USCHO.com poll, 2/17

[Shorts]

For the non-math-oriented out there (Grant) I think the essential difficulty is this:
Suppose, at the very bottom of the rankings, you have that both Ithaca College and Tompkins County Community College get 4 points, and you have the rest of the solutions all figured out so that you know they each got two votes, and between them, there is 1 13th place, 2 14th places, and 1 15th place.  There is absolutely no way to tell whether IC got the 13th & 15th, and T3C got the 2 14ths or vice versa.

Of course, you can do this for any pair of teams that, for some subset of the votes they got, have the same number of votes with the same number of points.  Even adding the constraint that each ballot can only contain at most one of each team doesn't help things, because it could well be that, in the example above, it was a single pair of Ithaca-loving voters that gave points to both IC and T3C, and so the switching within their ballots wouldn't make anything infeasible.  So, to know exactly how many of each type of vote each team got, is not generally doable.

Now, just because it's not generally doable doesn't mean that it's never possible.  For example, as Adam '03 said, it's obvious that Miami got 1 15th place vote, and no other votes.  Similarly, given that Colorado College got exactly 30 1st place votes, I'm pretty sure the breakdown given by KeithK '93 is unique.

Aside from special cases such as those, though, I think the only way to know for sure would be to apply some social engineering, such as flat-out asking USCHO, or doing your own re-poll of the voters.

[Greg Berge]

Hypothetical question.  If there was just one more number for each team: the total ballots on which that team was included, would there then be sufficient information?

Aside: the more I look at this (now a whole 3 minutes, up from 2), the more it looks like one of those best-fit multiple node problems.  Avaya has 300 national nodes and wants to maximize their screwing of the consumer.  What is the most efficient way for them to route their volume?  And then you start out with seed values and run some sort of iterative fitting algorithm a few million times until the pct change from case N to N+1 drops below .0000000000000001%.

Of course, this is the kind of thing that smartass Chaos Theorists win Nobels for by proving there are various systems that will never settle down, no matter how many iterations, instead oscillating wildly between random values.  But as those guys rarely have offspring, maybe they will iterate themselves out of the picture.

[jd212]

wow, for something that people say means absolutely nothing, you guys are going to an AWFUL lot of trouble to figure out what our votes were. Obviously it doesn't mean nothing. But who really cares how the votes break down? and if you do, why?

and if you say you are doing it for fun, I swear you need to find a better hobby.

[Keith K]

To be honest I started trying to figure this out 'cuz I was bored at work.  And I find solving mathematical problems - particularly those involving meaningless things like hockey polls - to be strangely intersting and fun.  But hey, PhD rocket scientists are supposed to be geeks, right?

[nshapiro]

Ok, if you guys want another math problem that many have considered and none have solved  try this:

Pennsylvania used to have a Super 7 lottery, 80 numbers, you pick 7, then they pull 11 numbers, and you win if your 7 are in their 11.  What is the minimum number of tickets that must be bought to guarantee at least 1 winning ticket?

If this crowd can't solve this then I might try submitting it to the math dept. and letting them try.

[Keith K]

But where's the obligatory hockey content?

[jtwcornell91]

I think it has to be at least (80!7!4!)/(11!69!11!).  That's the number you need if you can manage it with no "overlap" between your tickets, i.e., no possibility of the same drawing winning for two of your tickets.  But I don't know off the top of my head if that's possible.  (The reasoning is that there are 80!/(11!69!) possible drawings, but any given ticket will win the prize for 11!/(7!4!) different drawings.)