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punteray
05-08-2017, 06:13 PM
At one time there was a discussion on calculating the payout for an EXACTA using probability and percentages.

I had a copy of the procedure but have misplaced it or just plain lost it.

I have tried to search the forum but have come up empty.

Can anybody help on this?

whodoyoulike
05-08-2017, 07:41 PM
At one time there was a discussion on calculating the payout for an EXACTA using probability and percentages.

I had a copy of the procedure but have misplaced it or just plain lost it.

I have tried to search the forum but have come up empty.

Can anybody help on this?


Exacta payouts are provided or shown on my ADW and the track and OTB provides it on their monitors before each race is run.

Are you sure it wasn't for Trifectas or a Superfecta?

Poindexter
05-08-2017, 08:16 PM
At one time there was a discussion on calculating the payout for an EXACTA using probability and percentages.

I had a copy of the procedure but have misplaced it or just plain lost it.

I have tried to search the forum but have come up empty.

Can anybody help on this?

If you are talking about from your own oddsline, the math is fairly simple. Say you have the 1 horse 2-1 and the 6 horse 8-1

.333333 chance the 1 wins and should he win the 6 has a .111111111/.666666666 chance of beating the rest of the field

so the chances of hitting are 1/3*1/6 or 1/18 which is a fair value of 17-1.
Actually the formula to determine the exacta price needed is much easier than that. Simply put it is :

(win odds*place odds)+win odds. So in the example I gave (2*8)+2=$18 for a $1 exacta. So the good news is you can likely do this in your head.

Now if you are using the public's line you would have to adjust for the takeout, so assuming a 15% takeout if the public had the 1) 2-1 and the 6) 8-1 the 2 becomes .33333/1.15 or .29(around 5-2) and the 8-1 becomes .1111111/1.15 or .096. More like a 9.5-1 shot. EDIT-I think my math is probably wrong on the public line.

Now all this assumes that the place line matches the win line, which is seldom the case in racing.

FWIW, I do not use this model myself even when I do line races.

Poindexter
05-09-2017, 04:56 AM
Okay with the public odds if you want to convert to a fair exacta price I would convert the tote odds to odds if there were no takeout. So to do so take the total (win pool/$ bet on a horse)-1. So if a horse has $5000 out of a $15,000 pool he would be a true 2-1 shot even though his actual odds is 1.55 to 1. If another horse in the race is has $1000 to win on him he is a true 14-1 shot(even though he is a 11.75-1 on the tote board). The fair exacta price theoretically would be (2*14)+2 or $30 for a $1. Of course you will never see that price so it is a completely futile excercise. However if you think that horse A looks to be at fair value at his current price of 3-1 and horse b looks to be at fair value at his current price of 6-1 then you could just use the formula, (3*6)+3 or $21 for a dollar as the fair exacta price as a point of reference. Making a line is much more accurate, but of course a lot more time consuming.

HalvOnHorseracing
05-09-2017, 01:48 PM
If you are talking about from your own oddsline, the math is fairly simple. Say you have the 1 horse 2-1 and the 6 horse 8-1

.333333 chance the 1 wins and should he win the 6 has a .111111111/.666666666 chance of beating the rest of the field

so the chances of hitting are 1/3*1/6 or 1/18 which is a fair value of 17-1.
Actually the formula to determine the exacta price needed is much easier than that. Simply put it is :

(win odds*place odds)+win odds. So in the example I gave (2*8)+2=$18 for a $1 exacta. So the good news is you can likely do this in your head.

Now if you are using the public's line you would have to adjust for the takeout, so assuming a 15% takeout if the public had the 1) 2-1 and the 6) 8-1 the 2 becomes .33333/1.15 or .29(around 5-2) and the 8-1 becomes .1111111/1.15 or .096. More like a 9.5-1 shot. EDIT-I think my math is probably wrong on the public line.

Now all this assumes that the place line matches the win line, which is seldom the case in racing.

FWIW, I do not use this model myself even when I do line races.

Like much of your math, the calculation is interesting. The flaw is that once you put a horse in the win slot, the probability of that horse finishing second becomes zero and the probability of a horse from the rest of the field finishing second is 100%. Which is also why the place line can't match the win line. And of course the odds of any horse finishing second do vary based on which horse wins the race. It is almost a given that calculating the exacta odds has to parallel the win odds. I actually agree that you somehow have to make the calculation based on win probability for the sake of simplicity. I'm not criticizing the technique. It's actually better than people betting while not having a clue whether the exacta is underpaying or overpaying.

If you are using the public line - either the morning line or the tote board odds - that line has already the take computed. A good line maker makes a 115-120% line, and of course the tote board odds are after the take has been deducted.

I've made it easy for people by developing a table for determining if the exacta is a fair payoff based on your real odds, on the thinking that if you make people come up with a win line and a separate place line they'll never do it. It also has some margin of error built into it.

http://halveyonhorseracing.com/?p=4319

punteray
05-09-2017, 01:58 PM
Thanks for the replies! I forgot to say it was a method using EXCEL.

AltonKelsey
05-09-2017, 05:15 PM
"And of course the odds of any horse finishing second do vary based on which horse wins the race."

Interesting. What would be the calculation for that?

Would it be easier to run second if the favorite wins or a longshot?

