Horse Racing
 the Racing Form or "Program"

So we went to the races at Woodbine (Toronto), got us a racing form and tried to decipher it.
For each race, lots of info is available for each horse.
In race 4, for example, a horse called Almira Fawn had this info:

>Wait! What's that place and show?
They don't say a horse came in 1st or 2nd or 3rd; they say win, place and show.
When you bet, you say you'd like to bet \$2 on number 4 to win or place or show.

>Just \$2?
That's the minimum bet ... and that's what we always bet.
Also, when the race has ended, they identify the first three horses and what they'd pay for a \$2 bet.
No matter if you bet a \$jillion dollars, they tell you the "payout" for a \$2 bet.
If they say the winning horse pays \$4.50 to win, that means for every \$2 bet (on that horse, to win), you'd get back \$4.50.

>So you'd get \$4.50 when you go to collect?
Yes. That's NOT your "profit". It's your \$2.00 bet plus a profit of \$2.50.
You'd also get a payoff if you bet on that horse to place or show.
Further, if you bet on some horse to show and it came in 3rd, you get a payoff (but not if you bet on it to win or place).

In addition, there's information (in the racing form) on the historical performance of each horse.
For example, for each horse's last race, there are several lines in the form like so::

Also, for each past race that the horse has run, the winners are noted ... like so:

It gives the horses who came in 1st, 2nd and 3rd with the weight carried and the winning margins.

>Okay, after the race they show the odds ... how much do I win?
If the odds are, for example 5/3, then for a \$2 bet you'd get a payoff of (5/3)x(2) + 2 = \$5.33.
But they usually just give you \$5.30 (Rounded down to the nearest 10 cents.)

>That's a profit of (5/3)x2 = \$3.33, rounded to \$3.30 plus your original \$2 bet, right?
Right. It's always:
 [1]   payout = odds x (original bet) + (original bet)

If you bet \$100, it'd be (5/3)x100 + 100 = \$266.67 (rounded down).

>And how do they calculate the odds?
This is what I think happens ... to calculate the odds for the winning horse
 Suppose the amounts bet on each of 8 horses is shown in the table Further, suppose horse #5 wins. Determine the total money bet on this race to win, for all horses. (example: \$100,000 and that's the win pool) Subtract the racetrack's cut. (example: 20% of \$100,000 = \$20,000 leaving \$80,000 = the amount for the payout) Determine the total amount of the betting on the winning horse ... to win. (example: \$30,000 for horse #5) Calculate the payout per \$1.00.(example: \$80,000 / \$30,000 = \$2.67 for each \$1 bet) That's the original \$1 bet plus a "profit". (example: For a payout of \$2.67, the profit on a \$1 bet is \$1.67 which would probably be rounded to \$1.60 and that'd be a profit of \$3.20 for a \$2 bet.)

>And the odds are ... 1.6?
Uh, yes, but it'd be written as a fraction (for historical reasons ?) like 1.6 = 8/5.

>So, if I bet \$2 and use that magic formula [1], I'd get back a payout of (8/5)x2 + 2 = \$5.20
That's my guess.

>So how should I bet ... and win some money?
Gimme a minute; I'm still googling.

 Tips on Winning

Here's one thing I read about picking a winner by finding the losers:

• Look at each horse entered in the race and try to find horses that are obviously overmatched.
• Do their speed figures fit with the other horses?
• In looking at their last few races are they in good form or bad?
• Do they have recent workouts that suggest they are fit and ready?
• Are they racing at a distance or on a surface that they do not like?
• Are they making too big of a class jump?
• Throw out as many duds as you can and hopefully you are only left with a handful of contenders.
>That's it?
No.
• You should only have 3-4 horses that are candidates for winning the race.
• Make your selection based on their record at this track, distance, average speed ratings, trainer and jockey.
• If you still end up with too many horses that are logical contenders, pass the race.
• Watch the races you don't bet on with as much interest as those you do.
• See if the race unfolds the way you thought it would, look at the results and ask yourself if the results are consistent with what you knew going into the race.
• Go through your racing program and circle any horse who raced within the last 35 days and finished within two lengths of the winner at the same class.
• Now scratch out any horse who has a low percentage jockey (any rider who can't win more than 12%).
Of course, ya gotta consider the horse and how s/he usually runs:
1. Early speed types that want and need the lead.
2. Them that can either go on top or sit 1 to 3 lengths off the pace.
3. Them that don't want the lead, but stay in contact with the field, usually running 4 to 6 lengths back.
4. A closer that likes to run way off the pace and then make a late rally.
>How about just betting on the favourites?
They don't pay much, a little over \$2 for a \$2 bet ... and they apparently lose 2/3 of the time.

