BCS predictions/discussion/Knock Em' Off

[methomps]

Damn, hanging chads pervade the BCS!

Hanging chads will be nothing compared to the shitstorm if two teams ever tie for 2nd in the BCS.

 

[HailToMichigan]

I use the electoral analogy out of laziness. It is not exactly how the BCS works, but it illustrates the concept just fine. If anyone wants a concrete example using BCS numbers, I will put that together some other time.

Think of the electoral college. Ohio has 20 electoral votes. Under the traditional method, whichever candidate gets the most votes gets all 20 electoral votes. That is sort of how each individual computer works. Whoever is first gets 25 points. Whoever is second gets 24, whether they are .0001 or .1 away from 1st place.

Now, some people have proposed changing the electoral system so that votes are given out in proportion to how many actual votes a candidate gets. Thus, if Candidate A got 60% of the votes in Ohio, he would get 12 of Ohio's electoral votes. Say the other 40% of Ohio votes voted for Candidate B. Candidate B would get 8 electoral votes. This is sort of how the human polls are calculated.

Now imagine that you have a country with three states: Coaches, Harris, and Computers. Coaches and Harris proportion their electoral votes. Computers is all or nothing. Each has 20 electoral votes.

Candidate A gets 60% of the votes in Coaches and Harris, and 49% of the votes in Computers. 12 (Coaches) + 12 (Harris) + 0 (Computers) = 24

Candidate B gets 40% of the votes in Coaches and Harris, and 51% of the votes in Computers. 8 (Coaches) + 8 (Harris) + 20 (Computers) = 36

I give you your new President, Candidate B.

Your math is off, because the electoral college and the BCS aren't calculated quite the same. Here's what I mean:

In the coaches' poll, a "perfect" score is everyone voting for you, and I believe it ends up being 1,625 points, although the actual number isn't important. The Harris poll is the same only larger, between 3500-4500 points I believe. Your score for BCS purposes is the percentage of a perfect score, and every team is calculated this way. What this means is that in the electoral college, all the percentages add up to one. The BCS does not.

In your example, let's say there were 20 voters in each poll and only two teams. So first place is worth two points and second place is worth one, just as with the BCS polls, first place is worth 25 and so on. 12 voters in each poll vote for Team A and eight for Team B, and the computers all make Team B #1.

Thus Team A gets:

24/40 from the first poll,
24/40 from the second poll
20/40 from the computers

Team B gets

12/40 from the first poll
12/40 from the second poll
40/40 from the computers.

Team A gets 68/120 for a .567 BCS score. Team B gets 64/120 for a .533 BCS score. Team A is in first.

 

[BB73]

In your example, let's say there were 20 voters in each poll and only two teams. So first place is worth two points and second place is worth one, just as with the BCS polls, first place is worth 25 and so on. 12 voters in each poll vote for Team A and eight for Team B, and the computers all make Team B #1. Thus Team A gets:

24/40 from the first poll,
24/40 from the second poll
20/40 from the computers

Team B gets

12/40 from the first poll
12/40 from the second poll
40/40 from the computers.

Team A gets 68/120 for a .567 BCS score. Team B gets 64/120 for a .533 BCS score. Team A is in first.

Your example doesn't follow the BCS 25-24-23...01 points model, and is thus flawed.

The way the points work (25 for 1st, 24 for 2nd, etc., it's actually):
Team A (with 12/20 of the humans in each poll voting them #1, the rest of the voters and all the computers having them at #2) gets:

0.9840 - 492/500 from the first poll,
0.9840 - 492/500 from the second poll
0.9600 - 096/100 from the computers

0.9760 - average

Team B (with 12/20 of the humans in each poll voting them #2, the rest of the voters and all of the computers having them at #1) gets:

0.9760 - 488/500 from the first poll,
0.9760 - 488/500 from the second poll
1.0000 - 100/100 from the computers

0.9840 - average

Team B wins in your example of 12/20 of the poll voters ranking Team A #1 and Team B #2, and all 6 of the computers ranking Team B #1 and Team A #2.