The answer is not so simple, I suspect.

AltonKelsey
05-09-2017, 05:35 PM
Just thought of something, do the odds of the winner vary based on who runs second.......

Just a thought.

Poindexter
05-09-2017, 05:46 PM
"And of course the odds of any horse finishing second do vary based on which horse wins the race."

Interesting. What would be the calculation for that?

Would it be easier to run second if the favorite wins or a longshot?

The answer is not so simple, I suspect.

The Math is simple. Say in a race we have

horse 1) 1-1 (50%)
horse 2) 4-1 (20%)
horse 3) 20-1 (5%)

Horse 3 has a 5% chance of winning. No matter who wins, he theoretically would have a 5% chance(of the remaining % left) of beating the rest of the field. So if the favorite wins there is 50% probability left on the remaining horses. So the 3 should have a 5/50 or 10% chance of coming 2nd. So the chance of the 1-3 exacta is .50*.1 or .05 or 20-1. If the 2 wins the 5 has a 5/80 or 1/16 chance of coming 2nd.............................

The problem is this is not accurate because of things that happen in a race or the win or miss nature of some horses. For instance a horse that won for 25 k last week, now drops for 16. You factor in the suspicious drop and make him 8/5 or about a 38% chance of winning. Now amongst that 62% chance you give him of losing is a significant chance that he just went no good. Now it gets even more complex, because sometime even when they are no good, they still manage to run a well beaten 2nd. So it is a very inaccurate math with these types. A lot ot trainers really have a knack for knowing just how bad there horse went. Also, factor in the nature of racing, 2 horses figure to duel on the lead, can they both hang on to run 1-2, sometimes, but often someone will split them or one will stop worse. Then you have layoff horses. There is a siginificant chance that they will be out for excercise or not ready and if they lose, they have nowhere near the same chance of coming 2nd as winning, but many of them do come 2nd. Then you have 1st time starter, 2nd time starters, first time turf, horses stretching out, horses shortening.......Then of course you have horse that refuse to win or distances that give certain styles a much better chance of coming 2nd/3rd than winnin....Then you also have the live horses, horses that you can't see but the money is all over them. So many of these horses very well. It is just a very inaccurate estimation process mainly because of the variables involved in your typical horse race. Which is why I personally focus on the value horse(s) and not worry about the "value exacta".

But if this works for people, great. I found that it was too inaccurate and thus abandoned it.

AltonKelsey
05-09-2017, 06:24 PM
I like your answer, but think about the second question. Why is running second any different than winning? Or running last?

If you had to assign place ONLY odds in advance , wouldn't there be one set of odds, irrespective of who finished in the win slot? I think so.

So the odds of a horse running second , is independent of who wins.

Try wrapping your head around it before you say its crazy. :cool:

Poindexter
05-09-2017, 07:09 PM
I like your answer, but think about the second question. Why is running second any different than winning? Or running last?

If you had to assign place ONLY odds in advance , wouldn't there be one set of odds, irrespective of who finished in the win slot? I think so.

So the odds of a horse running second , is independent of who wins.

Try wrapping your head around it before you say its crazy. :cool:

I don't think it is crazy at all. I think you mean odds to run 2nd only and not place. Sure you can do that

So you would have a line that looked like this to win

1) 3-1
2) 4-1
3) 6/5
4) 20-1
5) 20-1

Then you might have a to run 2nd line that looked like this.

1) 5/2
2) 3-1
3) 2-1
4) 12-1
5) 20-1

So basically the Math would look like this. On Say a 1-3 exacta
.25 chance the 1 wins * .33/.714 on the 3 coming 2nd leaves you with or .1155 or fair price of $8.65 for a $1 exacta. As opposed to using the win line only which in the above case you would have a fair price of $6.60. You can easily program a spread sheet to calculate all of the payoffs for you. Or if you want to save time and effort just multiply win odds of the top horse by place odds of the 2nd horse+win odds of the top horse. It will be less accurate(It would come up with $9 instead of $8.65 in the example I just gave) but you do not have to go through the process of creating a spreadsheet.

It is a good idea and certainly will be more accurate. I think a lot of the factors I mentioned prior still make it a tough task, and it makes the handicapping process a lot more larborious as now you have to come up with 2 lines instead of one, but roll with it and see how it works for you.

HuggingTheRail
05-09-2017, 08:11 PM
And if the math decision isn't tricky enough, you could also consider running styles - odds of two deep closers finishing 1-2 is less likely...etc

TheRinger
05-09-2017, 09:31 PM
Formula for fair exacta payoff.

=BetSize/(WinProb*(PlaceProb/(1-WinProb)))

Robert Fischer
05-09-2017, 09:45 PM
http://oi64.tinypic.com/260qqro.jpg

lazy right now, after dinner. Not sure how this explanation will come out.

