Your friends really do have more friends than you.

Not on average for you specifically — for almost anyone. Line up every friendship in a network, pick one at random, then pick one of its two people at random: their expected number of friends always beats the network's true average, however evenly or unevenly popularity is spread. Drag the dial to see why — and pick a few random people to watch it happen to them personally.

What you're seeing

Everyone here starts on an equal footing: a ring where each person is friends with their four nearest neighbors, nobody more connected than anyone else. As you raise the dial, a small fixed set of six people — the amber hubs — steadily gain extra friends from across the whole network, while almost everyone else's count barely moves. That's all "popularity" is in this model: most people staying near the baseline while a few pull far ahead.

Now play the sampling game in the panel above: line up every friendship as a pair, grab one at random, and look at one of its two people. You are far more likely to land on a hub than a uniformly random pick of people would suggest — not because hubs are special or lucky, but because every one of a hub's many friendships is a separate ticket in the drawing. A person with 40 friends holds 40 tickets; a person with 4 friends holds 4. Add up everyone's tickets and average the friend-count you land on, and hubs — who by definition have lots of tickets — pull that average up. This is the entire mechanism. No one's friend count changed by being asked about; the sampling just favors well-connected people, every time.

Click "pick someone at random" a few times. Most picks will be an ordinary person whose friends, on average, are more connected than they are — not a trick, just because most of their friends' friendship tickets include a hub somewhere nearby. Every so often you'll land on a hub instead, and see the pattern reverse for that one person. Both outcomes are the same phenomenon, viewed from two different seats at the same table.

The rule, exactly. Take any network with people-degrees d₁…d_n and m friendships (m ≥ 1). Let μ be the population's true average friend count. Let Q be the expected friend count of the person you land on by picking a uniformly random friendship, then a uniformly random one of its two people: μ = (1/n)·Σd_i, Q = Σd_i² / (2m) Then, for any degree sequence whatsoever — this is Cauchy–Schwarz applied to the list of friend counts, not a special property of any one network — Q = μ + Var(d)/μ ≥ μ with equality only when every single person has exactly the same number of friends, i.e. Var(d)=0. Verified in node (improve/verify/87-friendship-paradox.js, 221 checks, on this exhibit's exact graph-building algorithm):

The closed form for Q matches an independently-enumerated average over every friendship's two endpoints.
A live Monte Carlo of the literal sampling procedure described above — not of the formula — lands on the closed-form value, within a statistically derived tolerance.
The gap Q − μ stays non-negative and non-decreasing across the whole hub-strength range.
The graph builder only ever adds friendships as the dial rises. It never removes or swaps one, so dragging the dial grows the network instead of reshuffling it.
Negative control at hub strength 0: the network is exactly regular, and Q equals μ to floating-point precision. No paradox when nobody is unusual.

Feld's original 1991 result, applied to any network: acquaintances, coauthors, follower graphs, even airports connected by routes.