Is Schoolyard Team Picking Fair?

Question

Two captains are picking teams for dodgeball. They both will pick players in order of their skill level, best to worst. John gets first pick, Sarah gets the next two picks, and John gets the 4th. If skill in dodgeball is normally distributed, are players 1 and 4 guaranteed to have a mean skill level equal to players 2 and 3?

Answer 1

This question has been here for a couple of days without a thorough answer, so maybe some comments, references and simulated ressults will help. (This post has been revised and describes a simplified selection procedure--after an exchange of Comments with @Tonyk.)

My interpretation is that the two team captains each select two players in the specified order, making two teams of three people, the minimal number for a game of dodgeball according to rules posted on several Internet sites.

Let $X_{(1)}, X_{(2)}, X_{(3)},X_{(4)}$ denote the order statistics of a random sample of size $n = 4$ from $Norm(\mu, \sigma).$

It seems you want to know whether the expected value of the average of $X_{(1)}$ and $X_{(4)}$ is the same as the expected value of the average of $X_{(2)}$ and $X_{(3)}$.

Without loss of generality, we can use a standard normal distribution. I think a complete analytic solution may be a bit tedious. A general discussion is given in Chapter 15 of Kendall, Stuart, & Ord (5e); particular values of $E(X_{(i)})$ are derived and tabled in Headrick and Pant, Table 3. In particular, $$E(X_{(4)}) = \frac{3}{\sqrt{\pi}} - \frac{6\tan^{-1}(1/\sqrt{2})}{\pi^{3/2}} = 1.0294$$ and $$E(X_{(3)}) = -\frac{3}{\sqrt{\pi}} + \frac{18\tan^{-1}(1/\sqrt{2})}{\pi^{3/2}} = 0.2970.$$

By symmetry, one must have $E(X_{(4)}) = - E(X_{(1)})$ and $E(X_{(3)}) = - E(X_{(2)})$. Thus $E[(X_{(1)} + X_{(4)})/2] = 0$ and $E[(X_{(2)} + X_{(3)})/2] = 0$ so that neither team has a systematic advantage.

In the simulation below we use 100.000 choices of four people to get approximate results (about 2 significant digits) and show figures. The averages of the two people chosen for each team seem similar: in expectations (essentially 0), standard deviations (both about 0.55), and distributional shapes (shown in histograms). Even though there is no systematic advantage to either team, the standard deviations and plots indicate that there can be considerable variability in overall ability each time four people are chosen.

 m = 10^5;  n = 4
 x = rnorm(m*n);  DTA = matrix(x, nrow=m)  # m x n: each row has n choices
 ORD = apply(DTA, 1, sort) # n x m: each row has m sim order statistics
 a.14 = (ORD[1,]+ORD[4,])/2;  mean(a.14);  sd(a.14)
 ## 0.002015859  # approx 0
 ## 0.5435122
 a.23 = (ORD[2,]+ORD[3,])/2;  mean(a.23);  sd(a.23)
 ## 0.002232982  # approx 0
 ## 0.5456485
 rowMeans(ORD) # approx the values from H & P
 ## -1.0258067 -0.2937697  0.2982357  1.0298384

Furthermore, captains are likely both to be relatively satisfied or both unsatisfied because the correlation between the two averages is about 0.67.

 cor(a.14, a.23)
 ##   0.676061

Here is a scatterplot of the two averages from the first 30,000 groups of four players selected. (Plotting all 100,000 gives an uninformative blob.)

Finally, we show histograms of the simulated values of the four order statistics, where vertical lines indicate means. Notice the symmetry for the first and fourth, and for the second and third order statistics.

Answer 2

Of course not, but this is as even as you can get