The best method to estimate our total wins this year

#51
#51
There is no such thing as a 100% guaranteed win projection. If there were some games wouldn't even be played.
 
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#52
#52
AboveAll, it has been a loooong time since my prob & stat classes, but I'm pretty sure your additive method is invalid. It may seem to give the right answer, but it's not based on good math.

To determine the chance of winning 9 and losing 3 next season, using the fictional probability estimates you came up with, you would have to calculate the chance of each possible 9-3 outcome first. So one calculation to determine the chance we lose to BGSU, Oklahoma, and W Carolina, then beat all others (.05 * .45 * 0 * .66 * .75 * .55 * .20 * .90 * .80 * 1 * .66 * 1 = 0.00). Another calculation for the chance we lose to BGSU, Oklahoma, and Florida, but beat all others...and so on and so on, until you've included all possible combinations of 9 wins and 3 losses.

This is a TON of calculations, but it's the only accurate way to do this. Your additive approximation is only that, a "poor man's approximation."

Like I said, that's if I remember my prob and stats classes right.
 
#53
#53
BearCat204 said:
mth.gif
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Rainman did it and he was a 'ruh-tard'
 
#54
#54
VFL-82-JP said:
AboveAll, it has been a loooong time since my prob & stat classes, but I'm pretty sure your additive method is invalid. It may seem to give the right answer, but it's not based on good math.

To determine the chance of winning 9 and losing 3 next season, using the fictional probability estimates you came up with, you would have to calculate the chance of each possible 9-3 outcome first. So one calculation to determine the chance we lose to BGSU, Oklahoma, and W Carolina, then beat all others (.05 * .45 * 0 * .66 * .75 * .55 * .20 * .90 * .80 * 1 * .66 * 1 = 0.00). Another calculation for the chance we lose to BGSU, Oklahoma, and Florida, but beat all others...and so on and so on, until you've included all possible combinations of 9 wins and 3 losses.

This is a TON of calculations, but it's the only accurate way to do this. Your additive approximation is only that, a "poor man's approximation."

Like I said, that's if I remember my prob and stats classes right.
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My "additive method" is explained here. Pay attention to steps 4 and 5 in the first method: 3 Ways to Calculate an Expected Value - wikiHow

In our case, the "value" we're talking about is the expected number of wins in a season.
 
#55
#55
AboveAllNations said:
Most of us on Volnation are optimistic about our chances to win the SEC East this year, but there's still a vocal minority who think we should be "content" to win 8 games. Other posters think we'll lose to a terrible Florida team this year because of games that happened years ago with totally different players and different coaching staffs.

However, if you put aside irrational superstitions about losing streaks, ignore obvious recruiting propaganda from Butch Jones about our alleged "lack of depth," and realize that our perceived shortcomings (backup QB, offensive tackle, and #2 LB) pale in comparison to those of our SEC East rivals, it becomes clear that we should be considered the SEC East favorites this season. We've got the most returning starters, the best QB, the best D-line, the best secondary, and we host the other team with the best odds to win the division (Georgia).

So how many games should we realistically expect to win this season? The most accurate way to get an estimate would be to look at our schedule and add up the probabilities that we win each game. Josh Dobbs is really the only truly irreplaceable player on our team this year, so as long as he stays healthy (and there's absolutely no indication that he won't), I'm confident in these game-by-game odds:

v. Bowling Green (95% win)
v. Oklahoma (55% win)
v. Western Caroline (100% win)
@ Florida (66% win)
v. Arkansas (75% win)
v. Georgia (55% win)
(bye)
@ Alabama (20% win)
@ Kentucky (90% win)
v. South Carolina (80% win)
v. North Texas (100% win)
@ Missouri (66% win)
v. Vanderbilt (100% win)

Multiply each game by the win probability percentage:

v. Bowling Green (.95)
v. Oklahoma (.55)
v. Western Caroline (1.0)
@ Florida (.66)
v. Arkansas (.75)
v. Georgia (.55)
(bye)
@ Alabama (.20)
@ Kentucky (.90)
v. South Carolina (.80)
v. North Texas (1.0)
@ Missouri (.66)
v. Vanderbilt (1.0)

Then add up the numbers to get the number of expected wins:

.95 + .55 + 1.0 + .66 + .75 + .55 + .20 + .9 + .8 + 1.0 + .66 + 1.0 = 9.02 wins

So even though we're arguably favored in 11 of our 12 games, we should reasonably expect to win about 9 games this year. If you agree with the odds I've estimated for each game, 8-4 would be a disappointment. By contrast, here's my expected win total for Georgia:

v. U. La-Monroe (100% = 1.0)
@ Vanderbilt (100% = 1.0)
v. South Carolina (80% = .80)
v. Southern Jaguars (100% = 1.0)
v. Alabama (33% = .33)
@ Tennessee (45% = .45)
v. Missouri (75% = .75)
- bye -
@ Florida (75% = .75)
v. Kentucky (90% = .9)
@ Auburn (20% = .20)
v. Ga. Southern (100% = 1.0)
@ Ga. Tech (50% = .50)

Adding up their win probabilities for each game, Georgia's expected win total is only 8.68.

