89 Miami (6-6)
37% at Ohio
95% vs. BGSUcks
99% vs. Akron
43% @ Ball State

90 Ohio (8-4)
63% vs. Miami
60% vs. WMU
85% @ BGSUcks
97% @ Akron

93 WMU (6-6)
60% vs. Ball State
40% @ Ohio
46% @ NIU

95 Ball State (7-5)
31% @ WMU
71% vs. CMU
57% @ Kent
57% Miami

103 Toledo (7-5)
72% vs. Kent
58% vs. NIU
29% @ UB
40% @ CMU

104 EMU (7-5)
55% vs. UB
94% @ Akron
35% @ NIU
72% vs. Kent

105 Buffalo (7-5)
45% @ EMU
58% @ Kent
71% vs. Toledo
94% vs. BGSUcks

107 NIU (5-7)
47% @ CMU
42% @ Toledo
65% vs. EMU
54% vs. WMU

108 CMU (7-5)
53% vs. NIU
29% @ BSU
60% vs. Toledo

114 Kent (3-9)
28% @ Toledo
42% vs. UB
43% vs. BSU
28% @ EMU

125 BGSUcks (3-9)
87% vs. Akron

129 Akron (0-12)

Minor correction... But I think the first game under Ohio should be vs. Miami instead of Ball State

Quote:90 Ohio (8-4)
63% vs. Ball StateMiami
60% vs. WMU
85% @ BGSUcks
97% @ Akron

5-7 looks about right for us at this point. Really hoping we don't do worse than that.

(10-30-2019 12:51 PM)BobcatEngineer Wrote:  Minor correction... But I think the first game under Ohio should be vs. Miami instead of Ball State

Quote:90 Ohio (8-4)
63% vs. Ball StateMiami
60% vs. WMU
85% @ BGSUcks
97% @ Akron

FIXED

I don't have any problem with the % estimates of the individual games, but the better estimate would be to add the 4 percentages up to get a total number of estimated wins. Then a total of 300% in 4 games would mean a likely 3 wins, 200% would mean a likely 2 wins. For example. Ball State has 57, 57, 31, and 71, which adds up to 216, which gives a most likely result of 2 wins, not the 3 which you are predicting. Ohio has a total of 305 which would be a most likely result of 3 wins,not 4. Miami has 274 which would be a most likely result of 3 wins, not 2. All 3 of those would be different estimated wins totals.

That is actually the way Vegas figures estimated win totals for a season, and is more accurate than just saying Ball State has two games with a 57% chance of winning so that is 2 likely wins when mathematically it is closer to 1 likely win than 2.

(10-30-2019 03:34 PM)DICK Wrote:  I don't have any problem with the % estimates of the individual games, but the better estimate would be to add the 4 percentages up to get a total number of estimated wins. Then a total of 300% in 4 games would mean a likely 3 wins, 200% would mean a likely 2 wins. For example. Ball State has 57, 57, 31, and 71, which adds up to 216, which gives a most likely result of 2 wins, not the 3 which you are predicting. Ohio has a total of 305 which would be a most likely result of 3 wins,not 4. Miami has 274 which would be a most likely result of 3 wins, not 2. All 3 of those would be different estimated wins totals.

That is actually the way Vegas figures estimated win totals for a season, and is more accurate than just saying Ball State has two games with a 57% chance of winning so that is 2 likely wins when mathematically it is closer to 1 likely win than 2.

But then lots of teams are playing four games without either team being granted the win.

If you were a 61% winner in all 4 games, that's 244%... 2 wins. While 39% for the opposing teams Is 156%, and that's one win. Who wins the last game?

Realistically if you play 4 games, 400% is impossible, therefore no one can win all 4.

57% x 2 games is 32.5% to win em both. While for the opposing team it's 18.5% to do the same thing. So it's still likely 2 wins for the favorite.

I think the Broncos should be favored over NIU, and vs BSU should be more of a toss-up. 6-6 reasonable, but I think the team has talent to snag 2 of the 3.

BSU favored 70% over CMU too high. That game could go either way.

(10-30-2019 04:18 PM)UofToledoFans Wrote:  

(10-30-2019 03:34 PM)DICK Wrote:  I don't have any problem with the % estimates of the individual games, but the better estimate would be to add the 4 percentages up to get a total number of estimated wins. Then a total of 300% in 4 games would mean a likely 3 wins, 200% would mean a likely 2 wins. For example. Ball State has 57, 57, 31, and 71, which adds up to 216, which gives a most likely result of 2 wins, not the 3 which you are predicting. Ohio has a total of 305 which would be a most likely result of 3 wins,not 4. Miami has 274 which would be a most likely result of 3 wins, not 2. All 3 of those would be different estimated wins totals.