 Our Picks

Okay, so here's how some of us picked our horses:
For the 8th race I saw this:
 Harlan's Panther: In the previous month, the horse started in post position 6, came from 6th position to win with jockey Wilson. Earlier that month, the horse started in post position 5, came from 4th position to place with jockey Contreras Contreras had an in-the-money percentage of 20.5% which is good. By "in-the-money", I mean he came in 1st, 2nd or 3rd. The horse had 5 races so far this year and was in-the-money twice. I ignored the race earlier in the month where the horse started in post position 11 and finished in 9th place with jockey Pizarro.
>So how did the horse do?
He came in last! I notice that, for this race, he had a different jockey: Ramsammy with an in-the-money percentage of just 9.4%
However, my wife saw this:
 Dixie Strike: In the previous month, the horse started in post position 9 and came from last place to place with jockey Husbands. Earlier this month, the horse started in post position 9 and again came from way behind to win with jockey Husbands. Husbands had an in-the-money percentage of 23.2% which is very good. The horse had 2 races so far this year and was in-the-money both times. For this race, the jockey was again Husbands.
In fact, after a while my wife just bet on the horse with Husbands. This jockey was in-the-money four times that day.

>So how did the horse do?
He came in 2nd, like so:

>So your wife made a \$2.50 profit for her \$2 bet on show?
No, she made a profit of \$0.50. Remember, she bets \$2.00 and gets back that \$2.50 payout ... and that's just \$0.50 profit.

>Why did "Rose and Shine" win?
Maybe because his jockey, for this race, was Contreras who was in-the-money four times that day!
He even brought in a long shot that paid \$42.30 to win!

>So you should bet on the jockey, not the horse!
Maybe ... however, if you enjoy watching and aren't concerned about winning lots of \$money, you could just bet the favourite to show.
Although they win only 1/3 of the time, they come in 1st, 2nd or 3rd about 2/3 of the time.

>Or bet the post position. The horse in P.P. 1 has a shorter distance to run.
Apparently the percentage of winners is about the same for each post posn.

>Then I repeat: Bet on the jockey, not the horse!
Maybe, but after playing with "betting on the favourite" I decided I didn't like profits of a few cents when it came in-the-money.
The last time we were at the track I picked 7 out of 8 "in-the-money", betting \$2 to Show on all eight races ... and a typical payout was \$2.20.
That means a profit of just 20 cents. That means I'd have to win about 5 races for every race where I lost money.

So now I think I'll bet on long-shots. They may not come in-the-money often, but when they do they pay for a lot of losses.
For example, I bet \$2 to Show on a horse whose ML odds were 20-1 and it paid \$17.10 and that \$15.10 profit paid for at least seven losses.

>Huh? ML odds?
Yeah, somebody guesses the odds of winning before there's even any betting (maybe a couple of days before racing day). It's called Morning Line odds.
It's a (sometimes lousy) prediction of how the public will bet on racing day.

>If that ML value of 20-1 were correct, you'd have made a profit of 20*\$2 = \$40, right?
Right (if I bet to Win), but I often find that an ML of 20-1 ends up being much less at race time. In the above example, the actual odds turned out to be about 7-1.
Remember that the ML odds, as well as the odds you see at post time, are odds to Win ... not to Place or Show.
'course, ya don't hardly know post-time odds the day before the race so I'm using ML odds as a measure of the betting activity for Win, Place or Show.

>I'm surprised you don't have a spreadsheet.
I do, actually ... for Woodbine. The results of downloading and identifying the "long shots", (based upon ML odds) looks like this:

The ODDS are actually the ML odds ... and note the following:
 Assuming you bet \$2 to Show, you'll also get a "profit" table like so Them's my profits for a day at Woodbine: June 7, 2012 as displayed in the spreadsheet image, above. I should also note that if you can dig up the appropriate URL for some other track (Belmont? Churchill Downs?)     and identify the Top Jockeys, then the spreadsheet should work. >Should? Well ... I do offer a money-back guarantee on the efficacy of the ... >zzzZZZ

 Strategies

Now we do a wee bit o' math and ...
>zzzZZZ

We suppose the following \$2 bets:

