Exotic Calculator

[The Judge]

I like to post this from time to time with new members and all .

After missing a $21,000 superfecta some years ago because I couldn't calculate the bet 5 horse box. I came across the "Exotic Calculator" it can be downloaded.http://www.kimstarr.org . Look under "utilities". Remember its .org not .com.

Its a neat little program.

[raybo]

I like to post this from time to time with new members and all .

After missing a $21,000 superfecta some years ago because I couldn't calculate the bet 5 horse box. I came across the "Exotic Calculator" it can be downloaded.http://www.kimstarr.org . Look under "utilities". Remember its .org not .com.

Its a neat little program.


You can use the wagering pad at Brisbet.com for free. Just enter your selections and it tells you the cost of the ticket. Other online wagering sites probably have something similar.

[Kelso]

Hypothetical (extreme) situation:
- 10 horse race
- 4 contenders ("C")
- 6 serious others ("O")


Betting assumption:
- at least two of the "contenders" will finish in-the-money
- each of the "others" has a reasonable chance of finishing ITM


:confused: Question:
Would someone kindly tell me the formula ... in terms of C and O ... to calculate the cost of a $1 trifecta ticket covering all possible manifestations of the betting assumption?


Thanks very much, and sorry if this issue is a lot more elementary than it seems to me. :rolleyes:

[Zaf]

Kelso,

Try Steve Crist's book "Exotic Betting" , I think it will answer all of your questions !

Z

[PriceAnProbability]

I like to post this from time to time with new members and all .

After missing a $21,000 superfecta some years ago because I couldn't calculate the bet 5 horse box. I came across the "Exotic Calculator" it can be downloaded.http://www.kimstarr.org (http://www.kimstarr.org/) . Look under "utilities". Remember its .org not .com.

Its a neat little program.

This stuff should be memorized.

Five horse super box would be 5 * 4 * 3 * 2 = 120

Five horse tri would be 5 * 4 * 3 = 60

[Pace Cap'n]

Hypothetical (extreme) situation:
- 10 horse race
- 4 contenders ("C")
- 6 serious others ("O")


Betting assumption:
- at least two of the "contenders" will finish in-the-money
- each of the "others" has a reasonable chance of finishing ITM


:confused: Question:
Would someone kindly tell me the formula ... in terms of C and O ... to calculate the cost of a $1 trifecta ticket covering all possible manifestations of the betting assumption?


Thanks very much, and sorry if this issue is a lot more elementary than it seems to me. :rolleyes:

Assume C's are 1, 2, 3, & 4

1-2-3-4 / all / all...........$288

[Anderon]

I hope this helps you, it should be alot easier than an exotic calculator....I have this built into my software program but I had to copy it into word to upload it as I can't upload an Html page on here. :)

[raybo]

Hypothetical (extreme) situation:
- 10 horse race
- 4 contenders ("C")
- 6 serious others ("O")


Betting assumption:
- at least two of the "contenders" will finish in-the-money
- each of the "others" has a reasonable chance of finishing ITM


:confused: Question:
Would someone kindly tell me the formula ... in terms of C and O ... to calculate the cost of a $1 trifecta ticket covering all possible manifestations of the betting assumption?


Thanks very much, and sorry if this issue is a lot more elementary than it seems to me. :rolleyes:

If you are assuming one of the 4 contenders will win then you put those 4 on the win line, if one of those same 4 will finish 2nd then you only have to have those 4 horses on the place line, if any of the other 6 serious others might finish 2nd then you would have to have all 10 horses on the place line. However if you only want the 4 contenders as considerations for the place then your ticket would be 4 for win, the same 4 for place, all 10 for show. So your ticket would look like this:

1 2 3 4

1 2 3 4

1 2 3 4 5 6 7 8 9 10

Now, draw a line from the win #1 thru the show # 3 (shown as bold text above). (This only works when you have a progressive box/wheel where there are the same common horses on win, place,and show lines). Now count the horses on each row starting with the horse that the line is drawn thru and counting to the right of that horse. The count for each row is then multiplied with the count of the other lines. In this case you would have 8 on the show line (3 4 5 6 7 8 9 10), 3 on the place line(2 3 4), and 4 on the win line(1 2 3 4). Result is 8 x 3 x 4 = 96 combinations @ $1 each or $96. If you think any of the serious others might finish 2nd then you would have:

1 2 3 4

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10



8 x 9 x 4 = 288 combinations @ $1 each or $288

If any of the serious others might win also, then you would have to box all 10 horses:

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

For: 8 x 9 x 10 = 720 combinations @ $1 or $720

[raybo]

Hypothetical (extreme) situation:
- 10 horse race
- 4 contenders ("C")
- 6 serious others ("O")


Betting assumption:
- at least two of the "contenders" will finish in-the-money
- each of the "others" has a reasonable chance of finishing ITM


:confused: Question:
Would someone kindly tell me the formula ... in terms of C and O ... to calculate the cost of a $1 trifecta ticket covering all possible manifestations of the betting assumption?


