ECAC Predictions

It's that time of year again where I start to think, what are the possible finishes in the ECAC? Everyone has six games to play which only leaves a fixed number of possibilities. Then I quickly realize that 3^36 = 1e17 possibilities, which is a bit much.

How 'bout a different tack. Whelan's page (http://slack.net/hockey) provides Bradley-Terry rankings for the ECAC and then uses that to predict the final standings. Essentially if the rankings predict that Colgate has a 65.4% chance of beating Yale then the (Red) Raiders are awarded 1.38 points for the game and the Elis 0.62. The final standings don't have any spread though or probabilities - it's just a fixed expected value.

Let's try something using the BT rankings. Since we can't look at all 3^36 possibilities at this point, lets assume everyone in league performs according to BT expectations in all of their games except those involving Cornell. That's 30 games. The remaining 6 games only leave 3^6 = 729 possibilities, which is manageable in a spread sheet. Actually, it's only 2^6 = 64, since I don't know how to predict ties with BT.

Anyway, i used the BT rankings of these 6 games to predict the likely of each of the 64 possible outcomes. What you find is Cornell can finish between 3 and 8th with these assumptions (reality is 1 to 11) with the following probabilities:

3rd:  69.6%
4th:   1.8%
5th:  11.5%
6th:  15.7%
7th:   1.4%
8th:   0.1%

Just food for thought. The low probability of 4th place probably results because of tie games not being considered (or a mistake in my spreadsheet). Feel free to poke holes in this "analysis".

The link to Whelan's page is: [slack.net]
The paran messes it up above (and since I never log in I can't edit and fix it).

One more point. I used the ECAC only Bradley Terry rankings for the above calculations. I could also have used the NCAA B-T (KRACH) rankings. Using the whole NCAA would provide more data but isn't self-consistent since it wouldn't exactly predict the current ECAC standings. But the extra data might result in a better prediction for the future. I don't understand the workings of BT well enough to say for sure. I may try it the other way anyway. It shouldn't make much difference, but predicted standings using NCAA KRACH rankings are a little different.

Keith, the low probability of 4th comes down to this:

Assume Cornell wins out: we get third. RPI and Dartmouth get 4th & 5th.
Assume Dartmouth wins out, too: We're tied in points. We still get third (by virtue of more wins).

Now assume Cornell loses to RPI but wins everything else: RPI finishes ahead of Cornell. If Dartmouth wins out, they get 4th, we get 5th. If Dartmouth loses 1, we're tied. Dartmouth gets 4th by virtue of fewer losses (better winning percentage).

If however Cornell and RPI Tie in their second game, RPI wins the standings tie-break. That means, you're right. Cornell can finish 4th if Dartmouth takes a loss. If Dartmouth wins out, too, we're 5th.

JH

Actually, if both Cornell and Dartmouth win out it's almost impossible for Cornell to win the tiebreaker. We split the series H2H with Dartmouth and the next tiebreaker is Record vs. Top 4. If we both win out then we finish at 31 points in no worse than a 3rd place tie. So Record vs. Top 4 becomes record against the other two top teams. If it's Brown and Colgate (likely) we lose 5-1. If RPI is involved we'd lose 3-2 (RPI only). We'd win Harvard by a 4-3 margin but Yale Clarkson and SLU would be ties. And Union/Vermont won't be top 4. So I can't picture a scenario where the right teams make top 4 so we could win that tiebreaker.

That said, none of that matters in my scenario above. I didn't consider tiebreakers. Because of the way points are allocated (the assumptions), Dartmouth is guaranteed to finish with 26.0044 points and no tie is really possible.

Other extrapolations:

Case (1): top 6 teams win all remaining games against non-top 6 teams. Home teams win games between top 6 teams.

1. 37 Brown
2. 32 Colgate
3. 29 Dartmouth
4. 29 Cornell
5. 27 RPI
6. 24 Yale

Case (2): top 6 teams win all remaining games against non-top 6 teams. top 6 teams tie all games against other top 6 teams.

1. 35 Brown
2. 32 Colgate
3. 29 Dartmouth
4. 29 Cornell
5. 27 RPI
6. 26 Yale

Case (3): home teams win all remaining games.

1. 33 Brown
2. 30 Colgate
3. 27 Cornell
4. 23 Harvard
4. 23 Dartmouth
4. 23 RPI



Post Edited (02-12-04 12:42)

Jeff Hopkins '82 wrote:
Assume Cornell wins out: we get third. RPI and Dartmouth get 4th & 5th.
Assume Dartmouth wins out, too: We're tied in points. We still get third (by virtue of more wins).

Wins is not a tie-breaker in the ECAC. If we tie Dartmouth, head-to-head is a wash, and the next tiebreaker is record vs the top four. If we were tied for third, that would actually be the top two, which is Brown and Colgate in this scenario. We have a total of one point against these teams and Dartmouth already has 3 points, not counting their upcoming game against Brown.

