Even in that case you increase your chances of selecting an L by using this algorithm as long as (number of F nodes) >> (number of L nodes).
tshaddox 4y ago
Well, no, it doesn't work like that. There's already not some guarantee that your friends will have more friends than you do. Obviously some of the people have to be the ones that have lots of friends, and you could very well be one of those people. The point is just that most people aren't those people, so most people who read these articles and check their friend graphs will go "whoa, it's true!"
adammarples 4y ago
No their friend's friends do not have more friends than them. Some people have an unusually large amount of friends, what this means practically is that they are more likely to be one of your very few friends, because they touch so many smaller groups, yours included. When they say most people's friends have more friends than them, it's the same very popular people being counted again and again by each unpopular person.

If my friends have more friends than me, do their friends have more friends than them? And if I'm my friend's friend, am I not included in their friends that have more friends?

lupire 4y ago

No, because the statement is about averages (normal people), not the outliers (celebrities) that average people are friends with.

tshaddox 4y ago
Celebrities are a much less counterintuitive example. The effect is quite obvious if you look at asymmetric relations like being aware of a famous person (or, say, Twitter follows). It's pretty obvious that a lot more people know Justin Bieber or Barack Obama by name than the pop star or U.S. President know by name. What's slightly less intuitive about the friendship paradox is that this also tends to be true with symmetric relations.

vikingerik 4y ago

The real title and the principle are "your friends more likely have more friends than you". It's not true in every case, and obviously not on both sides of a reciprocal pair.

taneq 4y ago
Exactly. The heavy lifting in this case is being done by the phrase “on average.” They could have just said “a small number of people have WAY more friends than usual.”
- tshaddox 4y ago
  > They could have just said “a small number of people have WAY more friends than usual.” They do say that, but that's a way to explain why the friendship paradox is true. It's not some trivially equivalent statement. If the statements were trivially equivalent, then people wouldn't be complaining that one is counterintuitive while the other is obvious. The whole point is that one fact is a counterintuitive result of another more intuitive fact.

micah94 4y ago

Whenever you start talking about averages, you have to be careful. The average nuclear family having 2.5 kids, for example.

Oxidation 4y ago
More like 2.499 once you consider the mass defect when you fission the third child.

aidenn0 4y ago

If you select a random person, and then select a random friend of that person, most of the time, the latter has more friends. This is because friendship graphs tend to have a small number of highly-connected nodes and a large number of less-connected nodes. Therefore most of the connections are between a less-connected node and a more connected node. So, if you just select a random node, you are highly likely to select a less-connected node, which also means that if you follow a random connection from that node, you are likely to reach a highly connected node. So to answer your original question: No.

taeric 4y ago
That summary is still confusing to me. Rather, "If you select a random person, and then average how many friends all of their friends have, most of the time the latter will be higher." I don't think it holds nearly as strongly in the case of selecting a single neighbor. (I can't say it doesn't hold.)
- aidenn0 4y ago
  Let's simplify this a bit. Divide people into "lots of friends" (L) and "few friends" (F). Most nodes are F, most edges are F<->L. Therefore if you randomly select a person, you probably end up with an F. Since most edges are F<->L, following an edge from an F is likely to get you to an L.
  - taeric 4y ago
    I could see a plurality of edges are actually F<->F, though? Such that most people you connect with are connected with everyone else you also connect with. You will just have a few people in any large group that are hyper connected outside of that group. Such that, as likely, you have the same number of friends as any given friend. Similarly, though, most likely you will have fewer friends than the average of all of your friends.
    - aidenn0 4y ago
      Even in that case you increase your chances of selecting an L by using this algorithm as long as (number of F nodes) >> (number of L nodes).

tshaddox 4y ago

Well, no, it doesn't work like that. There's already not some guarantee that your friends will have more friends than you do. Obviously some of the people have to be the ones that have lots of friends, and you could very well be one of those people. The point is just that most people aren't those people, so most people who read these articles and check their friend graphs will go "whoa, it's true!"

adammarples 4y ago

No their friend's friends do not have more friends than them. Some people have an unusually large amount of friends, what this means practically is that they are more likely to be one of your very few friends, because they touch so many smaller groups, yours included. When they say most people's friends have more friends than them, it's the same very popular people being counted again and again by each unpopular person.