Thanks very much, and sorry if this issue is a lot more elementary than it seems to me. :rolleyes:

As far as a formula for figuring all the possibilities, even if you had one, it would still result in the scenario I described above. Same number of possible combinations.

[raybo]

You can use the wagering pad at Brisbet.com for free. Just enter your selections and it tells you the cost of the ticket. Other online wagering sites probably have something similar.

Best way to do this would be to analyze all horses and decide on which ones are good enough to win, put them all on the win line, then analyze which ones are good enough to finish 2nd and put them on the place line then all others that you can't disqualify completely can go on the show line. Write it down and draw a line like I described above to get the combinations or use an exotics calculator or the wagering pad from a wagering portal, if you wager online. Keep in mind that putting too many horses on your ticket might not be profitable even if you hit a few, especially if you cant take the favorite off the win line and place line. If you're going to play multiple winners you better have some good odds in order to make enough money to cover your losses. Personally I only wager if I have only 1 horse to win or maybe 2 if both their odds are good, never include the favorite on the win line unless you have some very long odds horses with a good chance of hitting in the money. I play supers rather than tri's because it's just as hard almost to hit a tri as it is a super with good coverage on your ticket and they pay much more. Ideally, for me anyway, the standard ticket would be: 1 with 2 with 4 with 4 which is $12 on a $1 super.

[Kelso]

1 2 3 4
1 2 3 4
1 2 3 4 5 6 7 8 9 10

Now, draw a line from the win #1 thru the show # 3 (shown as bold text above). Now count the horses on each row starting with the horse that the line is drawn thru and counting to the right of that horse. The count for each row is then multiplied with the count of the other lines. In this case you would have 8 on the show line (3 4 5 6 7 8 9 10), 3 on the place line(2 3 4), and 4 on the win line(1 2 3 4). Result is 8 x 3 x 4 = 96 combinations @ $1 each or $96. If you think any of the serious others might finish 2nd then you would have:


Thank you very much Raybo and all who responded to my question.

It involved betting 4 contenders, at least two of whom (but perhaps three of whom) will finish 1st, 2nd or 3rd in a trifecta. The remaining six horses can't be counted out from any ITM position, but not more than one of them will finish ITM.

Using your graphic, Raybo, it seems that the cost of covering all permutations of my scenario would require 3 tickets at a total cost of $228 (96+72+60).

1234
1234
123....10
=4x3x8=$96

1234
123....10
1234
=4x9x2=$72

123....10
1234
1234
=10x3x2=$60

Does this look correct to you? If it is correct ($228), then I'll just play with the numbers to come with the formula in terms of C (contenders) and O (others) ... perhaps with the need for an X (number of ITM finishing positions bet - in this case all 3).

Thank you, as well, for the strategy advice .... and thank you, again, to all.

[Kelso]

Try Steve Crist's book "Exotic Betting" , I think it will answer all of your questions !
Z


Thank you, Zaf. I'm going to get it. (After nosing around online, I think his "Horse Traders" might be interesting, too.)

[raybo]

Thank you very much Raybo and all who responded to my question.

It involved betting 4 contenders, at least two of whom (but perhaps three of whom) will finish 1st, 2nd or 3rd in a trifecta. The remaining six horses can't be counted out from any ITM position, but not more than one of them will finish ITM.

Using your graphic, Raybo, it seems that the cost of covering all permutations of my scenario would require 3 tickets at a total cost of $228 (96+72+60).

1234
1234
123....10
=4x3x8=$96

1234
123....10
1234
=4x9x2=$72

123....10
1234
1234
=10x3x2=$60

Does this look correct to you? If it is correct ($228), then I'll just play with the numbers to come with the formula in terms of C (contenders) and O (others) ... perhaps with the need for an X (number of ITM finishing positions bet - in this case all 3).

