Well i guess my post about trifecta payouts was removed as punishment for expressing myself abount an arrogant member of this forum...but for anyone interested i did...copy the information that was given me

3[(3X5)-1 multiplied by (10 + 1)].
The result of the above is then divided by 3 + 1

This equals to 3 X 14 X 11, then divided by 4.

This equals to 115. According to D. Scott, this is the dividend. The odds are therefore 114/1.

A further example is the one that Don Scott provides:

Winner = 3/1
Second = 4/1
Third = 9/1

[3(3 X 4) -1], then multiplied by (9+1). The result of this is now divided by the product of 3 + 1

The calculation is 3 X 11 X 10, The result is then divided by 4. The answer is 82.5, which is the dividend, the odds being 81.5/1


[ This Message was edited by: quapi on 2004-08-02 21:32 ]

Hi, Shaun.

I thought that maybe the posting was removed because the formula had been copyrighted, even though one would wonder how that could be so.

well i am not sure what was in your post that was a problem but as you can see i have copied it as it was from where i saved it....and if it was the problem i am sure they would have removed this one aswell.....unless someone posted something after what i had writen....in that case i would think it better to remove that paost not the whole thread....but hey i only post here not own the place...lol

[ This Message was edited by: Shaun on 2004-07-29 18:52 ]

[ This Message was edited by: Shaun on 2004-07-29 18:53 ]

Shaun,
Have you compared your formula with actual results?
I normally just multiply the win prices of the three placegetters to get an indication of the likely trifecta dividend - obviously it over-states the result if you have very short-priced favourites winning but generally its in the ball-park.
Your example of horses at $4, $5 and $10 paying only $82 appears to be a very unfavourable result.
Or is the DS formula what he considered the true odds to be?

I did about 50 tests on past results...have not tested it on live bets as i have to write up a compleat excel sheet for that.....in all the test i did the trifecta paid more than the formulas prediction....with out doing major tests i would say this is much more accurate....i just had a look at one of those races that had a short priced winner and a couple of long priced placers
today Weribee race 1
10-1.80
4-1.90 (win 9.70)
12-5.80 (win 32.00)

using yor method this would be 1.80x9.70x32.00=$558.72

using the method outlined above and the way i wrote the formula up the results are

1.80-1=0.80
9.70-1=8.70
32.00-1=31.00

(0.80x8.70)-1=5.96
31.00+1
(0.8x5.96x32.00)/(0.8+1)=$84.76
trifecta paid $153.60

a couple of more tests
Weribee race 2 trifecta paid $253.40 formula estimated $206.46

Tamworth race 1 trifecta paid $726.70
formula estimated $490.59

Canberra race 1 trifecta paid $24.70
formula estimated $15.89


[ This Message was edited by: Shaun on 2004-07-30 12:59 ]

Shaun,
There's no doubt your formula is more logic-based than my quick-and-dirty approach. To get a more accurate answer probably involves looking at different calc's for different field-sizes.
One thing all our number-crunching shows is that trifectas in small fields are poor value.

Well not exactly my formula it was provided by michaelg who got it from one of Don Scotts books......using this type of formula you can see the poor value bets and not place them...or you could use a dutching system to return a set amount no matter how they come inif you would normally have 4 horses boxed this way you can decide on what you want to get back

Shaun, I have missed what was written, but all I know is I tried to answer your question.
The trifecta formula I had in mind, is not just a formula. It has a whole chapter explaining why it is so,(the formula) and the rest of the book is a series of intense research based on this formula.
I hope you have the name of the books I posted,as far as I know, very hard to get,good luck.



[ This Message was edited by: puntz on 2004-11-03 10:06 ]

Yeah i have the sites....yes i did have a copy of Don Scotts winning in the 90s...but left it with my ex not that she would know what the hell to do with it.....love to get another copy it had some goos stuff in it

Anyway,
the exmple I have it assumes if the odds
of the 1st,2nd and 3rd place getters are
1:2/1
2:3/1
3:4/1
it should equate to 15.67/1 (16/1 rounded)

There is a qbasic program that does this,do you have that program?