If I wanted a generic method:
1. I'd look at the win odds for the top slot.
2. I'd then reduce the field size by 1 and rank the contenders for the place position in order using my handicapping insight(as opposed to the remaining public odds).
3 Then I'd use my ranking with Trifecta Mike's Odds Rankings probabilities formula.
3. I'd multiply the win odds% x the odds ranking within the remaining field size, and convert that into an $exacta payout, and consider playing that if I felt that I had an edge.


in the picture above, i used a 10 horse field.
I used the 2-1 favorite:7:(entered in B14 and C14) on top.
Then I ranked the remaining horses for their chance of running 2nd. Since I thought the third public choice:3:, is more likely than the second public choice, I'll consider a play.
I use (C1) because :3: is my top-ranked horse(to fill out the exacta) of the remaining 9 horse field.
Their probabilities multiplied in E15.
then converted to odds in E17.
$17 is my crazy estimate for fair price on a straight :7:-:3: exacta.
Maybe I consider $17, or maybe I want a $5 or $10 cushion or whatever....



not what I personally use, but I don't automate my stuff like this

Robert Fischer
05-09-2017, 10:05 PM
forgot to add in the wager to the odds part

so it would be $17X2+2 or $36 not $17 for $2 exacta.

also meant D1, rather than C1.

too much thinking

HalvOnHorseracing
05-09-2017, 11:11 PM
"And of course the odds of any horse finishing second do vary based on which horse wins the race."

Interesting. What would be the calculation for that?

Would it be easier to run second if the favorite wins or a longshot?

The answer is not so simple, I suspect.

Say you have a 10-1 shot win the race. That would make the 3-5 choice a much higher probability of finishing second. Conversely, if the 3-5 favorite wins, the odds for every other horse to finish second get shorter. The point, and this is not where the calculations get complicated, is that if you take one horse out of the mix by putting it in the win slot, then the probability of that horse finishing second are now zero and the probability of one of the other horses finishing second is now 100%. So in essence you would need a "new" line for second, and perhaps the horse that was the second choice to win might become the favorite to "win" the place at lower odds than it was to win the race. But as I said, the calculation seems hardly worth the effort.

HalvOnHorseracing
05-09-2017, 11:13 PM
Just thought of something, do the odds of the winner vary based on who runs second.......

Just a thought.

No they don't. The odds for a horse to finish second after you assign one horse to the win slot represents a dependent calculation. The win probability is independent of which horse finishes second.

HalvOnHorseracing
05-10-2017, 12:34 AM
The Math is simple. Say in a race we have

horse 1) 1-1 (50%)
horse 2) 4-1 (20%)
horse 3) 20-1 (5%)

Horse 3 has a 5% chance of winning. No matter who wins, he theoretically would have a 5% chance(of the remaining % left) of beating the rest of the field. So if the favorite wins there is 50% probability left on the remaining horses. So the 3 should have a 5/50 or 10% chance of coming 2nd. So the chance of the 1-3 exacta is .50*.1 or .05 or 20-1. If the 2 wins the 5 has a 5/80 or 1/16 chance of coming 2nd.............................


Your calculation doesn't work mathematically. I won't make a protracted argument. I'll just use your example to make the point.

Let's make it a six horse race with the following odds (your odds line) which allows a 16% take.

A 1-1
B 7-2
C 5-1
D 7-1
E 10-1
F 19-1

Let's say Horse A wins. Based on your methodology, B has a 44% chance of finishing second, C 34%, D 24%, E 18%, and F 10%. Add those up and you come up to a 130% chance of one of those horses finishing second, and that is too high.

Still, I'm not against doing something that simple to make a ballpark calculation, as long as you have a margin of error built in and as long as you understand the limitation of the calculation.

AltonKelsey
05-10-2017, 03:50 PM
No they don't. The odds for a horse to finish second after you assign one horse to the win slot represents a dependent calculation. The win probability is independent of which horse finishes second.


Not talking about that. Talking about the odds of a horse RUNNING second, no matter the winner.

Each finish position, can theoretically be considered equivalent mathematically. In a 12 horse field there are 12 Win possibilities, 12 place possibilities, etc. Each have the same characteristics odds wise and must total 100% (before takeout).

In fact , you could run a parimutuel pool based on ANY finish position. (would be fun to bet on who will be last, no?)

Don't think you can understand the Exacta without appreciating that.

PS I figured out the 'fair value' long before Harville published, in case you think I'm off the wall on this.

Prof.Factor
05-10-2017, 04:33 PM
example
Field (based on true odds to win ... zero takeout)
-----
.20 (5-1)
.20 (5-1)
.20 (5-1)
.20 (5-1)
.10 (10-1)
.10 (10-1)
====
100%
Lets say winner was a 5-1 shot with 20% chance to win, then the rest of the field had 80%.
Adjust Place probability to 100% with remaining field.
Field (chances to place)
-----
winner removed
.20/.80 = .25 (4-1)
.20/.80 = .25 (4-1)
.20/.80 = .25 (4-1)
.10/.80 = .125(8-1)
.10/.80 = .125(8-1)
====
100%
If the winner was one of the 10-1 shots then the rest of the field had 90% win probability.
Adjust Place probability to 100% with remaining field, like we did before.
Field (chances to place)
-----
.20/.90 = .222 (9-2)
.20/.90 = .222 (9-2)
.20/.90 = .222 (9-2)
.20/.90 = .222 (9-2)
winner removed
.10/.90 = .111 (9-1)
====
100%
exacta probabilities based on true odds:
5-1 with 5-1 = (.20x.25) = 20-1
5-1 with 10-1= (.20x.125) = 40-1
10-1with 5-1 = (.10x.222) = 45-1
10-1with 10-1= (.10x.111) = 90-1


So formula for above would look like:
WinnerWin% * (SecondHorseWin% / (1.00 - WinnerWin%))

Poindexter
05-10-2017, 05:12 PM
Halv, I am not ignoring you, I just have hard time understanding a lot of your posts. I must be developing ADD or something.

