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January 6

Head coach award CFL

Doesn't CFL have a head coach award? —Preceding unsigned comment added by 65.95.106.139 (talk) 03:13, 6 January 2011 (UTC)[reply]

Yes, the Annis Stukus Trophy. Clarityfiend (talk) 04:06, 6 January 2011 (UTC)[reply]

Then How come they didn't give the award in 2010? —Preceding unsigned comment added by 70.31.18.54 (talk) 19:35, 6 January 2011 (UTC)[reply]

Looking at the references, it seems they announce the winner in late February or early March. Clarityfiend (talk) 21:47, 6 January 2011 (UTC)[reply]

Couple of NFL questions

1. Is it possible for a team to go 16-0 and not get first or second seed?

2. Why is it that several teams in the regular season face off against all the teams from a specific division? Examples:

New England Pariots: AFC North (Ravens, Browns, Bengals, Steelers), NFC North: (Vikings, Bears, Packers, Lions)

Vikings: NFC East (Giants, Redskins, Eagles, Cowboys)

Packers: NFC East (Giants, Redskins, Eagles, Cowboys)

Raiders: AFC South (Colts, Jags, Titans, Texans)

Saints: NFC West (49ers, Cards, Rams, Seahawks)

Cardinals: NFC South (Falcons, Saints, Panthers, Bucs)

Exactly why is this? Buggie111 (talk) 15:54, 6 January 2011 (UTC)[reply]

1. ~~Yes. If three teams go 16-0 in separate divisions of the same conference, then they will be seeds 1, 2, and 3 (one of them will have to be third seed).~~

Strike that. It is not possible for 3 teams to go 16-0 in the same conference. If two of the teams are in the same division, they must play each other. So, that cancels out the possibility of both being 16-0. If you have two teams in separate divisions of the same conference, they will play their own division and they could be on rotation to play the other two divisions. If they didn't place the same in the previous year, they won't play each other - so two teams can go 16-0. If you add a third team, the inter-division play will guarantee that one of the teams must play the another team. So, one will have to lose (or tie). Three teams in a single conference cannot go 16-0.

2. That is how the schedule is done. Every team in a division faces the other 3 teams twice (once at home and once away). That is 6 games. Another 4 are done as a tour of another division in the same conference (the other division rotates yearly). Similarly, they play 4 games against a division in the other conference. The last 2 games are played against the two divisions in their conference that they aren't playing against. They play the two teams that placed the same as they placed - if they placed 1st in their division, they play the 1st place teams in the other two divisions. See National Football League regular season. -- kainaw ™ 16:05, 6 January 2011 (UTC)[reply]

(edit conflict)1: Mathematically yes, but only BARELY. It is possible for three teams in the same conference to all not play each other. Each team has six teams from its own conference it does NOT play in a given year; so there will exist, for any team, atleast one grouping of two other teams where none of the three will play each other. In that case, it would be mathematically possible for all three of these teams to go 16-0; meaning that one of them would have to play in the first round. It would be broken by the NFL tie-breaking procedures (see National Football League playoffs for an explanation), which would require "Strength of Victory" (basically the records of the defeated teams) to break first. That being said, only one team in history has completed the regular season at 16-0 (a few other teams have had an undefeated regular season when the NFL had less games, and only one paired a perfect regular season with a League championship).ed: See kainaw's explanation below --Jayron32 16:23, 6 January 2011 (UTC)[reply]

2: See National Football League regular season. Since it is not currently possible for every team to meet every other team (prior to the AFL-NFL merger, every team in the NFL played every other team at least once), the NFL has developed a scheduling rubric which does a few things: 1) It aims to give each team a balanced schedule, that is, as close to the same strength schedule as everyone else. 2) It aims to see that every team plays every other team at least once every 3-4 years. 3) It makes sure that each team in the same division plays a "home-and-home" series against the others in its division, since that's the primary route to the playoffs. The National Football League regular season article explains in detail how it achieves these goals, and answers the rest of your question regarding scheduling (full disclosure: I had a lot to do with editing these articles, especially the parts I explained above). I hope that helps! --Jayron32 16:11, 6 January 2011 (UTC)[reply]

