Re: Hockey East - Who's in, who's out, who's home: by the numbers - 2013-14 edition
OK. Let's see if I can add anything here by putting all of the scenarios into one thread in an easy-to-read format.
As has been covered, the top seed (BC) and the three first round road teams (UMA, BU, MC) are all set except for the placement of BU within the bottom 3.
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With all of its games complete, UMA is in the clubhouse, drinking at the 19th hole. They are out of reach of MC, so those two will be in that relative order.
BU could be at the top or bottom of this group, or remain in the middle. In this case, both of the teams that could move up (BU over UMA and MC over BU) hold a 1-0-1 tb over the team above.
For MC, even though they only have six points, they got three of them over one weekend in early December against the only team that they can still catch. For both BU and MC, the point they each got in the tie that Saturday night was the last league point either would earn until they each got a win on the same weekend - seven weeks later. BU has added on a couple of ties since then to widen the gap from 1 to 3, but that's a pretty dismal stretch for both squads.
It's hard to remember that after beating Cornell in MSG, BU was actually over .500 at 7-6-1. Even after the MC weekend that followed, the Terriers were at .500 at 7-7-2. It was a pounding by a hot Bentley team (only a single loss between Nov 15 and Jan 31) in their final game of the first semester that put BU under .500 for good.
Regardless of how we all got here, it means that if BU gets two points, they catch UMA and pass them for 9th and would travel to the 8th place team for the first round. If there are no upsets in the other games, the winner of the 8-9 game (BU, UMA, or their opponent) will travel to The Heights to face BC in the next round.
If MC can out-point BU by three, they would catch them and move into 10th.
The simplest resolution to all of this is if BU were to win on Friday, which would put them out of reach of MC and ahead of UMA. Done and done. Anything short of that would at least leave the 8-9 outcome in doubt until Saturday.
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Now for the muddled middle...
Let's start at the top.
If UML wins out, they're out of reach at 27. 2nd seed.
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If UML takes three from UVM and NU sweeps BU, they tie at 26. NU holds the 2-0-0 tb. NU 2nd, UML 3rd.
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UML +2, NU +3 and ME sweep leaves a three-way tie at 25. In that case, we have:<table border="1">
<tr><th>RRRs</th><th>UML</th><th>NU</th><th>ME</th><th>Total</th></tr>
<tr><td>UML:</td><td></td><td>0-2-0</td><td>0-1-1</td><td>0-3-1</td></tr>
<tr><td>NU:</td><td>2-0-0</td><td></td><td>1-0-1</td><td>3-0-1</td></tr>
<tr><td>ME:</td><td>1-0-1</td><td>0-1-1</td><td></td><td>1-1-2</td></tr>
</table>NU / ME / UML go 2/3/4. Current-2nd UML's bad luck that 3 of their 5 current losses are against the current 3rd and 4th place teams.
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If NU and ME tie at 25 without UML, NU takes the higher seed (2nd or 3rd, depending on whether UML is above or below this pair).
If UML and ME tie here w/o NU, then ME takes the higher seed (same options as above relative to NU's placement).
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Moving down to 24, we could have any of five teams at this mark, with as many as four of them doing it at once. The same would be true at 23. As the schedule works out, the only interplay in this group this weekend is ME v PC. Since either one would need a sweep to get to be relevant here, they can't both be included. that leaves us with a maximum combination of a ME/PC sweeper, UML, NU, and ME. In this grid there will be two totals, one with ME involved, and one with PC. The ME/PC records are dropped, as they are moot for these calculations:<table border="1">
<tr><th>RRRs</th><th>ME</th><th>PC</th><th>UML</th><th>NU</th><th>UNH</th><th>ME Total</th><th>PC Total</th></tr>
<tr><td>ME:</td><td></td><td></td><td>1-0-1</td><td>0-1-1</td><td>1-1-0</td><td>2-2-2</td><td></td></tr>
<tr><td>PC:</td><td></td><td></td><td>1-1-0</td><td>0-1-1</td><td>1-1-0</td><td></td><td>2-3-1</td></tr>
<tr><td>UML:</td><td>0-1-1</td><td>1-1-0</td><td></td><td>0-2-0</td><td>2-0-0</td><td>2-3-1</td><td>3-3-0</td></tr>
<tr><td>NU:</td><td>1-0-1</td><td>1-0-1</td><td>2-0-0</td><td></td><td>0-2-0</td><td>3-2-1</td><td>3-2-1</td></tr>
<tr><td>UNH:</td><td>1-1-0</td><td>1-1-0</td><td>0-2-0</td><td>2-0-0</td><td></td><td>3-3-0</td><td>3-3-0</td></tr>
</table>So here's what happens:
I'll assume that anyone can figure out the two-way tbs out of this bunch on their own, so I'll look at the multi-way ties.
