Each player has three cards numbered '2,' '3,' and '4.' Players secretly each select one of the three cards, then all reveal their choices simultaneously. Players whose selection matched that of any other player(s) advance their track figure only one space along the track, whereas players who selected a unique number (one that no other player selected) advance a number of spaces along the track equal to the number on the card they selected. The process repeats until one player's track figure crosses the finish line to win the race.
This mechanic caught my attention because I had just been brainstorming about ideas in which players might secretly and simultaneously select from among a number of options whose benefit would be greater if fewer other players made the same selection; i.e. the less popular the selection, the more beneficial it would be to those who chose it. So here was a perfect example, better than anything I'd thought of so far.
Now, it occurred to me that whereas this "mini-game" wouldn't work for only two players, it makes a nice game theory problem for a multi-player case. In two-player game theory, a game is represented as a matrix of pay-offs in which the axes represent the discrete decision spaces of the two players. We can examine the two-player game as a matrix and see why this game is meaningless as a two-player game.
|Figure 1: Two-player|
game theory matrix
|Figure 2: Three-player|
game theory matrix
If all three players choose simultaneously to minimize their potential loss, all three will choose '4.' But there appears to be a new Nash Equilibrium here - and not the one you might think. If A anticipates that B and C will choose 4, then A will choose 3 so as to extend his lead over B, who will have made the same choice as C. But now if B anticipates that A will choose 3 and C will choose 4, then B will choose 3 so as not to lose ground against A. If C anticipates that A and B will choose 3, then C will stick with 4. Coming back to A, if A anticipates that B will choose 3, A will stick with 3 so as not to lose ground against B. So, theoretically, the first- and second-place racers will select '3' to keep pace with one another, and the third-place racer will select '4' and so gain ground on the others until he becomes the leader.
This analysis is flawed and incomplete; it doesn't take into account cases where two or all three racers are tied (which is likely). I've read that in a zero-sum three-player game, Nash proved that there is always an equilibrium. The way I've characterized it, this game is not zero-sum, so that theory, at least, doesn't apply. It would be interesting to continue this analysis more thoroughly, and I find myself wishing I were formally trained in game theory to give the problem the analysis it warrants.