Let me run a hypothetical 5 horse field for you and you tell me where I a wrong.

fair odds line and I will assume that win odds = odds to come 2nd

1) 3-1 (25%)
2) 4-1 (20%)
3) 6-1 (14%)
4) 16-1 (6%)
5) 9/5 (35%)

#1 to come 2nd.

if 5 wins 35% * .25/.65 or 13.46% he come 2nd to 5.
if 4 wins 6% * .25/.94 or 1.56% he comes 2nd to #4
if 3 wins 14% * .25/.85 or 4.07 % he comes 2nd to #3
if 2 wins .20 * .25/.80 or 6.25% he come 2nd to #2

So the #1 has a 25% chance of winning and a 25.34% chance of coming 2nd.

What exactly do you disagree with and why?

When you create your own odds line you cant base it off os a 16% take.You have to base it off a zero percent take.

HalvOnHorseracing
05-10-2017, 05:39 PM
Not talking about that. Talking about the odds of a horse RUNNING second, no matter the winner.

Each finish position, can theoretically be considered equivalent mathematically. In a 12 horse field there are 12 Win possibilities, 12 place possibilities, etc. Each have the same characteristics odds wise and must total 100% (before takeout).

In fact , you could run a parimutuel pool based on ANY finish position. (would be fun to bet on who will be last, no?)

Don't think you can understand the Exacta without appreciating that.

PS I figured out the 'fair value' long before Harville published, in case you think I'm off the wall on this.

How can you calculate the odds of any runner finishing exactly second without first knowing which horse finished first? You can construct an odds line in the win position, but how do you calculate the same sort of line for place except in a relative sense - in other words, make the assumption that the win line represents the idea that Horse A is one position better than Horse B which is one position better than Horse C and so on.

But even if you had your "place" line without knowing the winner, the place price is still dependent on which horse finishes first, meaning your line would be almost meaningless for calculating fair place value because you don't know the winner.

You could run a parimutuel pool based on any finish position, but in most cases it would be foolish to bet into it for most finish positions. Who in their right mind would say, I'm going to bet Ole Hay and Oats for 7th? And who in their right mind wouldn't book the bet?

I'm going to say what I said. The odds to place are going to be dependent on which horse you decide wins the race. And while Poindexter's calculations are novel, he makes the right point that the only realistic option is to use the odds from the win line. Because in a 12 horse race, there are actually 12 different place lines depending on who wins. Think about this. If Battle of Midway had won the Derby and Lookin at Lee had finished second, how much more than $26.60 would place have payed? If you assumed Battle of Midway wins the race instead of Always Dreaming, how much would you have changed your "odds line" for place on Lookin at Lee? I got your point. I'm still sticking with mine.

This is the calculation from Brisnet on Fair Value for the exacta.

$Bet x (Win Odds) x (Place Odds +1)

Here is how to read this formula. The Fair Payoff for an Exacta equals the size of the bet, such as a $2 Exacta, multiplied by the (win horse's odds-to-1) multiplied by the (place horse's odds-to-1 plus 1). The odds are each horse's going-off win odds.

For example, if the horse in the win position is going off at 3 to 1 and the horse in the place position is going off at 4 to 1, the Fair Payoff for a $2 Exacta is:

$2 x (3) x (4 + 1) = $30


The problem with this is three fold. One, the horse's win odds cannot also be the horse's place odds, but as I've noted, it's really the only odds line that works. Two, in the Brisnet calculation, you have a 25% horse on top of a 20% horse. In order for $30 to be the fair pay, you have to assume the horse with the 20% chance of winning has a 56% chance of finishing second. But moreover, if you reverse the positions so that the 4-1 horse wins and the 3-1 horse finishes second, the fair pay only jumps to $32. You buy that? Third, the amounts that the calculation comes up with don't have a sufficient margin of error built in.

Still, I'm willing to concede for the sake of simplicity, these calculations are fine. But if someone uses them, they really need to understand the limitation of the calculation.

HalvOnHorseracing
05-10-2017, 06:44 PM
Halv, I am not ignoring you, I just have hard time understanding a lot of your posts. I must be developing ADD or something.

Let me run a hypothetical 5 horse field for you and you tell me where I a wrong.

fair odds line and I will assume that win odds = odds to come 2nd

1) 3-1 (25%)
2) 4-1 (20%)
3) 6-1 (14%)
4) 16-1 (6%)
5) 9/5 (35%)

#1 to come 2nd.

if 5 wins 35% * .25/.65 or 13.46% he come 2nd to 5.
if 4 wins 6% * .25/.94 or 1.56% he comes 2nd to #4
if 3 wins 14% * .25/.85 or 4.07 % he comes 2nd to #3
if 2 wins .20 * .25/.80 or 6.25% he come 2nd to #2

So the #1 has a 25% chance of winning and a 25.34% chance of coming 2nd.

What exactly do you disagree with and why?

When you create your own odds line you cant base it off os a 16% take.You have to base it off a zero percent take.