So basically, in layman's terms, 4th place (and thus home ice and a bye) is ours to lose?

...do we want home ice?

LOL, and shouldn't Greg's predictions assume that home teams win all remaining games, except for Cornell games, in which road teams all win?

Well, even if we finish 5-8 we'd be home for the first round. Better to get the extra week of rest and get players healthy(er).

The low probability of 4th place calculated by Keith undoubtedly comes from the fact that using the B-T ratings, two other teams must end up with a very small differential in expected number of points. Thus, Cornell is likely to be either ahead of both of these two very close teams and in 3rd, or fall behind them to 5th. Of course, since the analysis is based on expected value and not probabilties of each outcome, the real expected probability of 4th place would be much higher.



Post Edited (02-12-04 18:48)

Agreed, the real probability of 4th is higher, presumably at the expense of the 3rd place chance. Another way to see that is to note that while we get 4th place is only 4 of 64 (6%) possible outcomes considering only wins and losses in CU games, the number rises to 191 out of 729 (26%) when ties are allowed. So I'm throwing out a lot of possible ways that we could finish 4th by assuming that the probability of a tie is negligible.

It isn't the lack of calculating the ties so much as the fact that you've calculated two other teams (Dartmouth and RPI perhaps) to be so close in expected points that there is little room for Cornell to slip in between them. However, although their expected points may be very close, there's a much greater probability of a larger differential between them, giving Cornell more chances to fall in between.

That's the weird thing - the two teams that are extrememly close are 4th and 5th, not 3rd and 4th. The expected points without Cornell games (should've included this earlier).

Brown      35.0
Colgate    31.0
Dartmouth  26.0+
Yale       23.8
RPI        23.7
Harvard    22.1
Clarkson   17.8
SLU        16.6
Princeton 14.9
Union     14.9
Vermont    7.2

Looking at it again I realize that you're essentially right. Cornell has games against Yale and RPI. For Cornell to finish 4th, we need to finish with 26, 25 or 24 points to slip between Dartmouth and Yale/RPi. But if Cornell loses to either Yale and/or RPI (by the numbers roughly a 3/4 chance) then 4th place requires exactly 26 points, which is impossible without ties.

So I was wondering who, out of the top 6 teams, has the most difficult remaining schedule. So for each of the top 6, I added up the current standings of all the opponents. For example, for us, we play Yale (#6), Princeton (#11), RPI (#3), Union (#10), Clarkson (#8), and St. Lawrence (#9), to give a total "score" of 47. Obviously, the lower the score, the more difficult the remaining schedule. So from the most difficult schedule to the least, and with the "score" in parentheses, I came up with...

1. RPI (31)
2. Yale (36)
3. Dartmouth (42)
T4. Brown (47), Colgate (47), Cornell (47)

To break that three-way tie for fourth (and obviously, there would be a tie between travel partners Colgate and Cornell), I used "record against remaining opponents from earlier in the season" as the first tiebreaker and "wins" as the second, to give...

1. RPI
2. Yale
3. Dartmouth
4. Cornell (3-1-2 against remaining opponents)
5. Colgate (4-2-0 against remaining opponents)
6. Brown (5-1-0 against remaining opponents)


Alright so I don't know if that accomplished anything , but it gives you an idea of how RPI and Yale have much tougher schedules than we do, perhaps strengthening the comment mentioned by someone else that fourth place is ours to lose.

Likewise, using a sum of Bradley Terry rating or opponents points to characterize remaining schedule:

             BT    Pts
RPI        1074.7  110
Dartmouth   803.9   96
Yale        751.3   96
Cornell     498.2   87
Colgate     498.2   87
Brown       480.4   82

...shows that Yale and RPI have tougher slates ahead although these measured put Dartmouth's schedule a little above Yale's. Naturally Cornell and Colgate have the same SoS remaining.

Forgive me for not understanding the Bradley Terry stuff too well, but why is Dartmouth's schedule considered more difficult, using those measures, than Yale's, despite Dartmouth facing teams lower in the standings?



Post Edited (02-12-04 20:14)

Position in the standings is a coarse measure that will miss any large gaps in "strength" between teams located close in the standings. Example: say there's a big dropoff between 6th and 7th place in the standings (top 6 are all very good and bottom six are very bad). The standings won't see much difference between the 6th and 7th place teams. Points or BT rating will see the differnce between the two teams.

Just glancing at the schedule, the obvious difference is that Yale plays Vermont while Dartmouth plays Brown.