(To Jayron) Note that I amended my answer above. It is not possible for 3 teams in 3 different divisions of the same conference to not play each other. Here's an example for divisions N, E, S, and W. Assume N is on rotation to play E. Then E must be playing N. Then S and W are playing against each other. A team in N or E can go 16-0 and a team in S or W can go 16-0. If you want another team to go 16-0, it must play a team that we already designated as going 16-0. The way it works out, you get two sets of division pairings in each conference each year. So, you guarantee that every team in those pairings play each other once. -- kainaw ™ 16:17, 6 January 2011 (UTC)[reply]

Good catch. It looks like it is NOT possible for a team with a 16-0 record to be a 3-seed in the playoffs. Under the rules, which you correctly noticed, there can be, at MOST, 4 undefeated teams, two from each conference, and they would be then guaranteed to be the first two seeds for each conference's playoff bracket. --Jayron32 16:23, 6 January 2011 (UTC)[reply]

A correction to something said above: While all NFL teams may have played each other when there were only 8 or 9 teams in the league, that certainly wasn't the case in the 1960s, when a team in the Western Division might have gone a few years without playing a team in the East. -- Mwalcoff (talk) 00:29, 7 January 2011 (UTC)[reply]

Correct. What I should have said was that, pre-merger, a team would be guaranteed to play every team in its (side of the playoff bracket) prior to the merger. Thus, pre-Super Bowl, every member of the NFL's (Eastern/American) (Division/Conference) would play each other every year; and only play the members of the (Western/National) (Division/Conference) every few years. After the merger, even members of one conference did not play each other even every year. Doing some research, it turns out my statement, as presented, was WAY off, the ONLY season (indeed, not even the last of several or the first of several, but the ONLY one) where every team played every other team at least once was the 1932 NFL season. Other than that, there would be at least one team in the league that each team would not face. In most years, the NFL was a ten or eleven team league, and after the introduction of divisional structure in 1933, each team had a ten-game schedule; this was accomplished by playing a home-and-away against their own division, that's 8 games for a five-team division, leaving only two games against the other division. When the season was expanded to twelve games in 1948 that still left each team playing only four of the five teams in the other division. The 14 game schedule came in 1961, but that came at a time when the league had 14 teams; again assuming each team played the other teams in its own division (now called conferences) twice, that's 12 total games; that only leaves 2 for the other conference. --Jayron32 19:51, 7 January 2011 (UTC)[reply]

To answer questions 1 and 2, the article National Football League regular season#Formula helps. It's not just "several" teams, but "all" teams that face off against all the teams from a specific division. That wasn't the old formula, but it's the new one, and it also means that under the current formula, it's impossible to have more than 2 teams in a conference finish 16-0. First, since each time plays its division rivals every year, there can only be one 16-0 team in any division. Second, since each team plays the four teams from another division within its own conference on a yearly rotation, then among the two divisions that are paired for that season, there can only be one 16-0 team. To put it concretely: let's say the Bills, Browns, Texans, and Raiders all went undefeated in their divisions and undefeated against NFC teams (it's a stretch, I know). Well, the inter-AFC rotation means that the Bills must face either the Browns, Texans, or Raiders, depending on what division the East is paired with that season. Let's say they played the Browns -- now one of those teams is doomed to be 15-1, or else they're both doomed to be 15-0-1. Well, if East is paired with North, then South must be paired with West, so the Texans and Raiders must face each other -- and so again one of those teams can no longer go 16-0 that year. So you might have a 16-0 #2 seed, but never anything lower than that under the current scheduling system. In other words, it may be that the answer to your second question is that the NFL schedulers were considering the scenario in your first question! --M @r ē ino 16:44, 6 January 2011 (UTC)[reply]

On the other hand, with a sufficient number of ties (which admittedly seldom happens nowadays), there could be more than two undefeated teams in a conference, at 15-0-1. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:13, 7 January 2011 (UTC)[reply]