My point is that place odds are dependent on win odds, or as I mentioned, in a 12 horse race there are 12 place odds lines - one for each horse winning. But I agreed that for the sake of simplicity you base the place odds on the win odds, if you come up with a formula like Brisnet used.

So in your previous example you said,

Horse 3 has a 5% chance of winning. No matter who wins, he theoretically would have a 5% chance (of the remaining % left) of beating the rest of the field. So if the favorite wins there is 50% probability left on the remaining horses.

I hope this doesn't sound complicated, but you can't assume the odds of a horse winning are the same as a horse finishing second. Obviously, even for the sake of simplicity, it is illogical because there is a dependent relationship with the winner. Once we take the favorite out of the picture, the odds of any of the remaining horses finishing second has to get better. Conversely, if a horse other than the favorite wins, the probability of the favorite finishing second gets better while the probabilities of the remaining horses finishing second actually may decrease.

In the example above, once the favorite is assumed first, the probability of the second choice beating the other horses logically has to go up. How can a horse be 20% to win, but 13% to place after you eliminate the favorite from the place position? Does that make sense? The total probability of the remaining horses finishing second after you assign one horse to the win is 100%, while the probability of the winning horse finishing second is zero.

Let's prove this for your calculation. If the 5 wins, you have the probability of the 1 finishing second at 13.46%. If you do the same calculation for the 2, 3, and 4, you get 10.77%, 7.54%, and 5.53% respectively. If you add them all up, you get 37.30%, which we know is off by almost a factor of three. If the 5 wins, ONE of the remaining four has to finish second, making the probability that the 1, 2, 3, or 4 finishes second 100%.

What you could do is take the percentage you assigned to a horse - let's just say the favorite - and distribute it proportionally to the remaining horses. So you would take the 35% assigned to the 5 (since if he wins his probability of finishing second is 0%) and distribute it to the 1, 2, 3, and 4.

So after the 5 wins, the probabilities for the remaining horses to finish second are as follows.

1 38.5%
2 30.8%
3 21.56%
4 9.25%

And that adds up to 100% as it should. And the MINIMUM fair pay for a 5 with 1 exacta is about $14.

You can do the same calculation for any horse winning simply by distributing his win percentage proportionally, but in every case the horse's probability of finishing second will be higher than his probability of winning after eliminating him from the win slot.

Does that make sense to you now?

AltonKelsey
05-10-2017, 07:13 PM
I understand the concept is not easy to grasp. That's why I'm posting it.

Let's see how long it takes before someone figures out I'm right .

Poindexter
05-10-2017, 07:24 PM
Let's prove this for your calculation. If the 5 wins, you have the probability of the 1 finishing second at 13.46%. If you do the same calculation for the 2, 3, and 4, you get 10.77%, 7.54%, and 5.53% respectively. If you add them all up, you get 37.30%, which we know is off by almost a factor of three. If the 5 wins, ONE of the remaining four has to finish second, making the probability that the 1, 2, 3, or 4 finishes second 100%.

NO that is not what I posted. I posted that there is a 13.46 % probability that the 5 wins and the 1 runs 2nd. If the 5 wins the 1 has a 25/65 chance of running 2nd. If the 5 wins the 2 has a 20/65 chance of coming 2nd. If the 5 wins the 3 has as 14/65 chance of winning and if the 5 wins the 4 has a 6/65 chance of running 2nd. The 4 horses combined have a 65/65 or 100% chance of running 2nd. Which closely matches the percentages you have in the next paragraph, I have
1. 38.46%
2. 30.77%
3. 21.54%
4. 9.23%

So other than a couple of decimal points we are in complete agreement

What you could do is take the percentage you assigned to a horse - let's just say the favorite - and distribute it proportionally to the remaining horses. So you would take the 35% assigned to the 5 (since if he wins his probability of finishing second is 0%) and distribute it to the 1, 2, 3, and 4.

So after the 5 wins, the probabilities for the remaining horses to finish second are as follows.

1 38.5%
2 30.8%
3 21.56%
4 9.25%

And that adds up to 100% as it should. And the MINIMUM fair pay for a 5 with 1 exacta is about $14.

The formula I presented say (1.8*3)+1.8 or $14.40 for a $2 exacta. So once again we agree.





See bolded above.

AltonKelsey
05-10-2017, 07:35 PM
Todays quiz.

Is it easier to win or run second?

HalvOnHorseracing
05-10-2017, 08:15 PM
See bolded above.

I'm just going to say wow. We don't agree. Saying the horse has a 38% chance of running second no matter who wins is not what I am saying. I said it has a 38% chance of running second if the 5 wins. The percentage changes depending on the winner. And if I did what you did - added them up based on the winner - I'd have come up with a much bigger number than the 38% you came up with.

I did your calculation for the 5 winning and the probability of each respective horse finishing second. It adds up to 37% and it is absolutely the case that it has to add up to 100%. Very simple question. If there are five horses, and one horse wins, what is the probability one of the other four finishes second? In other words, whichever horse wins, the probability of one of the other horses running second is 100%, and if I did what you did, add up the probabilities, it would appear as if the four horses had a 400% chance of finishing second.

