# Game of Friendship Paradox

# Game of Friendship Paradox

### The paradox is that your friends probably have more friends than you. We take a closer look at this head-scratcher and create some data visualization using code.

Join the DZone community and get the full member experience.

Join For Free**The open source HPCC Systems platform is a proven, easy to use solution for managing data at scale. Visit our Easy Guide to learn more about this completely free platform, test drive some code in the online Playground, and get started today.**

In the introduction of my course next week, I will (briefly) mention networks, and I wanted to provide some illustration of the Friendship Paradox. On network of thrones (discussed in Beveridge and Shan (2016)), there is a dataset with the network of characters in Game of Thrones. The word “friend” might be abusive here, but let’s continue to call connected nodes “friends.” The friendship paradox states that:

People, on average, have fewer friends than their friends.

This was discussed in Feld (1991) for instance, or Zuckerman & Jost (2001). Let’s try to see what it means here. First, let us get a copy of the dataset:

```
download.file("https://www.macalester.edu/~abeverid/data/stormofswords.csv","got.csv")
GoT=read.csv("got.csv")
library(networkD3)
simpleNetwork(GoT[,1:2])
```

Because it is difficult for me to incorporate some d3.js scripts in the post, I will illustrate this with a more basic graph:

Consider a vertex v ∈V in the undirected graph G=(V,E) (with classical graph notations), and let d(v) denote the number of edges touching it (i.e., v has d(v) friends). The average number of friends of a random person in the graph is:

The average number of friends that a typical friend has is:

But:

Thus:

Note that this can be related to the variance decomposition:

i.e.:

(Jensen inequality). But let us get back to our network. The list of nodes is:

```
M=(rbind(as.matrix(GoT[,1:2]),as.matrix(GoT[,2:1])))
nodes=unique(M[,1])
```

And we each of them, we can get the list of friends, and the number of friends:

```
friends = function(x) as.character(M[which(M[,1]==x),2])
nb_friends = Vectorize(function(x) length(friends(x)))
```

As well as the number of friends our friends have, and the average number of friends.

```
friends_of_friends = function(y) (Vectorize(function(x) length(friends(x)))(friends(y)))
nb_friends_of_friends = Vectorize(function(x) mean(friends_of_friends(x)))
```

We can look at the density of the number of friends, for a random node.

```
Nb = nb_friends(nodes)
Nb2 = nb_friends_of_friends(nodes)
hist(Nb,breaks=0:40,col=rgb(1,0,0,.2),border="white",probability = TRUE)
hist(Nb2,breaks=0:40,col=rgb(0,0,1,.2),border="white",probability = TRUE,add=TRUE)
lines(density(Nb),col="red",lwd=2)
lines(density(Nb2),col="blue",lwd=2)
```

And we can also compute the averages, just to check:

```
mean(Nb)
[1] 6.579439
mean(Nb2)
[1] 13.94243
```

So, indeed, people on average have fewer friends than their friends.

**Managing data at scale doesn’t have to be hard. Find out how the completely free, open source HPCC Systems platform makes it easier to update, easier to program, easier to integrate data, and easier to manage clusters. Download and get started today.**

Published at DZone with permission of Arthur Charpentier , DZone MVB. See the original article here.

Opinions expressed by DZone contributors are their own.

## {{ parent.title || parent.header.title}}

## {{ parent.tldr }}

## {{ parent.linkDescription }}

{{ parent.urlSource.name }}