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Around the NFL, Dec. 10 & 11, Week 14 of 18.


Ridgewaycynic2013

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1 minute ago, Einstein said:

 

If the Bills win, there will be 6 teams with a 7-6 record after today.

 

For each of the 6 teams to end up at 9-8 or worse, they would need to win 2 out of their last 4 games via any combination:

 

Win-Win-Loss-Loss

Win-Loss-Win-Loss

Win-Loss-Loss-Win

Loss-Win-Win-Loss

Loss-Win-Loss-Win

Loss-Loss-Win-Win

 

Each of these scenarios has a 1/2 * 1/2 * 1/2 * 1/2 = 1/16 chance of occurring.

 

Since there are 6 teams, you would multiply this probability by 6 to account for all 6 teams having the same chance of ending up at 9-8 or worse:

 

6 * (1/16) = 6/16 (or 3/8).

 

So, there is a 3/8 or 37.5% chance that at least one team out of the 6 will end up at 9-8 or worse, which would potentially open up a playoff spot for a 9-8 team.

 

For simplicity sake. we can multiply them together to find the probability that all 6 teams end up at 9-8 or worse:

 

(3/8) * (3/8) * (3/8) * (3/8) * (3/8) * (3/8) = (3/8)^6 ≈ 0.00390625

 

So, there is approximately a 0.39% chance that all 6 teams end up at 9-8 or worse, assuming they each have a 37.5% chance of winning 2 out of their last 4 games. 

 

But of course we only need 2 teams to get to 9-8 and the remaining teams to lose more than 8. So both of these scenarios would have to happen:

 

Scenario 1) 4 teams lose 8 games, and 2 teams lose more than 8 games.

Scenario 2: 5 teams lose 8 games, and 1 team loses more than 8 games.

 

 

 

Scenario 1: 15 * (3/8)^2 * (5/8)^4

Scenario 2: 6 * (3/8) * (5/8)^5

 

Add the probabilities of both scenarios:

(15 * (3/8)^2 * (5/8)^4) + (6 * (3/8) * (5/8)^5) ≈ 0.0949

 

So, there is approximately a 9.49% chance that 4 out of the 6 teams lose at least 8 games

 

That doesn’t even take into consideration the rams who currently have 7 losses and could lose just 1 more. Or tiebreakers or common games.

 

Just saying, the probability is low.

 

 

 

We're in the middle of a football Sunday, I ain't reading all that :)

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