Another good one from Scott Ogawa. It’s the Creampuff Dilemma. A college football coach has to set its pre-season non-conference schedule, thinking ahead to the end-of-season polling that decides Bowl Bids. A schedule stocked with creampuffs means lots of easy wins. But holding fixed the number of wins, a tough schedule will bolster your ranking.

Here’s Scott’s model. Each coach picks a number p between 0 and 1. He is *successful* (s=1) with probability p and unsuccessful (s=0) with probability 1-p. These probabilities are independent across players. (Think of these as the top teams in separate conferences. They will not be playing against each other.)

Highest s-p wins.

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February 8, 2013 at 8:36 am

AndyE[s-p]=p(1-p)+(1-p)(0-p)=0 for all p

February 8, 2013 at 8:58 am

jeffDoesn’t mean you are indifferent among all p. You are not maxiizing E(s-p) you are maximizing the probability of having the highest realized s – p.

Suppose two players and the other guy is playing p2. If you play greater than p2 you win only if you get s=1 AND he gets s=0. If you play less than p2 you win if you get s=1 OR he gets s=0.

February 8, 2013 at 9:19 am

AnonymousClaim: p* (symmetric Nash Equilibrium strategy) is 0.5 * ((n-1)-sqrt((n-1)^2-4), where n is the number of players.

Not sure if this is right, since it is imaginary for n=2. Otherwise, as n increases, the equilibrium choice for p falls toward zero.

February 8, 2013 at 11:01 am

DaveA PSNE cannot be optimal because a player’s best response to p* is p* – epsilon. Perhaps something like the following: for one game and n players, there is a randomized strategy F, such that for any p in F, p*(1-Pr(a player wins with choice s p)^(n-1) = 1/n. In Latex,

$p\left[1-\int_{\underline{p}}^{p}sf(s)ds\right]^{n-1}+(1-p)\left[\int_{p}^{\bar{p}}(1-s)f(s)ds\right]^{n-1}=\frac{1}{n}$.

February 8, 2013 at 11:06 am

Davetypo: p*(1-Pr(a player win with choice q p)^(n-1) = 1/n

February 8, 2013 at 12:43 pm

DRDRThe solution is to find a distribution f(p) such that you’re indifferent between the strategy q chosen by the opponent. The payoff is p(1-q) for p>q and 1-q(1-p) for p<q. The expected probability of winning is then q*E(p|p<q)+(1-q)*P(p

February 9, 2013 at 3:50 am

DRDROops, wordpress did something weird with brackets.

The density f(p) on (0,1) has to be such that q*E(p|p<q)P(p<q) + (1-q)*P(p

February 9, 2013 at 1:28 pm

EnriqueThe trade-off between total number of wins and strength of schedule (the harder the schedule, the lower the probability of wins) is a real cool problem, one with many possible applications

February 10, 2013 at 6:12 am

DRDRAgreed. Another interesting wrinkle in optimal scheduling here is when the ruling body does not accurately perceive the schedule strength. In this example here, the expected value of the selection criterion is always zero, but in actual college sports that’s rarely the case. In the case of college football, poll voters’ perceptions may not be accurate. In most other college sports, an atheoretical “ratings percentage index” (RPI) calculation is a rigid selection criterion (in basketball, it’s a criterion but the selection committee has more discretion). The RPI can easily be gamed.

To see a clear example of how the selection criteria could be gamed, look at college hockey. USCHO.com calculates both the RPI (an actual selection criterion) and a logit-based ranking model called KRACH. If you take the model to be accurate, it’s easy to show a very wide range in the expected RPI when playing each team (the RPI can be decomposed into a value for each game played).

I wonder how much the ability to game the selection criteria outweigh the kind of strategic motives inherent in the simple model here.

February 11, 2013 at 2:46 pm

NotoFor two teams, I find that the equilibrium is mixed strategy over p=[0,1] with f(p) = 0.5*(1-2p+2p^2)^(-3/2). The PDF looks like a bell curve: http://bit.ly/WffRsq. My guess is that the general shape is similar while the PDF skews to the left (i.e., teams choosing harder schedules on average) as you add more teams.

February 12, 2013 at 11:28 am

Scott (Ogawa)A solution! Awesome. Have not verified, but looks reasonable, and I have actually played a bit with friends and that roughly matches what happens. Any chance you could type up how you derived that and post?

February 13, 2013 at 7:41 am

NotoSure, here is my sketch: http://bit.ly/V90DZ6. Don’t tell my wife I let our daughter watch extra TV this morning so that I could type this up. It’s a neat question; you should solve it for N>2 teams.

February 14, 2013 at 11:18 pm

Will DeardenYou can set up a differential equation to solve the problem in general but the nonlinearity makes it difficult to solve even numerically. However, you can use a little trick to get the expected value. I sketch the argument here: https://dl-web.dropbox.com/get/Micro/creampuff.pdf?w=AAA0nEesETL8riaVBt8oniFV1AnWccbXAxbR7P8ylItA4A

Teams do choose harder schedules as n increases.

February 15, 2013 at 5:53 am

DRDRHi Noto, thanks for the writeup.

You don’t need the symmetry of f(p) to derive that E[p]=.5. From substituting 1 or 0 into equation (1), you have Pr(Win|1)=1-E[p] or Pr(Win|0)=1-E[p]. But you also have Pr(WINjq) = .5 for all q, so E[p]=.5.

(My notation seems to failed in my last two posts with notation, but hopefully this is simple enough to survive.)

February 15, 2013 at 10:15 am

Scott (Ogawa)Will, that link would not work for me. DRDR and Noto, I was able to get the constant in the differential equation simply by imposing the fact that f(p) must integrate to 1. That seems more general. Is that right?

February 15, 2013 at 10:24 am

Will DeardenScott, hopefully this works:

bit.ly/12pjXDw

If X_n is the mixed strategy for each player, then E[X_n] = 1 – n^(-1/(n-1)), which means the expected value of p decreases and approaches 0 as n increases.

February 15, 2013 at 10:46 am

Scott (Ogawa)Will, marvelous. At the very least, I agree with the first line; working through the rest. Could you email me: sogawa@u.northwestern.edu?

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