If we're likely to win the same number of games, and likely to actually beat Georgia head-to-head, why shouldn't we be favorites to win the division? I'll hang up and listen.
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What a joke!:banghead2:

This is based on your win percentages, nothing scientific or rational at all, just delusional pontification.
 
#56
#56
AboveAllNations said:
My "additive method" is explained here. Pay attention to steps 4 and 5 in the first method: 3 Ways to Calculate an Expected Value - wikiHow

In our case, the "value" we're talking about is the expected number of wins in a season.
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Yeah, I looked it up for myself before seeing your link, heh. You're right, you're 100% right, I'm totally wrong. It's what I get for trying to use knowledge last tapped in the 1980s. :stinker2:

Accept my apology, AboveAll, you schooled me. :)
 
#57
#57
This isn't really a method. It's just a basic rule of probability. I guess in order to have an accurate win projection you need to make sure to add correctly... but the accuracy lies with in assigning probabilities of each game and confidence level for each set of outcomes. Those probabilities might add up to 9 wins, but what level of confidence do you have that we will win 9 games? 30%? What are the odds we only win 8 games? 29%? You might be accurate in that respect but it was pretty much a tossup between the two...
 
#58
#58
On a side not I'd like to think UK has a better shot than 90% at UT. We gave arguably our worst performance of the year at Neyland and the next week after we were one play away from upsetting a ranked opponent on the road. Both teams will be be within 2-3 games off a bye. Both coming off very tough games, Bama and MSU. I'm hoping we can least keep it to a 10 point loss.

Either way we know nothing until football season and we see the games. I can't wait.
 
#59
#59
VFL-82-JP said:
AboveAll, it has been a loooong time since my prob & stat classes, but I'm pretty sure your additive method is invalid. It may seem to give the right answer, but it's not based on good math.

To determine the chance of winning 9 and losing 3 next season, using the fictional probability estimates you came up with, you would have to calculate the chance of each possible 9-3 outcome first. So one calculation to determine the chance we lose to BGSU, Oklahoma, and W Carolina, then beat all others (.05 * .45 * 0 * .66 * .75 * .55 * .20 * .90 * .80 * 1 * .66 * 1 = 0.00). Another calculation for the chance we lose to BGSU, Oklahoma, and Florida, but beat all others...and so on and so on, until you've included all possible combinations of 9 wins and 3 losses.

This is a TON of calculations, but it's the only accurate way to do this. Your additive approximation is only that, a "poor man's approximation."

Like I said, that's if I remember my prob and stats classes right.
Click to expand...

Typically you'd just run a monte carlo simulation
 
#60
#60
golfballs said:
Typically you'd just run a monte carlo simulation
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Ok, Golfballs, you are definitely beyond my limited prob & stats knowledge at this point in my life. Isn't Monte Carlo a (1) city in Europe, (2) car race, and (3) drink? :)
 
#61
#61
golfballs said:
This isn't really a method. It's just a basic rule of probability. I guess in order to have an accurate win projection you need to make sure to add correctly... but the accuracy lies with in assigning probabilities of each game and confidence level for each set of outcomes. Those probabilities might add up to 9 wins, but what level of confidence do you have that we will win 9 games? 30%? What are the odds we only win 8 games? 29%? You might be accurate in that respect but it was pretty much a tossup between the two...
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Give me a little time, I'll try to figure out the odds we win exactly 6, 7, 8, 9, 10, or 11 games. I already calculated the odds of us going undefeated at 1.3%.
 
#63
#63
VFL-82-JP said:
Ok, Golfballs, you are definitely beyond my limited prob & stats knowledge at this point in my life. Isn't Monte Carlo a (1) city in Europe, (2) car race, and (3) drink? :)
Click to expand...

Monte Carlo simulation is a computerized mathematical technique that allows people to account for risk in quantitative analysis and decision making
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keanu-reeves-whoa.jpg
 
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#67
#67
AboveAllNations said:
No, adding up the probabilities gives you the expected number of times you will likely get shot during a 6-shot game (assuming you can still shoot after you've already been shot). In a six-shot game of Russian roulette with one bullet in a six-round chamber, you would expect to get shot one time.

You calculate the likelihood of surviving a six-shot game without being shot by multiplying the probabilities. Your likelihood of survival is roughly 33%. Do you appreciate the difference?

In my OP, I added up the probabilities to determine the expected number of wins in a 12-game schedule. You can calculate our odds of going undefeated by multiplying the probabilities.
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The calculations for the expected number of wins is correct. The calculations for a certain record like 12-0, 11-1 have to make an assumption that the games are independent events. This means that the outcome of a game does not affect the outcome of another game. This could be questionable, as an example, if we beat OU, the team could gain confidence and make future victories more likely. Of course, the reverse is also true.
 
