Discounted Harville

[mwilding1981]

Is anybody able to tell me these formulas or point me somewhere I can get them?

Thanks

[banacek]

Basically, you have probabilities for each horse winning. Example: If Horse A is 30% and horse B is 20%, then the probability of an AB exacta is 0.30*(0.20/(1-.30))=.08571. The probability of BA would be 0.20*(0.30/(1-0.20))=0.075.

SInce there are horses who have a good prob. of winning, but little of coming second (like need to lead types) and horses with a little chance of winning, but a decent chance of coming second, I don't think many people use them. Theory vs. reality?

[mwilding1981]

Thanks for the reply. Isn't that the normal Harville formula though, I understood that the discounted Harville was a modification of this to help counter the problem that as you say a horse may have no chance of winningbut high of placing or high chance of winning and none for placing.

[saevena]

I believe Banacek is right. I recall reading the article written by Mr. Harville at the Georgetown U. Library many years ago. Harville described his use of his formulas for place and show wagering at an Ohio track and found his results to be "mediocre." The formulas are theoretical in nature, not based upon empirical data, and don't show a profit in practice, I've found.

[banacek]

Originally Posted by mwilding1981

Thanks for the reply. Isn't that the normal Harville formula though, I understood that the discounted Harville was a modification of this to help counter the problem that as you say a horse may have no chance of winningbut high of placing or high chance of winning and none for placing.

I hadn't heard of a variation, but you certainly could adapt them to fit a particular race. Many races might be such that the Harville formulae would work fine, but other races would need an adaption to the numbers that would be very specific to the horses in that particular race.

[mwilding1981]

Thanks for the replies, I have also been trying to get my head around the Henery formula, does anyone know this? Also the Shin probability formula for adjusting for fav-longshto bias.

Originally Posted by mwilding1981

Thanks for the replies, I have also been trying to get my head around the Henery formula, does anyone know this? Also the Shin probability formula for adjusting for fav-longshto bias.

"the "harville formula" is the straight mathematical probability of a horse running
a place given its win probability.

however in practise it does not apply as the shorter priced horses do not place as
often as the win probabilities would indicate.

"henery, stern" may have been the 1st to come up with an alternate formula which
approximately indicated the favourite would run 2nd only 80% as often as the win
prob would indicate and 3rd about 65%.

lo, bacon-shone & kelly" later published a more elegant formula.

using a number alpha = .89 for 2nd and .80 for 3rd. ( for public probabilities
would use .80 & .65 ).

for each horse prob for 2nd or 3rd recalculated as ---- prob ** alpha / ( sum of
prob ** alpha ) where ** means raised to the power.

[mwilding1981]

robert99 thats great thank you. So am I right in understanding that if you are using your own probabilities then use the .89 and .80 and if using probabilities derived from the win odds then use .80 and .65?

I am a little confused I admit. Where have we got the 'sum of probbilities' from are you talking about the place probs now or the win probs? and if we are talking about place which I think you are the have you used the harville to get the probs and are then adjusting for the bias?

I have the original henery formula which is very complex as I am sure you know but hadn't beena ble to get the revised one, so thank you very much.

[jfdinneen]

Robert is correct that Harville's theoretical formula has been superseded by later research into racing in Hong Kong, Japan, and USA done by Lo and Bacon-Shone (1993) - cited in Hausch, Lo, and Ziemba (1994), pp. 477-478 [EoRBM]. Their 'Discounted Harville' approach is relatively straightforward and can be adapted for use in Excel, for example.

Best wishes,

John

[arkansasman]

Mwilding,

If you have a database with public probabilities, you could have different coefficients than .80 and .65 for place and show as indicated by Lo, Bacon-Shone and Busche in 1994.

My coefficients for place and show in my model are different than the .80 and .65.


John

Originally Posted by mwilding1981

robert99 thats great thank you. So am I right in understanding that if you are using your own probabilities then use the .89 and .80 and if using probabilities derived from the win odds then use .80 and .65?

I am a little confused I admit. Where have we got the 'sum of probbilities' from are you talking about the place probs now or the win probs? and if we are talking about place which I think you are the have you used the harville to get the probs and are then adjusting for the bias?

I have the original henery formula which is very complex as I am sure you know but hadn't beena ble to get the revised one, so thank you very much.

