For those in the know, how is draft position determined for teams that have an equal number of wins and losses at the end of the season? I look around, and EVERYONE except the Bengals has at least 2 wins. This could be a season where the Texans win 3 or 4 games and still have pretty good draft status...but that might depend on how the ties are broken. Is it too much to hope for that the Texans will automatically win all tie breakers since they will only be a 2nd year team at that point?
Ties are broken using opponent's winning percentage; the team with the higher/highest opponent winning percentage gets the better pick. Failing that (i.e. all teams have equal opponent winning %s), they use a coin toss.
So there is no special clause that says 2nd year expansion teams can automatically win tie breakers? Not that I have heard or read this anywhere, but it just sounded like something that might be done since we are getting extra draft picks.
if anything, having extra picks would hurt us under those circumstances. anyway, no, lack of superiority isn't an issue (unfortunately). on the plus side, the texans' record against this year could be very strong, thus giving us many tie-breakers: PHI, SD, IND, BUF and PIT could all be division winners and teams like TEN, NYG, WAS, CLE, BAL and JAX should all finish at or above .500.
No, it's lower/lowest gets the better pick. The thinking here is that if two (or more) teams have identical win percentages, the worst of the teams is the one playing the easier schedule (i.e. the one with the lower opp. win %). After week 8, the Texans opp win % is the lowest among teams with their win %. See The Huddle Report's Draft Order for current ordering.
Uh, that's what I said in my post. The team that plays the better (i.e. tougher) schedule gets the better draft pick. The team that has the lesser (i.e. weaker) schedule gets the lower (worst) draft pick.
I think you are still saying the opposite of what he is saying. What he is saying is that if you had this situation: Team A -- Win percentage of .300 Team B -- Win percentage of .300 Team A's Opponent Win Percentage -- .500 Team B's Opponent Win Percentage -- .300 Then Team B would be picking before Team A, because they had the same win percentage but their opponents were worse than Team A's opponents...so Team B must be a crappier team.
That illustrates exactly what you pointed out, Raven Lunatic, and comes from my first post. If you have two .300 teams with opp. winning % of .350 and .650, the team with the .650 one will get the better pick, as it says above.
Ming Dynasty is saying that the team whose opponent winning % of .350 will be the one to get the better pick. Think about it. The better pick is supposed to go to the team that is the worst. Two teams have the same record, but one was against opponents who were crappy (.350) and one was against opponents who were good (.650). The worst team of the two is the one that played the easier teams and still had a bad record (.300 in this case). So that team will be the one to get the best pick.
Uh, no it isn't. What you said was wrong, and what I said was right. Did you click on the link I gave you? I was hoping a more visual representation would be helpful, but as the saying goes, you can lead an mfclark to water, but you can't make him drink it.
