Suppose there are two captains of equal skill, A and B, drafting two players each from four players, 1 2 3 and 4. 1 is the strongest and 4 is the weakest.
Clearly, the equitable way to divide the players is 1 with 4, and 2 with 3. But A and B decide to use a two-round EU draft. A will go first.
Here are the possible scenarios:
A picks 1. B will take 2 and 3.
A picks 2. B will take 3, then 1. A has made an error.
A picks 3. B will take 1 then have 4 left over, or 4 then 1. B will not take 2, because A will take 1.
A picks 4. B will take 1, and then be happy with either 2 or 3. A has made an error.
What if we made 4 a ‘booby prize’, such that also having 1 cannot compensate? Then A must pick 3 first. B will then have no way of winning the draft.
Let’s extend the draft to three captains, A B and C, choosing from six players, 1 2 3 4 5 and 6. Again, 1 is the best, and 6 will be a booby prize.
Suppose A chooses 1. B and C will end with 2 and 5, and 3 and 4, and A will be left with 6.
A wisens up and chooses 5, B chooses 4, and C is stuck with 6 and 1. A will then get 2, for 5+2=7, and B gets 4+3=7.
Suppose A still chooses 5, but B is clueless and chooses 1. C will now choose 2, since 1 is assured of getting 6. In the 2nd round, A gets 3, for 5+3=8, and C gets 4 for 2+4=6. C has gained from B’s cluelessness, and A has suffered. A could avoid this by starting with 4 instead of 5.
If everyone acts rationally, A will choose 4, B will choose 5, and C will be stuck with 6.
I debated between posting this now or waiting until after the draft. I figured I’d do a service to my readers and post it before. It will make draft-watching this Sunday more interesting.
I purposefully kept this analysis somewhat superficial to make it more accessible, but it should be clear that there are some theoretical problems with EU drafting.
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