Read what you wrote and follow it out to the logical conclusion. If the 5 wins, and you calculate the 1 has a 13.46% chance of finishing second, do the same calculation for the other three horses, add them up and you come up the aggregate probability of one of the four horses running second, and it adds up to 37% which is mathematically impossible. It has to add up to 100% unless the other four horses fall down. You say the 1 has a probability of 13.46% of running second if the five wins, I say the 1 has a probability of 38.5% of running second if the 5 wins. Two entirely different numbers.

You made an apples to oranges comparison and come to a spurious conclusion. I said I didn't want to argue, mainly because I figured you were married to your methodology and wouldn't see the flaw easily. You can believe what you want. At this point continuing the discussion would simply entail me repeating myself.

HalvOnHorseracing
05-10-2017, 08:44 PM
Todays quiz.

Is it easier to win or run second?

I'm guessing you're going to make what you figure is a sharp point once you get someone to bite.

Poindexter
05-11-2017, 12:21 AM
I'm just going to say wow. We don't agree. Saying the horse has a 38% chance of running second no matter who wins is not what I am saying. I said it has a 38% chance of running second if the 5 wins. The percentage changes depending on the winner. And if I did what you did - added them up based on the winner - I'd have come up with a much bigger number than the 38% you came up with.

I did your calculation for the 5 winning and the probability of each respective horse finishing second. It adds up to 37% and it is absolutely the case that it has to add up to 100%. Very simple question. If there are five horses, and one horse wins, what is the probability one of the other four finishes second? In other words, whichever horse wins, the probability of one of the other horses running second is 100%, and if I did what you did, add up the probabilities, it would appear as if the four horses had a 400% chance of finishing second.

Read what you wrote and follow it out to the logical conclusion. If the 5 wins, and you calculate the 1 has a 13.46% chance of finishing second, do the same calculation for the other three horses, add them up and you come up the aggregate probability of one of the four horses running second, and it adds up to 37% which is mathematically impossible. It has to add up to 100% unless the other four horses fall down. You say the 1 has a probability of 13.46% of running second if the five wins, I say the 1 has a probability of 38.5% of running second if the 5 wins. Two entirely different numbers.

You made an apples to oranges comparison and come to a spurious conclusion. I said I didn't want to argue, mainly because I figured you were married to your methodology and wouldn't see the flaw easily. You can believe what you want. At this point continuing the discussion would simply entail me repeating myself.

If you want me to address your points, quote me, explain your arguments and I will explain. There is just too much misunderstanding between us. That is the only way I can do this. If you don't want to do so, no problem, I completely understand and I promise I won't lose any sleep over the fact you think I am wrong.

HalvOnHorseracing
05-11-2017, 10:42 AM
If you want me to address your points, quote me, explain your arguments and I will explain. There is just too much misunderstanding between us. That is the only way I can do this. If you don't want to do so, no problem, I completely understand and I promise I won't lose any sleep over the fact you think I am wrong.

Simple questions. If in your example the 5 wins - and only the 5 wins - what is the probability the 1 finishes second? What is the probability of the 2 finishing second? What is the probability of the 3 finishing second? What is the probability of the 4 finishing second?

Now add those four probabilities and tell me what you get.

Remember, the scenario is only the 5 winning with one of the four other horses finishing second. I don't want to know what the probability of the 1 finishing second to the 2, 3, or 4 is. Just the four probabilities of the respective horses finishing second to the 5.

AndyC
05-11-2017, 10:50 AM
.... You say the 1 has a probability of 13.46% of running second if the five wins, I say the 1 has a probability of 38.5% of running second if the 5 wins. Two entirely different numbers.....

Yes they are 2 entirely different numbers. The 13.46% represents the probability of a 5/1 exacta and not the probability of the 1 running second if the 5 wins. There is no disagreement on your 38.5% calculation! Poindexter merely used that figure to calculate the exacta probability.

HalvOnHorseracing
05-11-2017, 12:34 PM
Yes they are 2 entirely different numbers. The 13.46% represents the probability of a 5/1 exacta and not the probability of the 1 running second if the 5 wins. There is no disagreement on your 38.5% calculation! Poindexter merely used that figure to calculate the exacta probability.

This is what Poindexter wrote:

#1 to come 2nd.

if 5 wins 35% * .25/.65 or 13.46% he comes 2nd to 5.
if 4 wins 6% * .25/.94 or 1.56% he comes 2nd to #4
if 3 wins 14% * .25/.85 or 4.07 % he comes 2nd to #3
if 2 wins .20 * .25/.80 or 6.25% he come 2nd to #2

So the #1 has a 25% chance of winning and a 25.34% chance of coming 2nd.


I can see the confusion. The 13.46% is the probability of the exacta 5-1. It's just stated awkwardly. I was thrown by the statement the 1 has a 25.34% chance of finishing second in total. We know it has at least a 38.5% chance of finishing second to the 5 alone.

The percentages of the other four horses running second to the 5 are:

1 38.5%
2 30.8%
3 21.56%
4 9.25%

and if you multiply those by the percentage of the 5 winning, you get the numbers Poindexter came up with, only it is the probability of a respective exacta.

5-1 exacta 13.46%
5-2 exacta 10.77%
5-3 exacta 7.54%
5-4 exacta 5.53%

Fascinating.

