He is now 38-23 as the HC of the Dallas Cowboys and has a good chance to become only the second coach in franchise history to win 12 games in three consecutive seasons.

Before his arrival Dallas hadn’t won 10 games in consecutive years since 1996.

  • OfficialBenReilly@alien.topB
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    10 months ago

    This is a cool stat, but it definitely does not paint the whole story. For one thing, he coached in Green Bay longer than he has coached for Dallas, so the amount in which a season would affect win percentage in Dallas is greater than that of Green Bay.

    Also, if this includes every year he was in Green Bay, then that would me it includes the year where Rodgers broke his collarbone and the team went 7-9 (which, while not as bad as I remember, that is still a season that would hurt his win percentage and help explain where that some of that gap came from). If you were to add the wins that season to the cowboys wins and the losses to the cowboys losses, then added the wins and losses together, you would get 77 games in total. Divide the total number of wins (45) by the total amount of games (77) you get a .508 win percentage. If you add another season like that to his record in Green Bay, he would have a .598 win percentage. That is a .115 difference compared to a .02 difference. I know it’s a bit unfair to add a season from Green Bay to the total record of his tenure in Dallas, I just thought it was a good way to show how much the difference in games played can affect the overall win percentage when the two teams had the same record in a season.

    While this stat is cool to look at, when there is a 127 game (which is the total amount of wins he had in Green Bay funnily enough) difference in the games played, win percentage is hard to compare.

    (I’m a 49ers fine by the way, so I have no horse in this race)