Quote:
Originally Posted by formula_2002
from my web page
based on a 15% win pool take out
THE PROBABILITY OF A HORSE WINNING IS (1/(ODDS+1))/ THE RACES TOTAL BOOK PERCENTAGE .
THE TOTAL BOOK PERCENTAGE FOR A TRACK WITH A 15% TAKE OUT SUMS TO ABOUT 1.18
THE PROBABILITY OF ANY HORSE WINNING IS (1/(ODDS+1))/1.18
TAG THE WIN HORSES WIN PROBABILITY WITH "A"
TAG THE PLACE HORSE WIN PROBABILITY WITH "B"
EXACTA PROBABILITY PROBABILITY FOR A/B IS:
A X (B/(1-A))
EXAMPLE: A HORSE ODDS =2-1, B HORSE ODDS = 3-1, TRACK TAKE OUT IS 15%
A HORSE WIN PROBABILITY IS (1/(2+1))/1.18 = .282
B HORSE WIN PROBABILITY IS (1/(3+1)/1.18 = .212
PROBABILITY FOR THE A/B COMBINATION IS,
.282 x(.212/(1-.282)) = .083
THAT EQUATES TO A $1 RETURN OF (1/.083)= 12.04
THE FAIR $1 PAYOUT IS $12.04
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why would you do a complicated thing like that when the win odds as displayed by the tote already have considered the takeout?
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