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cosmicway
07-30-2016, 11:52 AM
The placepot is an accumulator in which you have to predict the horses that are placed in consecutive races.
More popular is the winpot in which you have to predict the winners, but placepot is easier.

The problem is this:
I have a set of probabilities for each leg (race) of the placepot.
Make it two legs for simplicity and the horses in the first leg have place probabilities {P1, P2, P3 ....}, those in the second leg have {Q1, Q2, Q3 ...}.
The rule is first and second horses place, so P1 + P2 + P3 + ... = 2, Q1 + Q2 + Q3 + ... = 2.
My playlines are N different playlines made up of two selections, one horse from each leg and the line probabilities are say P(i) x Q(j) for the first one, P(m) x Q(n) for the second one and so on, where i, j and m, n are the numbers of the runners in the first and second legs respectively for each of the lines.

The problem is how to compute the probability of at least one catch among the N lines ?
If it was winpot then it's simple of course. N different playlines can have only one winner line among them, so the sum of the P x Q products gives me the win probability.
But in placepot ?

GameTheory
07-30-2016, 12:39 PM
First off, why would P1+P2... = 2? Seems like you'd have probabilities for each horse to come in first OR second (one probability for that), and that the whole field for each leg would sum to 1 as usual. Are you taking the win probability and just doubling it or something? (That wouldn't make sense.)

cosmicway
07-30-2016, 01:03 PM
The sum is 2 if two place, 3 if three place, 4 if four place (as happens in big fields of over 20).
To derive place probabilities is not as simple as doubling the win probability. It is "first or second" but not first x 2. But that has to do with the way a handicapping algorithm is constructed.
Here we have a logistics problem.

DeltaLover
07-30-2016, 01:14 PM
Before we continue the interesting conversation, we need to clarify that I can detect a pitfall in the way you are stating the problem.

Your { P1, P2, .., Pn} is not a probability distribution and this is of course why the tuple does not add up to 1 but to 2. The logical reason of course is that Pi is not independent from Pj.

I will continue later with a more detailed view

cosmicway
07-30-2016, 01:54 PM
Before we continue the interesting conversation, we need to clarify that I can detect a pitfall in the way you are stating the problem.

Your { P1, P2, .., Pn} is not a probability distribution and this is of course why the tuple does not add up to 1 but to 2. The logical reason of course is that Pi is not independent from Pj.

I will continue later with a more detailed view

Yeah but ...
Suppose I know the win probabilities {W1,W2,W3 ...}.
Those are pdf definitely and W1 + W2 + W3 ... = 1.
Now you say suppose the 2 won, what is the probability of 1 winning among the rest, come first that is in the depleted by one field.
You could say S2 = W1 / (W1 + W3 ...) + W1 / (W1 + W4 + ...) + ...
So P1 = W1 + S1.
This is valid is n't it ?
I don't like it because it's not the best way of doing it, but essentially that's how to get to the P values (the better formula is a bit more complicated but follows the same logic).

Or suppose that only speed is taken into account and in this case there is a closed integral solution to give you the probabilities:

http://www.untruth.org/~josh/math/normal-min.pdf

With this integral you compute the probabilities of 1sts, 2nds, 3ds ...
If you are not happy again with the results and it does n't make money it's yet another story, but it is a closed solution.

Nonetheless you will have the same thing with place probabilities, they add up to 2.

BCOURTNEY
08-10-2016, 11:11 PM
Depends on your model choice to model place probability.

Harville (1973). Stern (1990), Henery (1981), Bacon-Shone et al (1992) Lo and Bacon-Shone (1994)

Normal probability models best fit the data, yet suffer from favorite and longshot bias. i.e. the probability of finishing first or second or third is overestimate for horses which have a high probability and underestimated for horses having a low probability of winning

There has historically been some limited success with obscure gamma ranking models as well

Fast way would be to compute Harville probabilities then adjust the Harville values based on a regression with track data.

cosmicway
08-10-2016, 11:41 PM
Depends on your model choice to model place probability.

Harville (1973). Stern (1990), Henery (1981), Bacon-Shone et al (1992) Lo and Bacon-Shone (1994)

Normal probability models best fit the data, yet suffer from favorite and longshot bias. i.e. the probability of finishing first or second or third is overestimate for horses which have a high probability and underestimated for horses having a low probability of winning

There has historically been some limited success with obscure gamma ranking models as well

Fast way would be to compute Harville probabilities then adjust the Harville values based on a regression with track data.


You have links to those ?
Interesting, I agree with your observations.

But it's besides the point.
Suppose we have the crudest of probability estimates.
Like for example convert the morning line place prices shown on the tv screen to place place probabilities. Or if the place prices are not displayed use the win prices and crude reduction method to compute 2nd place probabilities and place probabilities.
Those will be "the probabilities". If somebody-somewhere has any better then good for him.
Again the problem is how to compute the probability of one catch in the set of lines.

Do you have placepots or only winpots in the States ?