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]

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.

[robert99]

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

[robert99]

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)))

[mwilding1981]

that's not quite right is it? I am assuming that the 1 is for the 100% book?

[jfdinneen]

Michael,

Apologies for not replying sooner but I have not logged on to the forum for a couple of days.

Using your example [H1 = 0.4 H2 = 0.2 H3 = 0.1 H4 = 0.1 H5 = 0.08 H6 = 0.06 H7 = 0.06] - except for H3 which I set to 0.1 so that the total probability of all runners in the race is 100% (Do not confuse probability of winning with pari-mutuel prices), I calculate both the example exacta and trifecta below.

In Excel, assuming that cell B7 = lambda (0.76) and cell B9 = rho (0.62) and that H1 to H7 are in cells E7 to E13 respectively, then
prob(H1 wins, H2 places) =PRODUCT(E7,PRODUCT(POWER(E8,$B$7),1/SUM(POWER(E8,$B$7),POWER(E9,$B$7),POWER(E10,$B$7), POWER(E11,$B$7),POWER(E12,$B$7),POWER(E13,$B$7)))) = 0.114929

and prob(H1 wins, H2 places, H3 shows) =PRODUCT(E7,PRODUCT(POWER(E8,$B$7),1/SUM(POWER(E8,$B$7),POWER(E9,$B$7),POWER(E10,$B$7), POWER(E11,$B$7),POWER(E12,$B$7),POWER(E13,$B$7))), PRODUCT(POWER(E9,$B$9),1/SUM(POWER(E9,$B$9),POWER(E10,$B$9),POWER(E11,$B$9) ,POWER(E12,$B$9),POWER(E13,$B$9)))) = 0.026556

NOTE: I inadvertently added an additional SUM term to both equations in my original repy - apologies - never post mathematical equations to forum late at night!

Best wishes,

John

[mwilding1981]

Hi John,

Thank you very much. Just had time to go through the top (place equation). I am getting a different result to you though when i do it manually and I can't work out why. This bit works fine:

1/SUM(POWER(E8,$B$7),POWER(E9,$B$7),POWER(E10,$B$7), POWER(E11,$B$7),POWER(E12,$B$7),POWER(E13,$B$7)

But when I add this on:

PRODUCT(POWER(E8,$B$7),1/SUM(POWER(E8,$B$7),POWER(E9,$B$7),POWER(E10,$B$7), POWER(E11,$B$7),POWER(E12,$B$7),POWER(E13,$B$7))

your sum gets 0.2873 and manually i am getting 0.3014 and I am not sure why. What I am doing is 0.2 to the power of 0.76 and then multiplying this total with the total from the first equation above? Any thoughts?

thanks

Michael

[jfdinneen]

Michael,

Did you remember to change the prob(H3, wins) value from 0.15 to 0.10, as per my instructions in the last post.

Best wishes,

John

[mwilding1981]

Hi John,

Yes I definately changed that because I used your equation and then I also did each stage in excel seperately. For some reason that I still don't understand, when I do it seperately I get the figure differently to you and it changes at the point above. I can't see any reason for this though!! Very odd

[jfdinneen]

Michael,

As you say, it does not make any sense.

I suggest you leave it for 24 hours and perhaps a fresh look at that stage might lead to a resolution.

Best wishes,

John

[Midas Pansy]

Sorry for bumping a very old thread to the top as my first post on the forum. I've been so impressed with the level of knowledge here, I've no idea how I hadn't found this forum before.

I'm not sure if jfdinneen or mwilding are still about, or if anyone else can help, but I decided to try in excel the formulas jfd contributed so generously below.

The exacta formula works a treat, but somehow I cannot get a sensible result out of the trifecta formula. Should the results of this add up to 1 ? If I sum all the suggested probabilities from the exacta formula I get 100%, but nowhere near for the trifecta. I suspect I have made a silly error somewhere, but any help would be greatly appreciated.

Looking forward to reading and contributing more here, some of the old threads hold a lot of value.

[jfdinneen]

Midas,

I have uploaded a Discounted Harville spreadsheet (Discounted Harville II (http://vendire-ludorum.blogspot.com/) ) with VBA Functions (DiscountedHarvilleExacta, DiscountedHarvilleTrifecta), which should remove the limitation on number of runners and make for easier usage.

Please check for errors.

John

[formula_2002]

I thought it worth bringing this thread back to the board

[Dave Schwartz]

Midas,

I have uploaded a Discounted Harville spreadsheet (Discounted Harville II (http://vendire-ludorum.blogspot.com/) ) with VBA Functions (DiscountedHarvilleExacta, DiscountedHarvilleTrifecta), which should remove the limitation on number of runners and make for easier usage.

Please check for errors.

John

Although I did not find the spreadsheet that JF Dinneen mentioned, I am much more interested in that blog site itself.

There is some pretty interesting stuff there.

[formula_2002]

I just clicked one on of the links "how not to be wrong......."
some interesting stuff.

[jfdinneen]

formula_2002 / Dave,

Thank you both for the kind words on the blog.
It appears that Google deleted the original "Harville" spreadsheet from the blog some time ago, without requesting my permission? I have recreated the spreadsheet and VBA functions (Excel 2007+) from first principles and have attached it to the original blog entry Discounted Harville v1.17 (https://vendire-ludorum.blogspot.co.uk/2011/03/discounted-harville-vba-functions.html).

Best wishes,

John

[formula_2002]

John

Thanks for the update.


Formula

[jfdinneen]

Formula_2002,

As ever, you are very welcome.

Best wishes,

John

[formula_2002]

I have included the "informed Factor" size into my bet size calculations.
I have included it in the last 35 races and it improves the results
let's see how long that last. :) the results are included in the power factor label.
see attached