Latest ESPN FPI has Vols favorites in nine games

#79
#79
Catbone said:
I dont think it is that simple. We went 2-1 vs those teams last year... and then lost to Vanderbilt and South Carolina.
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You have to beat FL and GA to get to ATL.. BAMA is the evil empire .. Vandy and SC traditionally are not a factor in the East
 
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#80
#80
Biopsyman said:
You have to beat FL and GA to get to ATL.. BAMA is the evil empire .. Vandy and SC traditionally are not a factor in the East
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While they don't factor as far as them ever really winning the East (SC won't be back often), they sure do factor in the East.
 
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#82
#82
My FPI has us winning 10 (we win the Fla. game)

Otherwise ESPN agrees with my calculations. :acute::acute::acute::acute::acute::acute:
 
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#83
#83
Roustabout said:
The Georgia game is because it's at home. A healthy team, sure.
7 wins or 10 are both real possibilities based on a few factors. Injuries again being at the center.
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BUTCH IS RESPONSIBLE FOR ALL THESE INJURIES !!!

FAHR BUTCH !!!

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#87
#87
They need to watch the Dormady montage Spring game throws 10 / 10 that Freak made. Then redo those probabilities.
 
#89
#89
goldvol said:
According to those probabilities, the probability of beating all 3 of GT, GA and uSCe is 26.6%

The odds of beating both Kentucky and Vanderbilt is 48.7%.

Gonna be a long year when we lose games our fanbase think we are "supposed" to win. :pilot:
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It doesn't work like that. Using the same method the probability that Tennessee loses to all three is just 3.8%. That does not add up.

The probability that Tennessee wins all three is actually 65.2%. Add the percentages and divide by 300.
 
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#90
#90
goldvol said:
I don't think it does, but I am open to discuss it. As an engineer, I've had enough math to get a basic understanding of probability.

So, feel free to enlighten me with your methodology for calculating the probability of beating Kentucky and Vanderbilt based on the probabilities given.
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Well I am a math major with statistics as the concentration - you really can't take these numbers and do what you did.

You can consider that each is a chance among 100 tries and add them together and divide by 300. That is still not statistically accurate.

But there is no way you go from predicting three wins to predicting what you did.
 
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#91
#91
volfan102455 said:
It doesn't work like that. Using the same method the probability that Tennessee loses to all three is just 3.8%. That does not add up.

The probability that Tennessee wins all three is actually 65.2%. Add the percentages and divide by 300.
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It does work like that you just have no concept of how probability is computed.

Using ESPN's FPI numbers as the base probability of success, UT has a 3.8% chance to lose all three of those games. That doesn't take into account CBJ's struggle with Muschamp, or any other factor, that's just using the FPI numbers, and they are correct. According the FPI projections as ESPN has calculated them, UT has a very low chance of losing all three of those games.

When determining the probability of independent events, P(A and B) = P(A) × P(B); you do not add the percentages and divide by the combined denominators.

The probability of you flipping a coin twice and getting heads both times, and rolling a six-sided dice and getting a 6 at the same time is 1/2*1/2*1/6= 4.16%. The fact that your thumb is 4 inches wide or that you toss the dice with your left hand doesn't matter, because that's not how probability works.

You can argue with ESPN's FPI calculations all day long, but but that's not going to change how probability is computed.

And it might be time to consider a different major IMHO.
 
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#92
#92
The probability of any specific combination of a bunch of things all happening is going to be pretty low, and also pretty irrelevant due to cherry-picking. I'd rather see the probabilities of different final records, based on those (probably incorrect, but still decent) numbers.
 
#93
#93
BeardedVol said:
It does work like that you just have no concept of how probability is computed.

Using ESPN's FPI numbers as the base probability of success, UT has a 3.8% chance to lose all three of those games. That doesn't take into account CBJ's struggle with Muschamp, or any other factor, that's just using the FPI numbers, and they are correct. According the FPI projections as ESPN has calculated them, UT has a very low chance of losing all three of those games.

When determining the probability of independent events, P(A and B) = P(A) × P(B); you do not add the percentages and divide by the combined denominators.

The probability of you flipping a coin twice and getting heads both times, and rolling a six-sided dice and getting a 6 at the same time is 1/2*1/2*1/6= 4.16%. The fact that your thumb is 4 inches wide or that you toss the dice with your left hand doesn't matter, because that's not how probability works.

You can argue with ESPN's FPI calculations all day long, but but that's not going to change how probability is computed.

And it might be time to consider a different major IMHO.
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The problem is that none of these calculations are statistically correct and cannot predict whether or not three separate events happen or don't happen.

Take the example of making free throws.

A person has made 3 of 4, so based on this so called math a person has only a 42.2 percent chance of making all of the next three.

The problem, when the person makes the first one - now their average is 80%. So is the chance they make the next two still 42% or has it increased to 51% since they made the first one.