This is all based on individual win probabilities as the starting point.
As later posters have indicated, the empirical correction figures are averaged over a number of races for a particular set of racing. So the corrections, for what they are worth, will vary. This subject theory is very academic, rather than practical or true to life - as is typical of economists - it is better to assess each race as it comes and just be guided by the average corrections - which you may judge too high or low for a particular race. Things happen due to a cause - you need to forecast that cause - not apply a formula that only looks at effects. Placing probability depends far more on the horses involved in a race than on the maths.

The sum of probabilities is from the individual horse win odds which when summed will not be 100%. Correcting this way is also a sweeping assumption and is not based on any evidence - just unproven economist theory.

[mwilding1981]

thanks everybody this has been incredbily helpful. jfidneen or anybody else, i was wondering if you would be able to put up 'the discounted harville' by hausch, lo and ziemba?

[jfdinneen]

Michael,

The "Discounted Harville" formulae, as requested (Excel-like syntax) - check accuracy:

In a seven horse (H1, H2, H3,...H7) race:

1. probability (H1 wins) = pi(H1)

2. probability (H1 wins, H2 places) - exacta =

PRODUCT(pi(H1),PRODUCT(POWER(pi(H2),lambda),1/SUM(POWER(pi(H2),lambda),SUM(POWER(pi(H3),lambda), POWER(pi(H4),lambda),POWER(pi(H5),lambda),POWER(pi (H6),

lambda),POWER(pi(H7),lambda))))

3. probability (H1 wins, H2 places, H3 shows) - trifecta =

PRODUCT(pi(H1),PRODUCT(POWER(pi(H2),lambda),1/SUM(POWER(pi(H2),lambda),SUM(POWER(pi(H3),lambda), POWER(pi(H4),lambda),POWER(pi(H5),lambda),POWER(pi (H6),

lambda),POWER(pi(H7),lambda))),PRODUCT(POWER(pi(H3 ),rho),1/SUM(POWER(pi(H3),rho),SUM(POWER(pi(H4),rho),POWER( pi(H5),rho),POWER(pi(H6),rho),POWER(pi(H7), rho))))

To approximate the original Harville model, let lambda = 1.0 and rho = 1.0. To approximate the Henery model, let lambda = 0.76 and rho = 0.62. The authors provide a table of intermediate values for both lambda and rho but, personally, I have found that the Henery values are satisfactory. This is not an exact science and, as such, an over-emphasis on precision is not warranted. A consistent, satisficing approach is sufficient.

Note that tractability issues arise because you have to sum over all exactas where H2 places in order to calculate the H2 place probability. Similarly, you have to sum over all trifectas where H3 shows to calculate the H3 show probability!

Best wishes,

John

[mwilding1981]

Okay this is how I understand the formulas below, can anybody correct me if I am wrong.

1) Probability H1 wins and H2 comes second

piH1 x (piH2 to the power of lambda) x 1/((pih2 to the power of lambda)x(piH3 to the power of lambda) etc.....)

2) Probability H1 wins and H2 comes second and H3 comes third

(piH1 x (piH2 to the power of lambda) x 1/((pih2 to the power of lambda)x(piH3 to the power of lambda)etc...)) x ((piH3 to the power of rho) x 1/((piH3 to the power of rho)x(piH4 to the power of rho)x(piH5 to the power of rho)etc....))

[mwilding1981]

or for example:
Probabilities H1 = 0.4 H2 = 0.2 H3 = 0.15 H4 = 0.1 H5 = 0.08 H6 = 0.06 H7 = 0.06

1) H1 and H2 come in first and second:

(0.5*(0.2 to the power of 0.76)) * 1/((0.2 to the power of 0.76) * (0.15 to the power of 0.76) * (0.1 to the power of 0.76) * (0.08 to the power of 0.76) * (0.06 to the power of 0.76) * (0.06 to the power of 0.76))

2)H1 and H2 come in first and second and third:

((0.5*(0.2 to the power of 0.76)) * 1/((0.2 to the power of 0.76) * (0.15 to the power of 0.76) * (0.1 to the power of 0.76) * (0.08 to the power of 0.76) * (0.06 to the power of 0.76) * (0.06 to the power of 0.76))) * ((0.15 to the power of 0.62) * 1/((0.15 to the power of 0.62) * (0.1 to the power of 0.62) * (0.08 to the power of 0.62) * (0.06 to the power of 0.62) * (0.06 to the power of 0.62)))