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What are the odds? (1 Viewer)

McGarnicle

McGarnicle

Footballguy
Look at the FF league stats below following week 1 action -- what's interesting about these numbers, and what are the odds that this outcome occurs in any given week?

Division A

Team 1: PF: 208 PA: 128

Team 2: PF: 193 PA: 111

Team 3: PF: 167 PA: 116

Team 4: PF: 128 PA: 208

Team 5: PF: 116 PA: 167

Team 6: PF: 111 PA: 193

Division B

Team 1: PF: 165 PA: 112

Team 2: PF:137 PA: 113

Team 3: PF: 135 PA:124

Team 4: PF: 124 PA: 135

Team 5 PF: 113 PA 137

Team 6 PF: 112 PA:165

 
Let me guess - all of the teams with more points than their opponent won?

Happened in my league too. Weirdest thing.

I assume the oddity is that the top-3 scoring teams in each division won
 
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ClownCausedChaos2 said:
This is McGarnicle. 10-1 says there's no pattern and he's just ####in' with the math nerds.
Click to expand...
Shuffle the deck completely, plenty of times 112 will play 111. Plenty of times 208 will play 193. I find it interesting that it happened like this, and I'm curious to see how the mathematical odds would be expressed and how odd this really is.
 
I hope the math nerds weigh in. My guess is you can't calculate the odds. Closest you could get would be to divide the top 6 and bottom 6 into two groups, assume 50/50 odds for each group to win, and multiply appropriately. So basically flipping a coin heads 6 times in a row. But, there are a lot more variables than that here.

 
On second thought...the only way it works is if each team in the top half is paired with a team from the bottom half. Any team in the top half that's paired with another team in the top half ruins it.

So the best team has a 6/11 chance of drawing an opponent from the "bottom" half.

Then the second best team has a 5/9 chance (since one bottom half team is already paired with the best team), and then 4/7, 3/5, 2/3, 1/1 for teams 3-6.

6/11 * 5/9 * 4/7 * 3/5 * 2/3 * 1/1 = 6.9%

 
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Six teams in each division. Assuming that each team always plays against a division opponent week one, the odds that the highest scoring team in division A faces a bottom three team are 3 in 5 or 60%. The second highest scorer must face either of the bottom two, out of three remaining teams. Multiple 2/3 by 60% and you get 40%. By definition if the top two teams face bottom three teams, the third highest scorer will too.

40% if division a doing it. 40% of division b doing it. Multiply it together and you get 16%.

So the answer is 16%, almost exactly the odds of rolling 7 in craps.

 
bostonfred said:
Six teams in each division. Assuming that each team always plays against a division opponent week one, the odds that the highest scoring team in division A faces a bottom three team are 3 in 5 or 60%. The second highest scorer must face either of the bottom two, out of three remaining teams. Multiple 2/3 by 60% and you get 40%. By definition if the top two teams face bottom three teams, the third highest scorer will too.

40% if division a doing it. 40% of division b doing it. Multiply it together and you get 16%.

So the answer is 16%, almost exactly the odds of rolling 7 in craps.
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Same idea, but his is right since I didn't realize it was two divisions.

 
wdcrob said:
On second thought...the only way it works is if each team in the top half is paired with a team from the bottom half. Any team in the top half that's paired with another team in the top half ruins it.

So the best team has a 6/11 chance of drawing an opponent from the "bottom" half.

Then the second best team has a 5/9 chance (since one bottom half team is already paired with the best team), and then 4/7, 3/5, 2/3, 1/1 for teams 3-6.

6/11 * 5/9 * 4/7 * 3/5 * 2/3 * 1/1 = 6.9%
Click to expand...
They're already in divisions. And you can see that the top three teams in division a all outscored the top three teams in division b, so you only have to work with half the league at a time.
 
In each division Team 1 has a 3/5 chance of playing a bottom team. In that situation, Team 2 has a 2/3 chance of playing a bottom team and then Team 3 1/1 chance so 2/5 multiplied by itself gives a 4/25 or 16% chance that the top 3 scoring teams in each division win. Determining the probability that each division has exactly 3 of the overall top 6 scoring teams is beyond my knowledge.

Using factorials would make it appear that 0 or 6 teams from division 1 = (1 * 2) situations, 1 or 5 teams = (6 * 2), 2 or 4 teams = (15 * 2), and 3 teams = (20 * 1). So 20/64 would be the likelihood of 3 teams from each division end up in the top 6. But thinking this over in my head, I don't think this is right...so my answer is between 5% and 8.43%.

 
Odds of top team playing a bottom 6? 6/11

Odds of 2nd team playing bottom 6? 5/9 (Two teams above no longer included)

Odds of third team? 4/7

4th? 3/5

5th? 2/3

6th? 1/1

So total odds are 6/11 X 5/9 X 4/7 X 3/5 X 2/3 X 1/1 = 6.926% Uncommon, but not all that special or rare.

ETA: I see someone above calculated the odds accurately almost exactly the same way.

 
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