space and games

March 24, 2008

Ordinal Performance Games

Filed under: Decision Theory — Peter de Blanc @ 9:56 pm

There are games (such as racing or DDR) in which players compete with one or more opponents, but the opponents do not interact. Each player performs individually, and receives some score for vis performance, and the player with the best score wins. I’ll call these “ordinal performance games.”

One might naively think that the best strategy is to ignore one’s opponents, and think only of maximizing one’s own score. After all, the players never interact, so your score depends only on your performance, right? That’s true, but the object is not to maximize your own score; it is to have a higher score than your opponents, and this depends on your opponent’s score as well.

For example, suppose we have a game with three player types, and possible scores of 1 through 5.

  • Alf is consistently mediocre, always receiving a score of 3.
  • Beth has adopted a risky strategy; she receives a score of 4 with probability 0.6 and a score of 1 with probability 0.4.
  • Gam may succeed spectacularly, but usually plays poorly. He receives a score of 5 with probability 0.4 and a score of 2 with probability 0.6.

Suppose Alf and Beth play. Alf will always score 3. Beth has a probability of 0.6 of scoring 4, and thus winning, and a probability of 0.4 of scoring a 1, and thus losing. So p(Alf < Beth) = 0.6.

Suppose Beth and Gam play. Beth has a probability of 0.4 of scoring 1, in which case Gam is guaranteed to win. The other way for Gam to win is for Beth to score a 4 while Gam scores a 5; the probability of this is 0.4 * 0.6 = 0.24. Thus p(Beth < Gam) = 0.4 + 0.24 = 0.64.

Suppose Gam and Alf play. Gam can only win by scoring a 5, with probability 0.4. So p(Gam < Alf) = 0.6.

So Alf usually loses to Beth, who usually loses to Gam, who usually loses to Alf. This shows that your optimal strategy in an ordinal performance game can depend on your opponents.


  1. That’s nifty. Reminds me of endlessly rising music.

    (Also, nice player names.)

    Comment by Jesse — March 24, 2008 @ 10:41 pm

  2. You’re assuming that there’s a sharp utility cutoff between “win” and “lose”. There’s nothing wrong with this assumption, but it should be stated explicitly; it bears some resemblance to the American presidential election, where you can lose even when you get a plurality of the popular vote.

    Comment by Tom McCabe — March 25, 2008 @ 4:31 pm

  3. Now I think figure skating is an even better example, because there are so many risky moves.

    Comment by Peter de Blanc — March 31, 2008 @ 1:32 pm

  4. This is a good proof. I think it means the ‘maximizing one’s own score’ is still right, but just not accurate. What you want to maximize is your expected score. To do this you have to take the risk of your strategy into consideration.
    I bet this gets more complicated depending on how many games you intend to play.

    The proof itself reminds me of the logic behind nontransitive dice:

    Comment by lorg — May 1, 2008 @ 4:09 pm

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