Randomized team drafting strategy

Number all the teams according to their original seeding order and suppose there are $K$ teams in total. Label the groups $1,2,\dots,T$, where Group 1 gets picked first, Group 2 second, and so on.

Let’s calculate team $N$’s expected draft position given that it was placed in Group $i$. For each other team $k$, define the random variable:
\[
X_k = \begin{cases}
1 & \text{if team }k\text{ finishes ahead of team }N\\
0 & \text{if team }k\text{ finishes behind team }N
\end{cases}
\]The final draft position of team $N$ is therefore $1 + \sum_{k\ne N} X_k$, where we sum over $k=1,2,\dots,K$ but exclude team $N$. The expected value of this sum is the sum of expected values (linearity of expectation), so we can focus our attention on a particular other team $k$.

• If $1 \le k \le N-1$, then $X_k = 1$ if team $k$ ended up in groups $1$ through $i$. The probability of this occurring is $\tfrac{i}{T}$.
• If $N+1 \le k \le K$, then $X_k = 1$ if team $k$ ended up in groups $1$ through $i-1$. The probability of this occurring is $\tfrac{i-1}{T}$.

The expected draft position of team $N$ is therefore:
\[
1 + (N-1)\frac{i}{T} + (K-N)\frac{i-1}{T}
\]Remember that this was conditioned on team $N$ ending up in Group $i$, which happens with probability $\tfrac{1}{T}$. So the final expected draft position of team $N$ is therefore:
\begin{align}
\sum_{i=1}^T \frac{1}{T} \left( 1 + (N-1)\frac{i}{T} + (K-N)\frac{i-1}{T} \right) \\
\end{align}After simplifying (make use of this identity), we find that the expected final position for team $N$ is given by the formula

$\displaystyle
\frac{2N+(K+1)(T-1)}{2T}
$

We can perform a few quick sanity checks to convince ourselves that this solution is correct.

• When $T=1$ (only one group), the formula yields $N$, the same as the original seeding, as it should.
• When $K=N=1$ (only one team), the formula yields $1$, as it should.
• More precisely, if the original seed is $N = \tfrac{K+1}{2} + \Delta$, where $\Delta$ is the offset from the middle, the formula yields an expected draft position of $\tfrac{K+1}{2} + \tfrac{\Delta}{T}$. In other words, if you started off better than average, you can expect to end up better than average and similarly if you started below average. The division by $T$ shows that as you add more groups, the randomized drafting strategy has an equalizing effect.

In the particular case mentioned in the problem statement, we have $K=30$, $N=10$, and $T=2$, which leads to an expected value of $\frac{51}{4} = 12.75$.

Finding the actual distribution of draft positions is more complicated that just finding its expected value. To find the distribution, we will answer the question: “what is the probability that team $N$ finishes with position $j$ given that it lands in Group $i$?”. We’ll compute the probability by counting the number of ways that $j-1$ other teams can end up ahead of team $N$ in the draft.

• Start by considering the teams $1 \le k \le N-1$, which are the teams that have a higher seed than team $N$. Suppose exactly $p$ of these teams end up in Group $i$. There are ${N-1 \choose p}$ ways this can happen, each with probability $\tfrac{1}{T}.$ Therefore we have a net probability of $\color{red}{P_1 = {N-1\choose p}\left(\tfrac{1}{T}\right)^p}$.
• Out of the remaining $N-1-p$ teams, suppose exactly $r$ of them end up in groups $1$ through $i-1$. There are ${N-1-p \choose r}$ ways this can happen, each with probability $\tfrac{i-1}{T}$. The net probability is $\color{red}{P_2 = {N-1-p \choose r}\left(\tfrac{i-1}{T}\right)^r}.$ If $i=1$, then $P_2=0$ for all values of $p$.
• The remaining $N-1-p-r$ teams must necessarily be in groups $i+1$ through $T$, each with probability $\tfrac{T-i}{T}$. The net probability is $\color{red}{P_3 = \left(\tfrac{T-i}{T}\right)^{N-1-p-r}}.$ If $i=T$, then $P_3=0$.
• There must be $j-1$ teams ahead of team $N$ in the final draft, and we have thus far accounted for $p+r$ of them. The rest must come from teams $N+1 \le k \le K$, which are the teams that have a lower seed than the team $N$. These must end up in groups $1$ through $i-1$. There are ${K-N \choose j-1-p-r}$ ways to do this, each with probability $\tfrac{i-1}{T}$. The net probability is $\color{red}{P_4 = {K-N \choose j-1-p-r}\left(\tfrac{i-1}{T}\right)^{j-1-p-r}}.$ If $i=1$, then $P_4=0$.
• The remaining $K-N-j+1+p+r$ teams must necessarily be in groups $i$ through $T$, each with probability $\tfrac{T-i+1}{T}$. The net probability is $\color{red}{P_5 = \left(\tfrac{T-i+1}{T}\right)^{K-N-j+1+p+r}}.$

This accounts for every possible arrangement, so to find the probability that team $N$ finishes with position $j$, we must sum over all possible values of $p$ and $r$, and then sum over possible values of $i$ while including an additional $\frac{1}{T}$ factor as we did in the solution to the first part. The final probability of interest is therefore:
\[
P(K,N,T,j) = \frac{1}{T}\sum_{i=1}^T \sum_{p=0}^{N-1} \sum_{r=0}^{N-1} P_1P_2P_3P_4P_5
\]Note that some terms will be identically zero. We use the convention that ${a \choose b} = 0$ if $b < 0$. Unfortunately, this messy formula doesn't appear to simplify to anything nice, so we have to be content with computing it numerically. To illustrate what the result looks like, here is a plot of $P(30,10,T,j)$ as a function of $j$ and for increasing values of $T$ (the number of groups).

• When $T=1$, the distribution is concentrated at $N=10$, since there is no probability involved; each team will always be drafted in seed order.
• When $T=2$, the distribution is bimodal. Even though the initial seed was $N=10$ and the expected value of the final draft position is $12.75$ as we computed earlier, the probability of being drafted anywhere near the expected value is very low! This makes sense, because whether we end up in Group 1 or Group 2 has a substantial effect on our final draft position.
• Similar multi-modal distributions occur for $T=2,3,\dots$ Ultimately, the distribution converges to the uniform distribution. Again, this makes sense; if you imagine a very large number of groups, the probability that two teams end up in the same group is vanishingly small. So the final draft position is essentially determined by the group number only.