## Answer: The Rockettes Return

Six months ago I asked a question about the rockettes.

The question was,

Suppose n Rockettes have a fixed order in which they’re supposed line up to perform, but due to backstage commotion they always come out in a random order instead. On average, how many Rockettes will be in the right place?

The answer is one. Remarkably, it does not depend on how many Rockettes there are. If there are two or two hundred, on average only one will be in the right spot.

A simple heuristic shows this makes sense. If I’m a Rockette (hey, a guy can dream, can’t he) going into the order at random, there’s a $1/n$ chance I’ll get in the right spot. There are $n$ Rockettes, and $1/n * n = 1$.

That’s actually it. It’s a statistical fact that $E(X+Y) = E(X) + E(Y)$, and so by induction, for $n$ Rockettes the expected number to be in order is the number of Rockettes times probability for any one Rockette to be in order.

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### One Response to “Answer: The Rockettes Return”

1. The Deranged Rockettes « Arcsecond Says:

[…] Deranged Rockettes By Mark Eichenlaub Previously, I noted that if a line of dancers comes out on stage in a random order, then on average one dancer will be […]