AndyC
05-11-2017, 01:50 PM
This is what Poindexter wrote:

#1 to come 2nd.

if 5 wins 35% * .25/.65 or 13.46% he comes 2nd to 5.
if 4 wins 6% * .25/.94 or 1.56% he comes 2nd to #4
if 3 wins 14% * .25/.85 or 4.07 % he comes 2nd to #3
if 2 wins .20 * .25/.80 or 6.25% he come 2nd to #2

So the #1 has a 25% chance of winning and a 25.34% chance of coming 2nd.


I can see the confusion. The 13.46% is the probability of the exacta 5-1. It's just stated awkwardly. I was thrown by the statement the 1 has a 25.34% chance of finishing second in total. We know it has at least a 38.5% chance of finishing second to the 5 alone.

The percentages of the other four horses running second to the 5 are:

1 38.5%
2 30.8%
3 21.56%
4 9.25%

and if you multiply those by the percentage of the 5 winning, you get the numbers Poindexter came up with, only it is the probability of a respective exacta.

5-1 exacta 13.46%
5-2 exacta 10.77%
5-3 exacta 7.54%
5-4 exacta 5.53%

Fascinating.

To be clear, the 1 has a 25.34% chance of running second not to the 5 but to the field and the percentage includes the races that the 1 wins.

AltonKelsey
05-11-2017, 01:53 PM
So Halv, are you ready to say I was right?

HalvOnHorseracing
05-11-2017, 09:16 PM
So Halv, are you ready to say I was right?

I'm ready to say I understand Poindexter's calculation and now that I understand it, yes, it is the same thing I calculated.

HalvOnHorseracing
05-11-2017, 09:31 PM
To be clear, the 1 has a 25.34% chance of running second not to the 5 but to the field and the percentage includes the races that the 1 wins.

I saw that as a valueless number. I only need the percentage chance of finishing second to any other horse in order to calculate the fair pay exacta. A calculation of the probability of a horse running second in the aggregate would only be important if I needed to decide if the horse was a good place price, and I'd still need to know which horse would run first. I'd still maintain that if the 1 wins the race, it's probability of running second is zero. But the argument is purely academic.

citygoat
05-13-2017, 03:10 PM
you might be better off hanging around the windows waiting to hear a strong straight exacta bet ($100 exacta 1-2)and then wager $5 on 2-1.;)

whodoyoulike
05-13-2017, 08:25 PM
I saw that as a valueless number. I only need the percentage chance of finishing second to any other horse in order to calculate the fair pay exacta. A calculation of the probability of a horse running second in the aggregate would only be important if I needed to decide if the horse was a good place price, and I'd still need to know which horse would run first. I'd still maintain that if the 1 wins the race, it's probability of running second is zero. But the argument is purely academic.

I still don't understand the purpose of this exercise when the the exacta payouts are being provided before each race.

And, who says this calculation is the "fair pay" value. It just provides another value. I will admit I don't understand the 2x win odds and the +1 to the place or second horse's win odds. Maybe if I did, it would be much clearer but I really doubt it.

Has anyone compared the estimates with actuals?

Poindexter
05-13-2017, 09:04 PM
I still don't understand the purpose of this exercise when the the exacta payouts are being provided before each race.

And, who says this calculation is the "fair pay" value. It just provides another value. I will admit I don't understand the 2x win odds and the +1 to the place or second horse's win odds. Maybe if I did, it would be much clearer but I really doubt it.

Has anyone compared the estimates with actuals?

If you do not create your own fair oddsline, just ignore the thread, it doesn't apply to you. If you do make your own oddsline, then it is just a way to tell you how much the exacta should pay based off of your oddsline to determine whether you are getting value on an exacta combination or not. But as I state prior it has it flaws.

HalvOnHorseracing
05-14-2017, 10:11 PM
I still don't understand the purpose of this exercise when the the exacta payouts are being provided before each race.

And, who says this calculation is the "fair pay" value. It just provides another value. I will admit I don't understand the 2x win odds and the +1 to the place or second horse's win odds. Maybe if I did, it would be much clearer but I really doubt it.

Has anyone compared the estimates with actuals?

What value are the exacta payoffs on the board unless you have an idea whether they represent an underlay or overlay? You're comparing your opinion to the crowd's opinion. If you're making that evaluation based on tote board odds or some gut sense, whether or not you are right is closer to speculative. The whole point of calculating the overlay price is that if you get any good at it and you are betting properly, in the long run you should be a winner, and that's true whether you are betting win or exacta.

whodoyoulike
05-14-2017, 10:44 PM
What value are the exacta payoffs on the board unless you have an idea whether they represent an underlay or overlay? You're comparing your opinion to the crowd's opinion. If you're making that evaluation based on tote board odds or some gut sense, whether or not you are right is closer to speculative. The whole point of calculating the overlay price is that if you get any good at it and you are betting properly, in the long run you should be a winner, and that's true whether you are betting win or exacta.

I don't remember if you're using that formula of 2x and +1 but as I mentioned .... How do you know that is a "fair value" of an exacta payout?

Any other values replacing the 2 and 1 will provide a different value.

And, I'll admit I just look at the exacta matrix for the horses whom I think will come in 1st and 2nd and evaluate whether it's worth the bet(s). Sounds like you guys are trying to only bet the exacta overlays which can be a number of them which will be expensive.