 

[methomps]

Your math is off, because the electoral college and the BCS aren't calculated quite the same. Here's what I mean:

In the coaches' poll, a "perfect" score is everyone voting for you, and I believe it ends up being 1,625 points, although the actual number isn't important. The Harris poll is the same only larger, between 3500-4500 points I believe. Your score for BCS purposes is the percentage of a perfect score, and every team is calculated this way. What this means is that in the electoral college, all the percentages add up to one. The BCS does not.

In your example, let's say there were 20 voters in each poll and only two teams. So first place is worth two points and second place is worth one, just as with the BCS polls, first place is worth 25 and so on. 12 voters in each poll vote for Team A and eight for Team B, and the computers all make Team B #1.

Thus Team A gets:

24/40 from the first poll,
24/40 from the second poll
20/40 from the computers

Team B gets

12/40 from the first poll
12/40 from the second poll
40/40 from the computers.

Team A gets 68/120 for a .567 BCS score. Team B gets 64/120 for a .533 BCS score. Team A is in first.

20 voters in each poll means 20 1st place votes (each worth 25 points), meaning a perfect score of 500.

If Team A gets 12 first place votes (25 points each) and 8 second place votes (24 points each), that is (12*25) + (8*24)= 300 + 192= 492 points. 492/500= .984

If Team B gets 8 first place votes and 12 second place votes, that is (8*25)+ (12*24)= 200 + 288 = 488 points. 488/500= .976

In the computer polls, if Team B is 1st in every single one, their computer average is 1.000. If Team A is second in every single one, their computer average is .960 (24+24+24+24= 96)

Team A: .984 (Coaches) .984 (Harris) .960 (Computers)= .976
Team B: .976 (Coaches) .976 (Harris) 1.000 (Computers= .984

Team A is 1st in two human polls, second in the computer polls, and 2nd in the BCS.

 

[BB73]

20 voters in each poll means 20 1st place votes (each worth 25 points), meaning a perfect score of 500.

If Team A gets 12 first place votes (25 points each) and 8 second place votes (24 points each), that is (12*25) + (8*24)= 300 + 192= 492 points. 492/500= .984

If Team B gets 8 first place votes and 12 second place votes, that is (8*25)+ (12*24)= 200 + 288 = 488 points. 488/500= .976

In the computer polls, if Team B is 1st in every single one, their computer average is 1.000. If Team A is second in every single one, their computer average is .960 (24+24+24+24= 96)

Team A: .984 (Coaches) .984 (Harris) .960 (Computers)= .976
Team B: .976 (Coaches) .976 (Harris) 1.000 (Computers= .984

Team A is 1st in two human polls, second in the computer polls, and 2nd in the BCS.

Slowpoke. And mine's also prettier.

 

[HailToMichigan]

True, I guess I got caught up in the electoral college thing....it didn't quite jibe, to me, and I set out to figure out why.

I'm still not convinced though. Now all the math is there, but I think the original premise is flawed. methomps, you're saying each computer assigns 25 points to the first place team regardless of how close the second place team is, while the polls take the percentages into account and include the margin between first and second. Do I have that right? But you're comparing one computer's vote to the entire poll. Think of the computers as individual voters and the computer section as just another poll. Coach Joe Blow (or Intern Joe Blow if Coach feels like sleeping in on Sunday) does the same as the computers. He assigns 25 points to the first place team, regardless of how close the second place team is.

So our little math exercises assumed that the voters were not unanimous and the computers were, so naturally that 1/3 weight, unanimously cast, will overcome dissension in the ranks in the other 2/3. If the computers unanimously pick Team B and the voters pick Team A by 75% instead of 60%, they overcome the computers.

