formula_2002
03-14-2003, 09:18 PM
“Exploiting the errors made by the handicapping public to an extent which provides us with the level of profit and consistency we require."
A quote from behind one of PA's posted banners.
Sounds cool. Now lets look behind the ice house.
We’ll look at the most prevalent odd group, >= 5/2 and < 7/2.
9850 horses 21.26 % winners and a loss of 17%
Average odds played = 2.95-1
Average winning odds =2.936
Actual winners were 2095
Expected winners (in order to break even) were 2506 horses.
(Expected winners = sum of 1/ (odds+1))
So just to outperform the public and break even, you would have to pick 411 more winners then the public, for a 25.44 win% (2095+411)/9850.
Verifying the above:
Edge=odds X win% - (1-win %)
Edge = 2.936 x .2544 – (1-.2544)
Edge =0
Our break even win % 25.44 / public win % 21.26 = 119%
So ALL WE HAVE TO DO is win 19 % more times then the public just to break even.
How realistic is that?
A quote from behind one of PA's posted banners.
Sounds cool. Now lets look behind the ice house.
We’ll look at the most prevalent odd group, >= 5/2 and < 7/2.
9850 horses 21.26 % winners and a loss of 17%
Average odds played = 2.95-1
Average winning odds =2.936
Actual winners were 2095
Expected winners (in order to break even) were 2506 horses.
(Expected winners = sum of 1/ (odds+1))
So just to outperform the public and break even, you would have to pick 411 more winners then the public, for a 25.44 win% (2095+411)/9850.
Verifying the above:
Edge=odds X win% - (1-win %)
Edge = 2.936 x .2544 – (1-.2544)
Edge =0
Our break even win % 25.44 / public win % 21.26 = 119%
So ALL WE HAVE TO DO is win 19 % more times then the public just to break even.
How realistic is that?