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

 Messages: [ Previous | Next ]
 root Posts: 57 Registered: 8/26/09
Re: Horse racing
Posted: Apr 1, 2013 4:24 PM

root <NoEMail@home.org> wrote:
> 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.
>
> I would appreciate any comments.
> Thanks.

Quick follow up: I can see W/Y does not represent
the chance that some Y rated horse will win a
given race. So I am back to thinking is the
determination of W/Y of any betting value?

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