I'm trying to figure out a way to calculate this.

In sports standings, it's normal when one team has a .900 winning percentage, that you'll have one at .100, if a team goes undefeated, you have a team that goes winless.

I was trying to find out the odds for this in the NFL. I tried to make it easier by saying "Assume that half the teams were within two games of .500, and that their winning percentages averaged out to .500."

The way I was calculating it was to consider a victory a tangible thing. There are 32 teams that play 16 games each, that leaves 256 victories to disperse. Take out half, 16 teams and 128 of the victories, and leave (at least) one team to be undefeated, so you have 112 victories to split between 15 teams, and you know that those teams are ranked, with each successive team having more than or equal to the amount of wins as the next team.

Beyond that, I don't know what to do...is there an equation or a way to calculate this?

I mean, I can literally try to compile the list

16 16 16 16 16 16 16 0 0 0 0 0 0 0 0

16 16 16 16 16 16 15 1 0 0 0 0 0 0 0

16 16 16 16 16 16 14 2 0 0 0 0 0 0 0

16 16 16 16 16 16 14 1 1 0 0 0 0 0 0...

Is there a program or a formula to tell me what the odds are that the last place team would win one game? Would win two games, or more? I know that they could win 7, but the odds would be miniscule, but I'm not even sure what the amount of possibilities are for the 15 teams.

## Standings probability problem

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- emlightened
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### Re: Standings probability problem

As far as I can tell, the odds of winning no games, assuming all (but one, the winning one) of the teams are equally skilled, the chance of one team loosing all of their games is approximately 1-(1-1/2

Your table is also very off. Remember that there are 16 teams, who play 15 other teams, for 16*15/2 = 120 games total. One team wins all 15 of their games, which leaves 105 matches. If one team looses all of their games, then (assuming no draws) all of the other teams have at lest one win, and at most 14 wins, and even the team which wins all of their matches only wins 15 matches.

I'm not particularly good at probabilities, so I can't say much about accounting for the interdependence of the cases, or winning at most 1 match (not 0), but Wolfram Alpha may contain an answer.

^{14})^{14}~ 0.085%. The actual probability is off by a bit due to the incorrect assumptions that probabilities are all independent, and that the teams are equally matched. (The latter is rather difficult to account for mathematically, in a general case like this.)Your table is also very off. Remember that there are 16 teams, who play 15 other teams, for 16*15/2 = 120 games total. One team wins all 15 of their games, which leaves 105 matches. If one team looses all of their games, then (assuming no draws) all of the other teams have at lest one win, and at most 14 wins, and even the team which wins all of their matches only wins 15 matches.

I'm not particularly good at probabilities, so I can't say much about accounting for the interdependence of the cases, or winning at most 1 match (not 0), but Wolfram Alpha may contain an answer.

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### Re: Standings probability problem

It's simultaneously easier and harder than that. In the NFL anyway, schedules are not created equally - not all teams play against each other, and some teams meet twice instead of once - so none of the lines on your table are actually possible. The schedule looks like this: Each conference of 16 teams is further split into 4 divisions of 4 teams each. Teams play 6 intradivision games, 2 per team, each year. They also play 4 games against a specific intraconference division, and 4 against a division from the other conference. The final 2 games are with teams from the other divisions in one's conference, chosen based on last year's rankings.

From this, certain facts emerge. For starters, there can never be more than 2 undefeated teams in a conference - the undefeated team must defeat all 3 intradivision opponents, and all 4 teams in another division. In addition, the other conference may have 2 undefeated teams as well, but only if the scheduling works out that all 4 of those teams don't play each other in interconference matchups. By the same logic, you can't have more than 4 winless teams. Additionally, for every undefeated team there are 3 division teams who can have at most 14 wins, and for each winless team there are 3 in the division with at least 2.

Even assuming a team has a sub-.500 win expectancy against all their opponents, it's hard not to win at least one. Historically, since the NFL season expanded to 16 games there has only been one team to lose them all. Other than listing all the possibilities, however, I don't see an easy way to calculate this. I'd recommend doing so with an actual schedule, however, so you can keep track of who plays whom and avoid the above pitfalls.

From this, certain facts emerge. For starters, there can never be more than 2 undefeated teams in a conference - the undefeated team must defeat all 3 intradivision opponents, and all 4 teams in another division. In addition, the other conference may have 2 undefeated teams as well, but only if the scheduling works out that all 4 of those teams don't play each other in interconference matchups. By the same logic, you can't have more than 4 winless teams. Additionally, for every undefeated team there are 3 division teams who can have at most 14 wins, and for each winless team there are 3 in the division with at least 2.

Even assuming a team has a sub-.500 win expectancy against all their opponents, it's hard not to win at least one. Historically, since the NFL season expanded to 16 games there has only been one team to lose them all. Other than listing all the possibilities, however, I don't see an easy way to calculate this. I'd recommend doing so with an actual schedule, however, so you can keep track of who plays whom and avoid the above pitfalls.

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