1. A horse with odds y will come in-the-money X times in 10 races. Odds of 15-1 means Y = 15 and odds of 5/2 means Y = 5/2 etc. etc.
Such a horse will generate a profit of \$2y for each race in-the-money, so \$2Xy for all X races. "Profit": is always Odds*Bet
2. Such a horse will then lose 10 - X times in 10 races.
Each is a \$2 loss so the total loss in 10 races is \$2(10 - X)
3. The Net Profit is then 2Xy - 2(10 - X) which we want to be positive.
That is, we require that: 2Xy > 2(10 - X) or Xy > 10 - X.
4. Let X = 10x, so x is the fraction of times the horse comes in-the-money. If it's in-the-money 25 % of the time, then x = 0.25.
That gives the following neat formula:
 Consider horses with odds y. If x is the fraction of races where they run in-the-money, then you expect a profit if: y > (1 - x) / x     or     x > 1 / (1 + y)
 >That's confusing! Okay, suppose we're talking about favourites. They're in-the-money 67% of the time, so x = 0.67. Further, if the payouts for favourites average, say, \$2.50, then the profit on \$2 is \$0.50. Remember: Profit = Odds*Bet so Odds = Profit/Bet = \$0.50 / \$2 so y = 0.50/2 = 0.25. That's like 1-to-4 odds and gives the blue dot in Figure 2, namely (0.67, 0.25). No money to be made there, eh? Now consider a long shot with odds 20-to-1, so y = 20. Suppose these guys are in-the-money 10% of the time, so x = 0.10. That gives the point (0.10, 20) which is the green point in Figure 1. >Does a 20-to-1 long shot come in 10% of the time? That'd be once every 10 races ... uh, I have no idea. Anyway, consider horses with 10-1 odds, so y = 10. To make a profit (in the long run), we need x > 1/(1 + y) = 1/11 = 0.091. That's the red line. Them thar horses gotta come in more than 9.1% of the time. Figure 1Green area is profitable

>Why did you say the payout for favourites is a measly \$2.50?
I was talking about favourites coming in 1st, 2nd or 3rd. That happens 67% of the time. If we bet \$2 to Show, I'd expect a payout of \$2.50.
Of course, you can stick in any numbers you like.

>I bet to Win and expect an average payout of \$5.
Okay, that's a profit of \$3 for a \$2 Bet, so y = Profit / Bet = 3/2 = 1.5 then we'd want x > 1/(1 + y) = 1/2.5 = 0.4 or 40%.
Your horses (with odds 1.5 or 3-to-2) should Win more than 40% of the time. Do they?
>That'd mean more than 4 times in every 10 races ... uh, I have no idea.
Welcome to the IDK club.
>Huh?
That's the I Don't Know club.

Note that this scheme relates to some game where:

1. You look for a particular situation to occur (a baseball team playing on home turf... or rain during a soccer match)
2. You bet on a particular result (the home team wins ... or an English team is playing in that rain)
3. You make a profit of \$y for a \$1 bet, each time that particular result occurs.
4. Of all situations descibed by #1 (above), the fraction of times when that particular result occurs is x.
>You mean it's not just for horse racing bets?
I don't think so ... but remember that I'm the president of the IDK club.

Today it's betting on a Top Jockey riding a longshot. Tomorrow? I don't know.

>That's 'cause your president of the I Don't Know club.

 Exactor / Exacta

 The Exacta or Exactor bet is where you pick a horse to Win and a horse to Place. That is, you pick two horses to finish in the first and second place positions in an exact order. The payouts can be quite respectable. For example, on July 1, 2012, at Woodbine >Wow! A Payout of \$3296, for a \$2 bet? Nice, eh? The winner paid \$57 to Win and the 2nd place horse paid \$81.40 to Place. If'n longshots come 1st and 2nd, you make big \$money!
In the spreadsheet, if I use the ML-odds to estimate the Exactor payouts, I get these:

The Exactor pair which paid \$3296 (11 to Win and 5 to Place) is shown in blue.

>But it's not \$3296.
Aah, but ML odds are guesstimates ... not actual post-time odds. Nevertheless, the Exactor Matrix shown does give some idea of the size of payouts.

>And that's now in the spreadsheet?
Yes ... however remember that calculating Exactor Payouts from ML-odds is pure guesswork.

Nevertheless, after extansive cogitation and cerebral machination, I figure the best strategy is this:

1. Pick a favourite for the first half of the Exactor (the one who you think will Win).
2. Pick a longshot for the 2nd half (the horse you think will Place).
The reason for picking a longshot (for Place) is to make the payout respectable.
Check out the following:

Note the following:
• In Races 1, 2, 3, 7, 8 and 9 a favourite came in first
• In Races 1 and 2 a longshot came in second
• Betting \$2 in each race we'd have paid out \$20 and got back \$30.90 + \$7.90 = \$38.80.

>And if you'd have bet \$200 on each ...
Yes, yes ... I'd have paid our \$2000 and got back \$3880. But what if the horses were all favourites and I didn't win anything? Or what if ...

