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


Ridgewaycynic2013

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1 minute ago, Warcodered said:

As a certain rat faced weaselly Chiefs fan political commentator calls him "The Prince that was Promised"

 

I really hope people start throwing it back in Nick's face that Herbert & "The Prince" haven't done as much as Josh had by the same points in their careers. 

 

He talked Herbert up all off-season prior to 2022, then quietly dropped it as he kept hyping up Trevor. Now Trevor is in year 3, and to match Josh, he has to finish runner up in MVP voting & lead his team to the AFCCG (both too late for Herbert).

 

So if that hasn't happened, he should have people constantly remind him of that energy he held against Josh in year 3, 4 & 5, and do the same for them.

 

But at least 49ers fans are realizing how stupid Nick can be once he decides he doesn't like a QB. His Purdy hate this year has put him on the Niners sh*t list 😆

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4 minutes ago, JohnnyBuffalo said:

So Browns and Bengals can put up points with second/third/semi retired QBs and yet the QBs they replaced are somehow stars in the NFL?   Maybe the teams around them help make them look good.  
 

You know sort of like when the pasties went 11-5 and 3-1 without their cheating QB.   
 

Yet media talks Allen turnovers vs the fact that he carries the entire team.   

Thought about this myself 

 

is burrow - dare I say it - is he THAT ELITE?!?

 

id take Josh Allen over everyone including Patty 

 

 

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20 minutes ago, TheFunPolice said:

 

We won't get in at 9-8 but one of those other teams will I'll bet 

 

Too many inconsistent AFC teams with backup QBs

 

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.

 

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. 

 

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

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

 

15 * (3/8)^2 * (5/8)^4 ≈ 0.1398

So, there is approximately a 13.98% chance that 2 out of the 6 teams lose 8 games, and the remaining 4 teams lose more than 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.

 

 

Edited by Einstein
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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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