The computers only have more power than the voters because there's 6 computers and 60 coaches voting. So each coach cannot have as much effect on the rankings as an individual computer. But the human polls do outweigh the computers, because the computers are just voters in a poll as well.

 

[BB73]

True, I guess I got caught up in the electoral college thing....it didn't quite jibe, to me, and I set out to figure out why.

I'm still not convinced though. Now all the math is there, but I think the original premise is flawed. methomps, you're saying each computer assigns 25 points to the first place team regardless of how close the second place team is, while the polls take the percentages into account and include the margin between first and second. Do I have that right? But you're comparing one computer's vote to the entire poll. Think of the computers as individual voters and the computer section as just another poll. Coach Joe Blow (or Intern Joe Blow if Coach feels like sleeping in on Sunday) does the same as the computers. He assigns 25 points to the first place team, regardless of how close the second place team is.

So our little math exercises assumed that the voters were not unanimous and the computers were, so naturally that 1/3 weight, unanimously cast, will overcome dissension in the ranks in the other 2/3. If the computers unanimously pick Team B and the voters pick Team A by 75% instead of 60%, they overcome the computers.

The computers only have more power than the voters because there's 6 computers and 60 coaches voting. So each coach cannot have as much effect on the rankings as an individual computer. But the human polls do outweigh the computers, because the computers are just voters in a poll as well.

No matter how close the teams are in the 6 computers, they each individually slot teams into a ranking order from 1 to 25. Those values are then flipped (1 becomes 25, 2 becomes 24, etc.), the highest and lowest are tossed out, and the total of the 4 remaining numbers is divided by 100 to get the computer 'average'. So as long as at least 5 of the 6 computers have Team B at number 1, Team B will get a perfect score of 1.0000. If Team A is first in 1 computer, and just a hair behind Team B in the other 5 computers, Team A will get a score of .9600.

Because, in each individual computer, the closeness between the spots doesn't matter to the BCS.

But if Team A can sneak ahead of Team B by just a hair in a second computer, the values become .0099 for Team B and .097 for Team A. If they are each #1 in three computers and #2 in at least two of the other 3 computers, they will both get a computer acverage of .0098. That's why a late game from Hawaii can change the BCS even though a top team isn't playing in it.

 

[methomps]

True, I guess I got caught up in the electoral college thing....it didn't quite jibe, to me, and I set out to figure out why.

I'm still not convinced though. Now all the math is there, but I think the original premise is flawed. methomps, you're saying each computer assigns 25 points to the first place team regardless of how close the second place team is, while the polls take the percentages into account and include the margin between first and second. Do I have that right?

Correct.

But you're comparing one computer's vote to the entire poll. Think of the computers as individual voters and the computer section as just another poll. Coach Joe Blow (or Intern Joe Blow if Coach feels like sleeping in on Sunday) does the same as the computers. He assigns 25 points to the first place team, regardless of how close the second place team is.

So our little math exercises assumed that the voters were not unanimous and the computers were, so naturally that 1/3 weight, unanimously cast, will overcome dissension in the ranks in the other 2/3. If the computers unanimously pick Team B and the voters pick Team A by 75% instead of 60%, they overcome the computers.

The computers only have more power than the voters because there's 6 computers and 60 coaches voting. So each coach cannot have as much effect on the rankings as an individual computer. But the human polls do outweigh the computers, because the computers are just voters in a poll as well.

The problem is that the common understanding of the system is that "human polls control 2/3, computer polls control 1/3." That's why there is going to be a lot of headscratching (from the neutral observers; implosion from the jilted fans) when one team is a solid #2 in both human polls but on the outside looking in (or even #1 in one or both human polls). People don't really think of needing a 75% edge in both human polls.

 

[BengalsAndBucks]

Would winning out ensure tradition-rich programs such as Oklahoma or No. 4 LSU get a shot at the national title? Or will college football fans be treated to a riveting Sugar Bowl showdown between Ohio State and South Florida? The Buckeyes and Bulls occupy the top two slots in the first poll.

