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Topic: Horse racing
Replies: 7   Last Post: Apr 6, 2013 9:02 AM

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 root Posts: 56 Registered: 8/26/09
Horse racing
Posted: Apr 1, 2013 4:18 PM

A friend of mine enjoys horse racing and is a regular small
time bettor. He has roughly broken even over several decades
of betting. I do not bet, but I am interested in the processes
he uses to determine his bets.

Along those lines I came up with an idea to possibly increase
his handicapping success. I asked him, for every race, to
divide the horses into three groups, group Y included horses
which had a chance of winning the race. Group M were those
horses that could, maybe, win. And group N were horses he
thought had no chance of winning. Of course, some races were
won by horses in groups M and N.

After some time we computed the ratio of horses in group Y
which went on the win the race. That ratio is W/Y
where Y is the total number of horses ranked Y, and W is
the number of those that did win. Some, even most, races
have more than one Y candidates. It turns out the ratio
of W/Y was 3/8 for my friend.

What I was hoping to achieve is a measure of what odds
a Y rated horse should pay before that horse would
be bet. It isn't enough to reckon that a horse is the
best choice to win, the odds that horse pays must be
high enough to overcome the uncertainty.

I have stumbled on to a conceptual error in my thinking.
In some ways, W/Y does represent a probability but I
am unable to rationalize that interpretation given
that some horse must win every race, and that the
a-priori probabilities for the various horses must
add up to 1. Given that most races have more than one
Y candidate, their chances of winning are not independently
related to W/Y. Clearly W/Y does not represent the
probability that a Y candidate will win.

Everything becomes simpler if I interpret W/Y as the
chance (probability) that in a given race with at
least one Y rated horse, that a(any) Y rated horse
will win. Given that interpretation I can say that
if there are m Y rated horses in a given race
then the probility that one of them will win
is, on average, W/Y. Given that, and assuming that
all Y rated horses are equivalent, then each Y rated
horse has a probability of W/(m*Y) of winning.
To be a good bet, then, the Y rated horse must
pay better than m*Y/W.

Thanks.

Date Subject Author
4/1/13 root
4/1/13 root
4/1/13 Richard Ulrich
4/4/13 Malcolm McLean
4/4/13 Richard Ulrich
4/4/13 root
4/4/13 Richard Ulrich
4/6/13 Malcolm McLean