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#68
#68
zqvol said:
What a joke!:banghead2:

This is based on your win percentages, nothing scientific or rational at all, just delusional pontification.
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Hey, I freely admit that the win probabilities I've assigned to each game are subjective personal opinions. But for purposes of estimating the likely number of wins any team would win in a season, that kind of guess work is unavoidable. Accounting for estimated game-by-game probabilities is more accurate than simply assigning a win or loss to each game, because - as I said before - we should be favored in 11 of our 12 games. My method accommodates for upset losses (or, in the case of the Alabama game, a potential upset win).

While subjective, my game-by-game percentages aren't arbitrary. They roughly correspondent to the point spreads I anticipate for each game:

favored by 21+ = 100% win
favored by 17-20 = 90% win
favored by 14-16 = 80% win
favored by 10-13 = 75% win
favored by 7-9 = 66% win
favored by 4-6 = 60% win
favored by 1-3 = 55% win
even point spread = 50% win
underdog by 1-3 = 45% win
underdog by 4-6 = 40% win
underdog by 7-9 = 33% win
underdog by 10-13 = 25% win
underdog by 14-16 = 20% win
underdog by 17-20 = 10% win
underdog by 21+ = 0% win

This accounts for the fact that games don't just fall into "locks" and "toss-ups," it's a sliding scale. A team favored by 7 points should, in simulation where the game is played hundreds of times, win about twice as much as they lose (2/3, or 66%). A team favored by 17 points should win nine times out of ten.

So, having explained that, do my estimated point spreads really look so "delusional?"

v. Bowling Green (-20 pts)
v. Oklahoma (-3 pts)
v. Western Caroline (-28 pts)
@ Florida (-7 pts)
v. Arkansas (-10 pts)
v. Georgia (-3 pts)
(bye)
@ Alabama (+14 pts)
@ Kentucky (-17 pts)
v. South Carolina (-14 pts)
v. North Texas (-24 pts)
@ Missouri (-7 pts)
v. Vanderbilt (-21 pts)

I don't think this is so crazy. Other people may fear the Arkansas game more than I do, but I'm not worried about a home game against a team that can't pass the ball.
 
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#69
#69
OldandStillaVol said:
The calculations for the expected number of wins is correct. The calculations for a certain record like 12-0, 11-1 have to make an assumption that the games are independent events. This means that the outcome of a game does not affect the outcome of another game. This could be questionable, as an example, if we beat OU, the team could gain confidence and make future victories more likely. Of course, the reverse is also true.
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Fair point. More importantly, an injury in one game would obviously carry over and have an effect (however small) in subsequent games. But, given that all teams experience injuries and emotional highs & lows during the course of a long season, I think it's okay to consider each game independently for the purposes of getting a pre-season estimate for the likelihood we win a specific amount of games. I'm still thinking about it, but I'm positive it can be done.

As others have pointed out, the key is having accurate game-by-game probabilities.
 
#70
#70
I just don't know how accurate the numbers can be taken as when one touchdown swings it from tossup to a heavy win/loss percentage. It goes back to the idea that while these numbers might be statistically correct, how do you factor in the probability that it actually happens. Golfballs worded it much better than I can.
 
#71
#71
I'm honestly laughing at this point. The discussion has become whether or not the OPs math is correct based off of given percentages. The part where those percentages are completely made up, no longer a factor..

Seriously, how can anyone not love this place? You just can't put a price on these things.
 
#73
#73
LittleCat said:
On a side not I'd like to think UK has a better shot than 90% at UT. We gave arguably our worst performance of the year at Neyland and the next week after we were one play away from upsetting a ranked opponent on the road. Both teams will be be within 2-3 games off a bye. Both coming off very tough games, Bama and MSU. I'm hoping we can least keep it to a 10 point loss.

Either way we know nothing until football season and we see the games. I can't wait.
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Reading this made my mouth taste like I licked an old battery
 
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#74
#74
LittleCat said:
I just don't know how accurate the numbers can be taken as when one touchdown swings it from tossup to a heavy win/loss percentage. It goes back to the idea that while these numbers might be statistically correct, how do you factor in the probability that it actually happens. Golfballs worded it much better than I can.
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Predicting what a bunch of college kids are going to do with a football in their hands... it prone to quite a bit of volatility..
 
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#75
#75
AboveAllNations said:
As others have pointed out, the key is having accurate game-by-game probabilities.
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There are more complicated statistical models which incorporate the uncertainty of your probabilities into the estimate of total number of wins. As it turns out, the simple additive method does ok as long as your subjective probabilities are reasonably unbiased (i. e. the probabilities aren't uniformly higher or lower than the true unknown probabilities).
 
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