Thank you, as well, for the strategy advice .... and thank you, again, to all.

That would be correct.

[singunner]

I find that thinking about the possible permutations in betting on exotics hurts my head if I've had anything to drink in the three weeks prior to thinking about it. And of course, I haven't gone without drinking for more than a week in 7 or 8 years, depending on whose fuzzy memory you use.

What I want to know is if there is any mathematical basis for the various techniques suggested. Where's the studying being done on exotics? I never see the programming people talk about them, which makes me wonder if it's just a pain in the ass to code or if there's some practical benefit to leaving them out. Anyone have answers for me? Week 3 starts the 29th if I make it. Hopefully someone can get me a decent answer by then.

[hcapperjag]

Actually, wouldn't the cost for the problem as stated be $240?

let 1234 be the contenders
let ABCDEF be the 'others'

bets would be:

1234/1234/ABCDEF cost $72

1234/ABCDEF/1234 cost $72

ABCDEF/1234/1234 cost $72

1234/1234/1234 cost $24

:cool:

[hcapperjag]

The formula you were asking for would be:

C=no of contenders
T=serious oThers

C^3 + 3(T-1)C^2 + (2-3T)C

using ^ to mean raised to the power (3=cubed 2=squared)

:jump:

[raybo]

That would be correct.


From what you say, that the 4 contenders could finish in any position and any of the others could also finish in any position, you are saying that any horse in the race could finish in any position. If that is so you would have to put all 10 horses in every position, or: 10 w 10 w 10 or 10 x 9 x 8 which would be 720 possible combinations or $720.

[raybo]

Actually, wouldn't the cost for the problem as stated be $240?

let 1234 be the contenders
let ABCDEF be the 'others'

bets would be:

1234/1234/ABCDEF cost $72

1234/ABCDEF/1234 cost $72

ABCDEF/1234/1234 cost $72

1234/1234/1234 cost $24

:cool:

Not from the way I read his scenario. He says that he cannot rule out any of the non-contenders from finishing 1st, 2nd, or 3rd, which tells me he has to bet all 10 horses to win, all 10 horses to place, and all 10 horses to show, which would be 720 combinations at $1 each. $720. This is an all, all, all ticket, and is ridiculous unless you know for sure the tri will pay at least $721, otherwise, what's the point in betting at all?

[TimesTheyRAChangin]

Hypothetical (extreme) situation:
- 10 horse race
- 4 contenders ("C")
- 6 serious others ("O")


Betting assumption:
- at least two of the "contenders" will finish in-the-money
- each of the "others" has a reasonable chance of finishing ITM

Raybo,
You are still incorrect.You're forgetting to take into account the fact that at least 2 of the 'contenders' will come in on the tri ticket,while only 1 of the 'others' does.

[Kelso]

take into account the fact that at least 2 of the 'contenders' will come in on the tri ticket,while only 1 of the 'others' does.



That's correct. I think the answer is either Raybo's $224 or Jag's $240 .... but can't begin to understand how/why the techniques return different answers.

Jag and Ray, if I'm not imposing by asking, would you please describe the math logic behind your formulae?

Thank you to all, again, for your assistance.

[TimesTheyRAChangin]

That's correct. I think the answer is either Raybo's $224 or Jag's $240 ....

It is $240.00.

[46zilzal]

It is $240.00.
good handle. Bobby Zimmerman would be proud.

[hcapperjag]

kelso:

for the case with 2 C's and 1 T:

6 ways of choosing 2 C's from a pool of 4
x
6 ways of choosing 1 T from a pool of 6
x
6 ways of ordering 3 contenders for the tri
=
$216

for the case of 3 C's:

4 ways of leaving 1 C out
x
6 ways of ordering 3 contenders for the tri
=
$24

I don't remember the nomenclature for writing a formula in terms of combinations, permutations and factorials but that is what it boils down to.

;)

[Kelso]

I don't remember the nomenclature for writing a formula in terms of combinations, permutations and factorials but that is what it boils down to.




Plenty good enough for this mathematical Neanderthal. Thanks very much, Jag.

[raybo]

Raybo,
You are still incorrect.You're forgetting to take into account the fact that at least 2 of the 'contenders' will come in on the tri ticket,while only 1 of the 'others' does.


Yes, but which one? That's why you can't avoid putting them all on the ticket on all 4 lines.