That is actually the way Vegas figures estimated win totals for a season, and is more accurate than just saying Ball State has two games with a 57% chance of winning so that is 2 likely wins when mathematically it is closer to 1 likely win than 2.

But then lots of teams are playing four games without either team being granted the win.

If you were a 61% winner in all 4 games, that's 244%... 2 wins. While 39% for the opposing teams Is 156%, and that's one win. Who wins the last game?

Realistically if you play 4 games, 400% is impossible, therefore no one can win all 4.

57% x 2 games is 32.5% to win em both. While for the opposing team it's 18.5% to do the same thing. So it's still likely 2 wins for the favorite.

In your example, 1.56 rounds to 2, not 1. Vegas uses half numbers, so they would call those two teams 2 and a half and 1 and a half, and you could bet over or under.

In your 400% example, since we are rounding to the nearest whole number, you would have to have at least 350% to round up to 4. Again Vegas uses halves, so you might get a 3 and a half. Look at Ohio's numbers, 3 is a far more likely total, those percentages only add up to 305.

Your last statement 57% x 2 = 32.5% chance of winning both games is correct. That means that it is UNLIKELY that they would win both games. 67.5% of the time they would not win both games. That is why 2 is the wrong number for expected wins.

When you multiplied .43 time .43 to get your 18.5 %, please understand that that is the likelihood of winning both games. If you want to know how many games they would be likely to win, you just add together the probablility of each game, 43% plus 43% = 86%, which would round up to 1

(10-30-2019 06:33 PM)DICK Wrote:  Your last statement 57% x 2 = 32.5% chance of winning both games is correct. That means that it is UNLIKELY that they would win both games. 67.5% of the time they would not win both games. That is why 2 is the wrong number for expected wins.

When you multiplied .43 time .43 to get your 18.5 %, please understand that that is the likelihood of winning both games. If you want to know how many games they would be likely to win, you just add together the probablility of each game, 43% plus 43% = 86%, which would round up to 1

But just look at the single game numbers to tell me if a team is favored. Ohio over Miami 2 games in a row at 63%, means in the long run 63/100 times Ohio wins.

The single game projection is still 63% the second game (assuming they replay at the same place). Not anything higher, Because they already won the first game.

Basically Ohio wins 30-23 when they are favored 63% by Massey.

There is still a 63% they win the next game. Not now 62% because in a 100 game sample they already got a win. It's 63% every time.

Football doesn't really work on this principal because it's 1 game sample sizes anyways.

But if a team has two games at a certain percentage in a row, the second game is not the added inverse of game one and game two of 86% in your example.

(10-31-2019 12:02 AM)UofToledoFans Wrote:  

(10-30-2019 06:33 PM)DICK Wrote:  Your last statement 57% x 2 = 32.5% chance of winning both games is correct. That means that it is UNLIKELY that they would win both games. 67.5% of the time they would not win both games. That is why 2 is the wrong number for expected wins.

When you multiplied .43 time .43 to get your 18.5 %, please understand that that is the likelihood of winning both games. If you want to know how many games they would be likely to win, you just add together the probablility of each game, 43% plus 43% = 86%, which would round up to 1

But just look at the single game numbers to tell me if a team is favored. Ohio over Miami 2 games in a row at 63%, means in the long run 63/100 times Ohio wins.

The single game projection is still 63% the second game (assuming they replay at the same place). Not anything higher, Because they already won the first game.

Basically Ohio wins 30-23 when they are favored 63% by Massey.

There is still a 63% they win the next game. Not now 62% because in a 100 game sample they already got a win. It's 63% every time.

Football doesn't really work on this principal because it's 1 game sample sizes anyways.

But if a team has two games at a certain percentage in a row, the second game is not the added inverse of game one and game two of 86% in your example.

You can't just assume they're going to win the first game because it's at 63% though. IF they do, then of course the win projection changes.

You're simply wrong on this, mathematically speaking. It's why it's so rare to go undefeated, even when a team is the favorite every game.

Everyone but NIU in the West sits at .500 or better. My guess is a team or 2 in the West are going to really struggle down the stretch.

Will it be CMU or BSU, who fall back down to last year?
Will it be WMU, who has a tough home game and tougher road games?
Will it be UT or EMU, who need to show more consistency?

(10-31-2019 06:49 PM)Bronco14 Wrote:  Everyone but NIU in the West sits at .500 or better. My guess is a team or 2 in the West are going to really struggle down the stretch.

Will it be CMU or BSU, who fall back down to last year?
Will it be WMU, who has a tough home game and tougher road games?
Will it be UT or EMU, who need to show more consistency?

It will be interesting. For CMU, 2 of our last 3 are at home, where we haven't lost yet, so I'm hoping that means at least two more wins. The BSU game for sure will be a toughie, especially the way we've played on the road this season.