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In a three-way tie at 24 without either of ME or PC, UML, NU, and UNH would all be 2-2-0 in RRR. We'd move to the 2nd tb (conf wins). To be at 24, UML would have to get a single tie and have a HE record of 10-6-4. UNH would have to sweep, and so would be 12-8-0. NU would split, but at either 1-1-0 or 0-0-2. In the former, they're 11-7-2, the latter puts them at 10-6-4. Given the new league declaration in writing ("reduce the number of teams tied <i>from the top</i>"), that puts UNH in front, however NU gets to 24. At that point, we go back to H2H tb, giving NU over UML. Whether this is 2/3/4 or 3/4/5 depends on whether ME sweeps PC to pass into 2nd at 25, or PC gets 2 or 3 to keep both ME and PC below 24.
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Although either or ME or PC has to hit at least 23 (either ME gets at least 2 or PC gets at least 3), UML, NU, and UNH could still be in a three-way tie at 23 if they are passed by the winner of the ME/PC series. In that case, UML and NU would each have 10 wins and UNH would have 11, which gives us the same UNH/NU/UML seeding as before, but this time below the ME/PC winner for sure - placing them 3/4/5.
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Now let's include ME and PC.
NU has the same results in their games against ME and PC, as does UNH - although those records are different from each other (NU is 1-0-1 against both and UNH is 1-1-0). Tying four ways with either ME or PC puts NU at 3-2-1 and UNH at 3-3-0. ME would be 2-2-2 and PC would be 2-3-1.
For UML, their RRR depends on the match-up. With ME in the mix, they are 2-3-1, but with PC as part of the foursome, they move to 3-3-0.
So, in either foursome, NU ends up a half game over .500, but that's the best anyone gets, so NU would be at the top, leaving a three-way tie behind.
Resetting the other three looks like this:<table border="1">
<tr><th>RRRs</th><th>ME</th><th>PC</th><th>UML</th><th>UNH</th><th>ME Total</th><th>PC Total</th></tr>
<tr><td>ME:</td><td></td><td></td><td>1-0-1</td><td>1-1-0</td><td>2-1-1</td><td></td></tr>
<tr><td>PC:</td><td></td><td></td><td>1-1-0</td><td>1-1-0</td><td></td><td>2-2-0</td></tr>
<tr><td>UML:</td><td>0-1-1</td><td>1-1-0</td><td></td><td>2-0-0</td><td>2-1-1</td><td>3-1-0</td></tr>
<tr><td>UNH:</td><td>1-1-0</td><td>1-1-0</td><td>0-2-0</td><td></td><td>1-3-0</td><td>1-3-0</td></tr>
</table>----
With ME, both ME and UML are 2-1-1 with UNH at 1-3-0. Since they now say "from the top", I believe that they would isolate ME and UML H2H (rather than move to 2nd tb) and then throw the loser back into a H2H with UNH. That puts ME next at 1-0-1 and then UML tops UNH. So that means that four way would be NU/ME/UML/UNH.
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With PC, UML tops at 3-1-0 to PC's 2-2-0 and UNH's 1-3-0. Then PC/UNH goes H2H at 1-1-0. Since UNH and PC are currently both at 20, either they both need to sweep in the 24 scenario, or they both need a win and a tie if this is all happening at 23. In either case, their records match this weekend. Since UNH comes into the weekend with one more win than PC, the 2nd tb in both cases goes to UNH and PC falls to the bottom. Here we get NU/UML/UNH/PC.
At either 24 or 23, the ME/PC loser is below the four-way group, so both seedings are for 2/3/4/5.
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The RRR matrix just above also applies for three-way ties in this group that don't include NU. The only difference is whether the tbs are deciding 2/3/4 ahead of NU or 3/4/5 below them.
(to be continued)