[raybo]

Hypothetical (extreme) situation:
- 10 horse race
- 4 contenders ("C")
- 6 serious others ("O")


Betting assumption:
- at least two of the "contenders" will finish in-the-money
- each of the "others" has a reasonable chance of finishing ITM

:confused: Question:
Would someone kindly tell me the formula ... in terms of C and O ... to calculate the cost of a $1 trifecta ticket covering all possible manifestations of the betting assumption?


Thanks very much, and sorry if this issue is a lot more elementary than it seems to me. :rolleyes:

These 2 statements force you to put all horses on all betting lines. "at least 2 of the contenders will finish ITM" means that any of the 4 could win, place or show, so all must be in each position. "each of the "others" has a reasonable chance of finishing ITM" means that any of the "others" could win, place or show, so all of them must be put in each position. 720 possible combinations. The only way to reduce the number of possible combinations is to reduce the number of horses that might finish ITM. This scenario is too vague to reduce the combinations. For example, if you could reduce the number of horses capable of winning, then you reduce the number of combinations by 25% with each horse taken off the win line.

[singunner]

This scenario is too vague to reduce the combinations. For example, if you could reduce the number of horses capable of winning, then you reduce the number of combinations by 25% with each horse taken off the win line.
Hmm... I was with you just up until then end, where I got a little lost. I don't know where you came up with the 25% number. I'm sort of guessing that you were referring to his scenario when he had 4 horses on the Win line. I could be wrong there, but I don't see how (with your horses across the board scenario) that 25% number comes up. Could you please clarify for me?

[hcapperjag]

Raybo:

what Kelso meant in the second condition when he said that 'each of the others has a reasonable chance of finishing ITM' was that 'each of the others has a reasonable chance of finishing out the ticket (and in any position) once the two contenders take their place on the ticket'.

you must toss any trifecta permutation that includes less than two contenders on the ticket due to the first condition.

hope that clears it up.

[raybo]

Hmm... I was with you just up until then end, where I got a little lost. I don't know where you came up with the 25% number. I'm sort of guessing that you were referring to his scenario when he had 4 horses on the Win line. I could be wrong there, but I don't see how (with your horses across the board scenario) that 25% number comes up. Could you please clarify for me?

Yes, my mistake, I'm getting confused myself. The reduction would be 10% rather than 25%. The ticket would be:

win 1 2 3 4 5 6 7 8 9
place 1 2 3 4 5 6 7 8 9 10
show 1 2 3 4 5 6 7 8 9 10

or 8 x 9 x 9 = 648 possible combinations

[raybo]

Kelso, in a subsequent post after his original one said: "It involved betting 4 contenders, at least two of whom (but perhaps three of whom) will finish 1st, 2nd or 3rd in a trifecta. The remaining six horses can't be counted out from any ITM position, but not more than one of them will finish ITM.

In this scenario 3 of the contenders could finish ITM at the same time, which changes what he had said earlier, "at least 2 of the contenders will finish ITM". So, you have 24 possible finishes involving only the 4 contenders:

1 2 3___1 2 4___1 3 4___2 3 4
1 2 3___1 2 4___1 3 4___2 3 4
1 2 3___1 2 4___1 3 4___2 3 4

We also have the possiblility that only 2 of the comtenders will finish ITM along with 1 of the "others":

1 2
1 2
1 2 5 6 7 8 9 10

12 combinations multiplied by 6 possible combinations of only 2 contenders (12, 13, 14, 23, 24, 34), so we have 72 possible combinations with the others finishing 3rd. Multiply this by 3 to put them on the place line and the show line and we have 72 x 3 = 216 possible combinations with only 2 contenders ITM and only 1 other ITM.

Add the 24 combinations for contenders finishing 1st, 2nd, and 3rd with no "others" ITM and we have 240 possible combinations.

So the formula: C^3 + 3(T-1)C^2 + (2-3T)C is correct.

Hcapperjag, thanks for the info.

[raybo]

In retrospect, I think the word "contenders" caused all the confusion, at least for me. In my world, "contenders are those horses capable of winning. "Others", would be horses not capable of winning.

[Kelso]

In retrospect, I think the word "contenders" caused all the confusion, at least for me. In my world, "contenders are those horses capable of winning. "Others", would be horses not capable of winning.



Sorry for the confusion ... and thank you for contributing to my understanding of the solution to my problem.