 Exactor Odds

We might imagine that if a horse's odds of winning go UP, the probability of that horse winning goes DOWN.
Let's assume that the probability of a horse winning is proportional to 1/oddsk:
[1]       Pj = λ/Ojk   with 0 < Pj < 1 and λ is some constant.
The sum of all the probabilities must equal 1 (since one horse WILL win), so:
ΣPj = λ Σ 1/Ojk = 1   so   λ = 1 / Σ 1/Ojk   where Σ means you add 'em all up.

Okay, now suppose that, in 100 races, horse#m wins x times (so x is the expected percentage of wins). We set
Pm = x/100 so 0 < Pm < 1. In the remaining 100 - x races, horse#m loses.

Now consider the number of times another horse#n wins in those 100 - x races. We'll assume it's y, so the probability of horse#n winning is y/(100 - x).
Then the probability of horse#m winning in the remaining races is:
y/(100 - x) = y / (100 - 100 Pm) = ( y/100 ) / (1 - Pm) = Pn / (1 - Pm).

Okay, so we'll set the probability of horse#m winning (that's Pm) and horse#n coming in 2nd (we put that as Pn / (1 - Pm)) as:
 [A]   Probability of an m-n Exactor is Pm Pn / (1 - Pm)     This is Harville's formula

Now stick in the stuff from relation [1] and get (after inverting to get the "odds" which we assume to be 1/probability)):
 [B]   Odds for an m-n Exactor is: (Onk/λ) (Omk/λ - 1)       where Oj are the odds on horse #j, λ = 1 / Σ 1/Ojk and k is some constant.

>I think that's pretty fuzzy math.
Uh ... yes, but Harville wanted desperately to get a probability based only upon winning percentages, like Pm and Pn.
 For example, suppose we assume k = 2 so the probability of winning is proportional to 1/odds2     then the odds for an Exactor, given the odds of winning for each of 8 horses, is >Huh? Look at the Exactor where horse #1 wins and horse #7 comes in 2nd. Their odds of winning are 3 to 1 and 2.5 to 1 (or 5 to 2). That Exactor would have odds of 0.3 so (presumably) a \$2 bet would provide a payout of \$2.60 since payout = 2*odds+2.

>And you believe that?
Well, of course. After all, we haven't made any unreasonable assumptions have we? Just Harville's formula and probability = λ/Oddsk and ...

>You're dreamin'
Yes. In fact, the actual Exactor payout (using the real-live numbers after the race had run) was (gasp!) \$33.50.
However, in support of our fuzzy theory, had we taken k = 1 (so probability is proportional to 1/odds), we'd have predicted a payout of \$30.70.
Remember that we're predicting based upon the ML odds which are guesstimates.

 Okay, let's look at the Exactor Payouts for a couple of hundred races (at Woodbine). If there are N horses in a race then there are N (N-1) possible Exactors. If'n you bet 'em all (at \$2 each), it'd cost \$2 N (N-1). How does the \$cost of the bet depend upon the number of horses? See the table It says (for example) that over 269 races, if you bet on every possible exactor for 6-horse races, your average payout would have been \$40. Alas, for 6 horses, it'd cost \$60 to make all 30 bets. Aaah, but suppose we could justify eliminating one long,long shot. Then, with just 5 horses, there'd be just 20 possible Exactors. A possible strategy, right? >And how would that work? I have no idea.

Somewhere, we talked about an Exactor payout matrix based upon the ML Odds for each horse.

Figure 3: Exactor probability matrix
 We could do this: Order the Exactor probabilities from largest to smallest. Place a bet on the largest, then the second largest, then the third largest etc. etc. Stop when the cumulative probability exceeds some value, like 65%. We ignore low probability Exactors. For example, in Figure 3, we include only probabilities greater than 1% (shown in blue). Sometimes we can't get a cumulative probability greater than, say, 65% ... then we don't bet. In Figure 4 we bet on the Exactor (6-4) which has a 4.1% probability, and also on (4-6) with 3.9% probability etc. etc. We get to 65% cumulative probability after 23 Exactor bets. (See Figure 4.) That'd cost us \$2*23 = \$46. In that particular race, the Exactor that actually came in was (5-6) ... shown in blue, in Figure 4. It paid (gasp!) \$235 for a \$2 bet. That was race#6 on July 27. >Did you lose money on July 27? Uh ... yes. Did I say this strategy was foolproof? Ya win some, ya lose some. Figure 4: Ordered probabilities

Note that we're using the ML Odds (which are guesstimates) and Harville's formula (fuzzy math) ... and we pray a lot.

>So what about historical results, for hundreds of races?
Patience.