Then if they make the second of the first three the percentage is now 83.3% so is the percentage that they make the next one still 42 percent or 69.4%?
 
#94
#94
All you can say from the FPI numbers is that Tennessee has a > 50% to win 9 of the 12 games they play.

You can't predict the chance to win 3 in a row or lose three in a row from these numbers.
 
#95
#95
volfan102455 said:
The problem is that none of these calculations are statistically correct and cannot predict whether or not three separate events happen or don't happen.

Take the example of making free throws.

A person has made 3 of 4, so based on this so called math a person has only a 42.2 percent chance of making all of the next three.

The problem, when the person makes the first one - now their average is 80%. So is the chance they make the next two still 42% or has it increased to 51% since they made the first one.

Then if they make the second of the first three the percentage is now 83.3% so is the percentage that they make the next one still 42 percent or 69.4%?
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Seriously, how are you a "math major" and you don't understand basic probability?:eek:hmy:

What you are describing is computing a percentage of success, that is not the same as computing the probability of success.

If you are a basketball player, who shoots free throws with a perfect 30% success rate, the probability of you making a single free throw, is 30%. 30% of the time, you should be successful in a single free throw. If you have to go up and shoot two free throws, that doesn't change the fact that you are still shooting with a probability of success at 30%. Hence, the chance of you making two free throws in a row, is P(A and B)=P(A)*P(B), .3 *.3= 9%. Because you only make free throws with a probability of success at 30%, when you have to make two shots, 9% of the time you will make both of them, and 91% of the time you will either make only one or miss both.
 
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#97
#97
BeardedVol said:
Seriously, how are you a "math major" and you don't understand basic probability?:eek:hmy:

What you are describing is computing a percentage of success, that is not the same as computing the probability of success.

If you are a basketball player, who shoots free throws with a perfect 30% success rate, the probability of you making a single free throw, is 30%. 30% of the time, you should be successful in a single free throw. If you have to go up and shoot two free throws, that doesn't change the fact that you are still shooting with a probability of success at 30%. Hence, the chance of you making two free throws in a row, is P(A and B)=P(A)*P(B), .3 *.3= 9%. Because you only make free throws with a probability of success at 30%, when you have to make two shots, 9% of the time you will make both of them, and 91% of the time you will either make only one or miss both.
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You do realize there are different views on probability calculations.

The purist version would imply that a player shooting a free throw, regardless of past history, always has a 50% change of making a free throw because there are only 2 outcomes and one of the two can happen. And for every game played every college football team would have a 50% chance of winning no matter how many games played. The team can win or lose the game. Only 2 outcomes are possible for 1 event.

But most factor in what has happened over time for situations that are identical - what they are really using is an average of success based on repeating the same event over and over again and use that to predict future outcomes. The thought is that over time the true probability emerges.

Even Frequentist probability does not align with your definition as that would use "number of made free throws" divided by "number attempted" to predict the probability that the next one is made. So the probability associated with the next event is based on all past events.

There are 5 different mathematical calculations for probability - not just one.
 
#98
#98
And yes as a math major I know that unless you understand the underlying information that created a percentage, you can't make assumptions based on it other than what it was intended to be used for.

And the FPI is just a predictions of who would win a game between two teams if they played 1, 2, ... 10, .... 100 times.
 
#99
#99
As for the original question - I see at least 10 wins as I am including LSU.

I am hopeful for Florida and would love to see the team exceed expectations and beat Alabama - but Saban is still at Bama and it is hard to win in Gainsville. One could hope for a rain out with a reschedule in Knoxville. :)

The 2017 team, at least at this point, seems to be all business and all about getting stronger and better.
 
#100
#100
volfan102455 said:
You do realize there are different views on probability calculations.

The purist version would imply that a player shooting a free throw, regardless of past history, always has a 50% change of making a free throw because there are only 2 outcomes and one of the two can happen. And for every game played every college football team would have a 50% chance of winning no matter how many games played. The team can win or lose the game. Only 2 outcomes are possible for 1 event.

But most factor in what has happened over time for situations that are identical - what they are really using is an average of success based on repeating the same event over and over again and use that to predict future outcomes. The thought is that over time the true probability emerges.

Even Frequentist probability does not align with your definition as that would use "number of made free throws" divided by "number attempted" to predict the probability that the next one is made. So the probability associated with the next event is based on all past events.

There are 5 different mathematical calculations for probability - not just one.
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Absolutely none of what you said changes the fact that the probability of independent events is computed as P(A and B)=P(A)*P(B). Whether it's the probability of flipping a coin and it landing on heads 3 times in a row, or rolling dice and coming up 6 100 times in a row, or the liklehood of making 20 free throws in a row, it's all computed using the same equation.

You can argue that the FPI projections are wrong, but you can't argue the way they are used when determining the probability of winning multiple games based off of the FPI projections.
 
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