I’ve created a fun and simple game-theory problem that I thought you might enjoy …  This is the sort of problem you could give undergrads to find out who are the really bright ones.  It might also be fun to mention (or play) in class.
Problem: Find the (unique) symmetic equilibrium of “The World’s Simplest Poker Game”, played as follows:
**0** two players
**1** each player pays ante of \$100
**2** each player receives ONE card, which we can think of as independent random numbers on [0,1]
**3** each player SIMULTANEOUSLY decides whether to “raise” \$100 or “stay”
**4A** if one player raises and the other stays, the raiser wins the pot, for net gain +\$100
**4B** if both raise, the players show their cards and whoever has the highest card wins for net gain +\$200
**4C** if both stay, the players show their cards and whoever has the highest card wins for net gain +\$100
If you decide to solve this problem, please let me know how long it takes you … I’m curious how immediately obvious the answer is to you 🙂  I have solved it myself and, I can tell you, the answer is simple and elegant.
Cheers,
David

N.B. My answer based on 5 minutes of thinking was wrong.  I will post David’s solution over the weekend.

Update:  As promised, here is David’s solution.  Looks like Keith was the first to post the correct answer in the comments and thanks to Nicolas for pointing out that this example appeared in von Neumann and Morgenstern.