But, again how can you be certain it's an overlay?

HalvOnHorseracing
05-14-2017, 11:56 PM
I don't remember if you're using that formula of 2x and +1 but as I mentioned .... How do you know that is a "fair value" of an exacta payout?

Any other values replacing the 2 and 1 will provide a different value.

And, I'll admit I just look at the exacta matrix for the horses whom I think will come in 1st and 2nd and evaluate whether it's worth the bet(s). Sounds like you guys are trying to only bet the exacta overlays which can be a number of them which will be expensive.

But, again how can you be certain it's an overlay?

Conceptually, you are assigning a win probability to each horse. After you assign the win probability, you can put each horse into the win slot of the exacta. Once the horse is in the win slot, it's probability of running second is 0%. Using the same odds that you calculated for win, redistribute the percentage of the winning horse proportionally. I did this above.

The assumption you are making is that no matter which horse you stick in the win slot, the odds of any of the remaining horses finishing second is in the order of best original odds to worst.

In other words, if horse A has a 36% chance of winning, as you calculate the exacta with horse A finishing first (because once he finishes first they probability of any other horse winning is zero and the probability of Horse A finishing second is zero) you take that 36% and redistribute it proportionally the the horses that can still finish second to get the percentage probability of a respective horse finishing second. This is simply the easiest way to assign place probability. So if Horse A has a 36% chance of winning, and Horse B has a 33% chance of finishing second with Horse A in first, then the A with B exacta has a 12% chance of coming in, making the fair pay 7-1. If the exacta is paying $30 on the board, that's an overlay.

If that sounded complicated, it's not. You can use Excel to construct a matrix, and see which exactas are overlays. Most people would not bet all the overlays - besides being expensive, obviously you lose your edge if you do. You might look at your top two or three choices, evaluate those exactas and bet if there is an overlay. The horse that I assign the lowest odds to, I still think is most likely to win the race. The exactas with him on top are the ones I'm most interested in.

Same exact concept with win. If I think a horse is 3-1 and it's 6-1 on the board, that would be a bet. If I have three overlays, and I bet them all, I've seriously lost my edge. Maybe I can bet two if the odds justify it (I have two horses, one at 3-1 on my line, one at 4-1. The tote board odds on the horses are 6-1 and 15-1. I can't pass either.)

Overlay. When my odds are less than the tote board odds, that is an overlay.

I believe most people at the track bet intuitively. "I think the 5 horse is going to win and it's 5-1. I think that's a good bet."

formula_2002
05-15-2017, 01:02 AM
THE PROBABILITY OF A HORSE WINNING IS (1/(ODDS+1))/ THE RACES TOTAL BOOK PERCENTAGE .


THE TOTAL BOOK PERCENTAGE FOR A TRACK WITH A 15% TAKE OUT SUMS TO ABOUT 1.18

THE PROBABILITY OF ANY HORSE WINNING IS (1/(ODDS+1))/1.18

TAG THE WIN HORSES WIN PROBABILITY WITH "A"

TAG THE PLACE HORSE WIN PROBABILITY WITH "B"

EXACTA PROBABILITY FOR A/B IS:


A X (B/(1-A))

EXAMPLE: A HORSE ODDS =2-1, B HORSE ODDS = 3-1, TRACK TAKE OUT IS 15%


A HORSE WIN PROBABILITY IS (1/(2+1))/1.18 = .282


B HORSE WIN PROBABILITY IS (1/(3+1)/1.18 = .212





PROBABILITY FOR THE A/B COMBINATION IS,


.282/(.212/(1-.282)) = .083


THAT EQUATES TO A $1 RETURN OF (1/.083)= 12.04


THE FAIR $1 PAYOUT IS $12.04

betovernetcapper
05-15-2017, 01:56 PM
If the race surface is very biased or the race shape favors a certain running style, the best two horses with that style may run one-two. So on a speed favoring track, the top two early may run one-two. If it's a closing track it might be the best closer followed by the 2nd best. In those cases you could use the odds to create some kind of formula.
In a lot of cases the model for the place horse is different then the model for the win horse. The model for the win horse might suggest he be within a length of the leader at second call & the model for the place horse might indicate having the fastest final quarter. In this case the place horses odds to win would be very slight but very good to place.
The you have the sucker horses that for one reason or another will finish second or third much more often than win. I think Quirin's formula was to multiply the number of the horses wins and the deduct the number of places & shows. If the resulting number was 3 or higher the horse was much more likely to finish 2nd or 3rd then win regardless of other factors. Maiden and non winners of two races are cluttered with such horses.
Sartin had a notion he called the contrarian horse. This notion suggests that if you think the winner is the best early speed runner, then the place horse was likely to be the best off the pace or closer. This theory works when it works.
At any rate IMO usually the win odds are only a factor in determining the place hose & other things should be considered.

AltonKelsey
05-15-2017, 06:06 PM
In theory a separate model for place would be ideal, but since 99.99% of players don't even have a decent model for win, where would they get that ?

punteray
05-16-2017, 01:54 PM
Thank all of you for posting, very interesting BUT what happened to my ORIGINAL request?

Ray

betovernetcapper
05-17-2017, 05:02 PM
HTR has a place model.