The election with no winner

Question

Three friends rank three restaurants, X, Y, and Z, from favorite to least favorite.

Friend 1: X > Y > Z.
Friend 2: Y > Z > X.
Friend 3: Z > X > Y.

They decide to settle it with head-to-head majority votes: for each pair of restaurants, whichever one more friends prefer wins that matchup.

Is there a restaurant that beats both others in a two-way majority vote, a clear winner?

Accepted Answer

Run each head-to-head vote by counting how many friends rank one restaurant above the other. - X versus Y: Friend 1 prefers X, Friend 3 prefers X (Z X Y), Friend 2 prefers Y. So X beats Y, 2 to 1. - Y versus Z: Friend 1 prefers Y, Friend 2 prefers Y, Friend 3 prefers Z. So Y beats Z, 2 to 1. - Z versus X: Friend 2 prefers Z, Friend 3 prefers Z, Friend 1 prefers X. So Z beats X, 2 to 1. The majority prefers X over Y, Y over Z, and Z over X. The preferences form a cycle, so no restaurant beats both of the others. The answer is no. This is the Condorcet voting paradox: perfectly rational individ…

The election with no winner

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