Keith K '93 wrote:
Likewise, using a sum of Bradley Terry rating or opponents points to characterize remaining schedule:

Adding the BT ratings is not really the right thing to do with them, since they're on a multiplicative scale. Also, for opponents ranked a lot higher or lower, it doesn't really matter how much stronger or weaker they are. Imagine two teams with ratings of around 100. Team A has opponents with ratings of 50 and 6000, while Team B has opponents with ratings of 500 and 5000. Team A's opponents have a total rating of 6050 to Team B's 5500, so by that measure they're facing stronger opposition. But it's likely Team A will do better than Team B in the two games, since neither team has much chance to beat their stronger opponent, and Team A is a 2:1 favorite against their weaker opponent, while Team B is a 5:1 underdog.

What you want is something like the expected winning percentage of some canonical team vs each team's remaining schedule. Or the weighted average we use to define SOS on the KRACH pages. (But there the weighting depends on the team's own rating, so it's hard to compare schedules for opponents with different ratings. Again, you'd want to use some typical rating, like the geometric mean of the teams being compared.)

Keith K '93 wrote:

Position in the standings is a coarse measure that will miss any large gaps in "strength" between teams located close in the standings. Example: say there's a big dropoff between 6th and 7th place in the standings (top 6 are all very good and bottom six are very bad). The standings won't see much difference between the 6th and 7th place teams. Points or BT rating will see the differnce between the two teams.

In psychophysics, we'd summarize this by noting that standings only provide ordinal level data whereas the BT ratings provide at least interval (and possibly ratio) level data. The classic example used to explain this to a bunch of psych students is a marathon; if I know 3 runners finished 1st, 2nd and 3rd, then I only know the order in which they finished. That is, I know *who* is better but not *how much* better. Conversely, if I know they took 2:15, 2:26 and 2:27 respectively to finish, I can conclude that #1 was much better while 2 and 3 are closely matched.

For the math types in the room, the 4 levels of measurement (nominal, ordinal, interval and ratio) proposed by SS Stevens have different allowable mathematical operations. Although I've never seen it, I've heard that much of the 70s was spent arguing that you couldn't apply parametric methods to analyze interval level data and that you needed to use non-parametric methods instead.

Hmm. I should stop rambling and start my stats homework.

This pedantry was brought to you by the number π and the letter Q

[Q]This pedantry was brought to you by the number n and the letter Q[/Q]

At least it wasn't brought to us by "n with property p." I hated that little f*cker.

(edit: fixed quote)



Post Edited (02-12-04 23:10)

"Let X be a set Y we call Z ..." (Just an example of what's wrong with mathematicians.)

Hayes...you're definitely not a facetimer, but maybe an outlier...

Now prove that Z, which somehow got property p, when combined with X-1, will mean that n-1 will have property p. And as always, label your work.

(edit: I realized I had capitalized that first property 'p,' and as we all know property 'P' is ENTIRELY different than property 'p.')



Post Edited (02-12-04 23:17)

Are we talking about obstruction penalty variability?

dss28 wrote:
Now prove that Z, which somehow got property p, when combined with X-1, will mean that n-1 will have property p. And as always, label your work.

But can you prove it by induction?

(edit: I realized I had capitalized that first property 'p,' and as we all know property 'P' is ENTIRELY different than property 'p.')

I have a colleague who once gave a lecture (with overhead transparencies) where p written in red ink was a different variable from p written in black ink, and they appeared in the same equation.

Did the students with blue pens try to shove them through their own eyes? I'd have cried.

Most people would cry after shoving a pen through their own eyes.

Gives a whole new meaning to that song "Black Eyes, Blue Tears," doesn't it....

Post Edited (02-12-04 23:39)

[Q]
But can you prove it by induction?
[/Q]

I never understood how or why we'd use induction for deductive logic.

Mike Ack wrote:

Hayes...you're definitely not a facetimer, but maybe an outlier...

While you were in sitting, I was out lying.... oh wait. Nevermind.

(And just for the record, instead of doing my stats, I watched L&O on tivo.)

dss28 wrote:

[Q]This pedantry was brought to you by the number n and the letter Q[/Q]

At least it wasn't brought to us by "n with property p." I hated that little f*cker.

Actually, that was a pi, not an n.

You must be using some crippled substandard microsoft browser or something.

Eh... bygones. I'll reassert:

At least it was brought to us by '[symbol for pi that my browser doesn't support but turns it into an 'n']' and not "n with property p." I hated that little f*cker.

Still works



Post Edited (02-12-04 23:58)

After reading all of this, it absolutely confirms my decision to become a chemical engineer. I haven't used statistics, ever, and I've only used calculus once in 22 years. Thank God!

JH

Well, I assumed head to head was the first tie-breaker, but record versus the top four?!?

BTW, if there's a three way tie in points for third, who ARE the top four?

JH

Jeff Hopkins '82 wrote:

Well, I assumed head to head was the first tie-breaker, but record versus the top four?!?

BTW, if there's a three way tie in points for third, who ARE the top four?

The ECAC actually cleared all this up several years ago:

[elf.elynah.com]

Actually it was a facetious question. I was sure it was figured out.

JH