I must have missed like 30 years or something, I could have swore that tOSU was one of the most tradition rich programs in college football history. Especially compared to LSU, but what do I know.

 

[HailToMichigan]

No matter how close the teams are in the 6 computers, they each individually slot teams into a ranking order from 1 to 25. Those values are then flipped (1 becomes 25, 2 becomes 24, etc.), the highest and lowest are tossed out, and the total of the 4 remaining numbers is divided by 100 to get the computer 'average'. So as long as at least 5 of the 6 computers have Team B at number 1, Team B will get a perfect score of 1.0000. If Team A is first in 1 computer, and just a hair behind Team B in the other 5 computers, Team A will get a score of .9600.

Because, in each individual computer, the closeness between the spots doesn't matter to the BCS.

But if Team A can sneak ahead of Team B by just a hair in a second computer, the values become .0099 for Team B and .097 for Team A. If they are each #1 in three computers and #2 in at least two of the other 3 computers, they will both get a computer acverage of .0098. That's why a late game from Hawaii can change the BCS even though a top team isn't playing in it.

But that's exactly what a human voter does, isn't it? They slot each team from 1 to 25. They might know for sure they're putting Boston College 3rd but agonize all night over OSU vs. USF....but in the end, they'll either slot OSU or USF to the 25 points and the decision process won't be reflected in the poll.

When you think about it, except for the removal of the top and bottom scores for each team, the computers go through the exact same process as the human polls. They each vote, the votes are tallied, and the teams are ranked accordingly.

And a late Hawaii game has the chance of influencing the human voters too. (I don't think college games should be played on Sunday, by the way. At all.) Say Hawaii played Washington yesterday instead of whoever, and utterly demolished them. A voter might then wish he could rethink his vote and put USF on top, based on Washington not looking as strong as he thought they were.

The problem is that the common understanding of the system is that "human polls control 2/3, computer polls control 1/3." That's why there is going to be a lot of headscratching (from the neutral observers; implosion from the jilted fans) when one team is a solid #2 in both human polls but on the outside looking in (or even #1 in one or both human polls). People don't really think of needing a 75% edge in both human polls.

Yes, but you're assuming the computers will be unanimous, voting in lockstep with each other. If they don't, and if they vote with roughly the same proportion as the human polls, then the 2/3 to 1/3 ratio will hold true.

 

[BB73]

But that's exactly what a human voter does, isn't it? They slot each team from 1 to 25. They might know for sure they're putting Boston College 3rd but agonize all night over OSU vs. USF....but in the end, they'll either slot OSU or USF to the 25 points and the decision process won't be reflected in the poll.

That's true. Many people look at a single computer poll as the same as the Harris or Coaches poll, but in terms of teams potentially changing positions, it's better to compare it to an individual voter.

 

[methomps]

Yes, but you're assuming the computers will be unanimous, voting in lockstep with each other. If they don't, and if they vote with roughly the same proportion as the human polls, then the 2/3 to 1/3 ratio will hold true.

I'm not assuming anything. We're talking hypotheticals.

 

[Zurp]

That's true. Many people look at a single computer poll as the same as the Harris or Coaches poll, but in terms of teams potentially changing positions, it's better to compare it to an individual voter.

Good point. But for consistency, shouldn't the average of each of the individual computer voters be calculated the same way as the average of the human voters?

 

[kinch]

Wow, lots of computing here.

At this rate, I bet that within about ten years or so the BCS computers will become sentient and just vote for who they think is best in their own poll.

 

[Zurp]

Wow, lots of computing here.

At this rate, I bet that within about ten years or so the BCS computers will become sentient and just vote for who they think is best in their own poll.

I think in order for me to agree to that bet, you need to more adequately define "about ten years or so."

I think that IF the BCS computers become sentient in that amount of time, they'll also decide that they have more important things to do than to fill out their ballots. They'll program a graduate assistant formula to fill out their ballots for them.