2015 ECAC Playoff Permutations

[Trotsky]

Huh, I've never read your table that way before.  I assumed that the tie-breaker was just assuming {x, y}, with any larger n-tuples being potentially too complicated to calculate.

Correct me if (inevitably) I am wrong, but if you have a 3+ team tie they apply the tiebreaker to the whole set, resolve the top team, then re-apply the tiebreaker to the n-1 set, and so on?  I've never really liked that, preferring to resolve the whole n-tuple on the first go.  Can anybody make a compelling case for why they do it their way?  I guess there's the idea that reapplying the criteria against smaller sets resolves those lower ties by a more "local" condition, but I'm bugged by the idea that teams which are all tied are being resolved by different (and likely contradictory) procedures.

With 3+ teams, the tie breaker last until there is ANY differentiation, then the split happens.  So you could end up with a top team(s) or a bottom team(s).  I think it's the right procedure, if team A has differentiated itself from team B & team C, why should the relationship between team B & team C be based on team A?

If you didn't do it this way couldn't you have an infinite loop?

The infinite loop is actually caused by re-differentiating.  If you just started from #1, resolving all ties by single pass, and allowing lower ties to stand provisionally (in John's cited example, allowing all the teams tied for 8th to be counted in record vs top without nesting the tie resolutions, there would be no return arrow and no infinite loop.

Say you begin with a 3-way tie for 4th and a 3-way tie for 8th.  To resolve the tie for 4th, you allow all three teams tied for 8th to remain in the "record vs top 8" set.  After a one-pass resolution, you now have just one 4th place team, so when you get to the 8th place tie you can use a 4-team "record vs top 4" set in that one-pass resolution.

So to pick team 4 you use all 8th teams, but to pick the 8th you only use team 4? As you said, it's arbitrary.

But at least you are using the same rules to resolve all the teams tied for a given slot.

[Jim Hyla]

We can't go lower than 7, nor higher than 5. The keys are the Harvard and Colgate games. If we win and both lose, we get 5. If only 1 loses, 6. If both win, we're 7. In fact we're 7 win or lose to Yale. We're also 7 if we all lose. So you could say the overwhelming odds are for us to be 7.::bang:: And I have to decide whether I want to drive to New Haven and back tomorrow.

Edit: and in 7 we'd play Union or RPI.

[Icy]

All permutations aside, I was impressed with the comeback tonight.