Another hint: If you write the number of vertices as a sum of 0,1-valued random variables (aka indicator random variables), the expected value of the sum is expressed as the sum of a bunch of probabilities, each of which is a ratio of areas. Jim On Monday, November 6, 2017, James Propp <jamespropp@gmail.com> wrote:

I wrote:

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I actually know two different proofs of the result. One follows the approach to such problems that I learned from George Hart (see his email to math-fun from October 18), and the other follows the approach I learned from Warren Smith (see my email to math-fun from October 19). If you look at their solutions to the random-slices-of-a-cube problem you might be inspired, as I was, to see how to solve the random-intersecting-parallelog rams-puzzle.

George's method can also be used to prove that if you take a random nonempty intersection of a regular hexagon with a translate of itself, the expected number of sides is exactly 5.

Hint: instead of counting sides, count vertices.

I'll sketch the proof tomorrow or Wednesday, if nobody beats me to it.

Anyway, two days have passed since I posted the puzzle, so I think it's

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appropriate to have an unbridled conversation, with or without spoilers.

Let me say, by way of motivation, that the "George-style" proof is really sweet. You'll feel good when you find it.

Jim