Randomized team drafting strategy

This Riddler Classic puzzle explores a randomized team drafting strategy designed to prevent teams from throwing games.

You are one of 30 team owners in a professional sports league. In the past, your league set the order for its annual draft using the teams’ records from the previous season — the team with the worst record got the first draft pick, the team with the second-worst record got the next pick, and so on. However, due to concerns about teams intentionally losing games to improve their picks, the league adopts a modified system. This year, each team tosses a coin. All the teams that call their coin toss correctly go into Group A, and the teams that lost the toss go into Group B. All the Group A teams pick before all the Group B teams; within each group, picks are ordered in the traditional way, from worst record to best. If your team would have picked 10th in the old system, what is your expected draft position under the new system?

Extra credit: Suppose each team is randomly assigned to one of T groups where all the teams in Group 1 pick, then all the teams in Group 2, and so on. (The coin-flipping scenario above is the case where T = 2.) What is the expected draft position of the team with the Nth-best record?

Here is my solution to the general case:
[Show Solution]

While the expected draft position is not that difficult to characterize, one can also ask about the distribution of draft positions:
[Show Solution]

9 thoughts on “Randomized team drafting strategy”

  1. Hi Laurent,

    I think this solution fails a different sanity check. Take the case of a 30 team league and two groups, like Ollie posed, but consider the worst team (which would be N=1 in your formula, because that’s the way you defined N even though the problem said “N-th best record” not “N-th worst record”.)

    Anyway, the worst team will always pick first in its group. It has a 50-50 chance of being in either group. Therefore its expected draft position is (1/2)(1+16) = 8.5.
    Your formula gives 8.25

    Either I’m missing something really basic, or Ollie has published an incorrect solution. If you can point out my mistake, great! Otherwise, I believe I have the correct general solution and I’d be happy to share it with you to check it out.

    Thanks,
    Greg

  2. Never mind. I solved the wrong problem. I misread the instructions on how the teams are divided into groups as “divided equally into groups” rather than “divided with equal probability into groups”.

  3. Laurent,

    Nice work calculating the distribution. I had never worked that out. While the probabilities don’t simplify nicely, there is a neat formula for the variance of the position of team $N$ when there are $K$ teams and $T$ groups. Remarkably, it does not depend on N: $$\frac{(K^2-1)(T^2-1)}{12T^2}$$ I can send you the proof if you’d like.

  4. You missed another sanity check: when T=K you should have a purely random assignment, but your general answer depends on the team’s position N still.

    A small mistake you made is that the probability of one of the other teams lands in group i given that one of the slots is taken by team N is a little bit lower than the probability it falls in any other group. I think this is enough to fix your solution.

    1. In the randomization step, each team rolls a $T$-sided die to determine which group they fall in. So it’s definitely possible that two teams end up in the same group. If we have as many teams as groups ($T=K$), the final ordering shouldn’t be random. For example, every time the 1-seed team lands in a group with another team, they will always come out ahead. So we should expect the solution to depend on $N$.

      The case where it becomes random is when $T\to\infty$. (infinitely many groups). In that scenario, the probability that two teams end up in the same group is vanishingly small, and the final ranking is simply a random permutation of the teams. I added this extra case to my write-up.

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