Math Question

[Robert Goren]

I have a question I have thinking about for years. When a horse scratches lot of a race, does another horse chances of winning increase proportionately to his chances winning in regard with other horses chances of winning or do does his chances of not winning decrease with regard his chances of not winning compared the other horses chances of not winning proportionately. I know that is as clear as mud. The only thing I can say they are not the same thing and will get different answers depending on which way you calculate it.

[OTM Al]

I have a question I have thinking about for years. When a horse scratches lot of a race, does another horse chances of winning increase proportionately to his chances winning in regard with other horses chances of winning or do does his chances of not winning decrease with regard his chances of not winning compared the other horses chances of not winning proportionately. I know that is as clear as mud. The only thing I can say they are not the same thing and will get different answers depending on which way you calculate it.

Do you mean actual chance of winning? It could go either way. Here's a simple case. If there were only 2 speed horses in the race and one scratches, the other has its chance of winning go up and the chance of others winning would go down. Scratch a horse with no chance and no pace, it probably changes nothing.

[Jeff P]

When one or more horses scratch from a race, the probabilities for all of the other horses change.

The degree to which the prob values change for each horse depends on which horse(s) scratch out of the race.

The best analogy I can think of to explain it in a straightforward way is that each time a horse is removed, the makeup of the race changes - essentially making the race at hand a new race.

-jp

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[Overlay]

For me, the removal of the scratched horse will change the relative probability and position (in comparison to the rest of the field) of all horses below it with regard to the various ranking-type factors that I use, as well as the overall composite probability on which I base the fair odds of every other horse. So I figure a whole new line when a scratch occurs.

[Some_One]

When one or more horses scratch from a race, the probabilities for all of the other horses change.

The degree to which the prob values change for each horse depends on which horse(s) scratch out of the race.

The best analogy I can think of is that each time a horse is removed, the makeup of the race is changed so that the race at hand now becomes a new race.

-jp

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Great Answer, totally agree. If a speed horse scratches out of a race, the E types probability of winning probably went up proportionally more than the closers in the race.

[Cratos]

I have a question I have thinking about for years. When a horse scratches lot of a race, does another horse chances of winning increase proportionately to his chances winning in regard with other horses chances of winning or do does his chances of not winning decrease with regard his chances of not winning compared the other horses chances of not winning proportionately. I know that is as clear as mud. The only thing I can say they are not the same thing and will get different answers depending on which way you calculate it.

Before and after the scratched horse is removed from the race we can in such cases define random probability subjectively as a measure of strength of belief.

Therefore we could conclude randomly that the winning probability of each horse left in the race after the scratched horse is removed gets better because n the size of the probability set has gotten smaller.

For example a race with 10 horses in it gives each horse a random probability of 1/10 chance of winning, but with 9 horses left in the race, the random probability of each horse winning the race becomes a 1/9 chance of winning.

However the conditional probability calculation is different because the conditional probability is the probability of one event if another event occurred.

Therefore in our example of the horse being scratched we can conclude that the conditional probability of Horse A winning is X given that Horse B is scratched.

In other words, the conditional probability of Horse A winning given Horse B is scratched is the probability of event A, if event B occurred; and if Horse A is independent of Horse B the calculation changes.

[Robert Goren]

I knew I had a problem explainlng it. Leaving aside such things as pace scenarios let look at it this way. a 4 horse race
:1: 40 % win
:2: 30% win
:3: 20% win
:4: 10% win
Now the :3: horse scratches What are the new chances? Is :1: chances now 50% gotten by taking :1: 's chances divided by the total of chances left or 40/80? :2: 30/80 or 37.5% :4: 10/80 or 12.5%. Or do you look at it by prorating the gain in win % by each horse's chances of not winning (1-Win%). In :1: 40%+[20%*(60/220)] or 45.4% :2: 30% +[20%*(70/220)] or 36.4% :4: 10% +[20%*(90/220)] or 18.2% The 220 is the sum of the :1:'s ,:2:'s, and :4:'s (1-win%) numbers. Which way is the correct math way of looking at. I have been pondering this for years and am leaning toward the later method.

[Greyfox]

In general, when a horse scratches, every horse in the race's chance increases.
However, the scratched horse impacts the race dynamic.
For example, if the horse that scratched was a pure early speed type and would have won the early pace battle, had it stayed in the race any other need to lead horses would have been toasted.
With that horse out of the race, if there was another need to lead horse who now will be the clear speed, his chances are enhanced in the new dynamic.
Without the scratch, this horse would have likely finished up the track.
In the new dynamic, it might win.
Every race, has to be examined very carefully with respect to the race dynamics.
So a Pick 3 player for example, might not play the same picks, if he knew that in the final leg a particular type of runner was going to be scratched just before the gate.
But, in general, a scratched horse, increases everyone else's chance and at the same time the tote board odds lower accordingly.

[Overlay]

I knew I had a problem explainlng it. Leaving aside such things as pace scenarios let look at it this way. a 4 horse race
:1: 40 % win
:2: 30% win
:3: 20% win
:4: 10% win
Now the :3: horse scratches What are the new chances? Is :1: chances now 50% gotten by taking :1: 's chances divided by the total of chances left or 40/80? :2: 30/80 or 37.5% :4: 10/80 or 12.5%. Or do you look at it by prorating the gain in win % by each horse's chances of not winning (1-Win%). In :1: 40%+[20%*(60/220)] or 45.4% :2: 30% +[20%*(70/220)] or 36.4% :4: 10% +[20%*(90/220)] or 18.2% The 220 is the sum of the :1:'s ,:2:'s, and :4:'s (1-win%) numbers. Which way is the correct math way of looking at. I have been pondering this for years and am leaning toward the later method.
If the outcome of the race were a random event (which it is not), and the chances of the horses were independent of each other (which they are also not), I would say that your first calculation would be correct, redistributing a 100% probability of winning among the three remaining horses in proportion to their relative chances prior to the scratch.

[Ray2000]

I believe Overlay is correct

Apply the math to an equal chance match race to see the fallacy of summing 'not winning' chances

[TrifectaMike]

I knew I had a problem explainlng it. Leaving aside such things as pace scenarios let look at it this way. a 4 horse race
:1: 40 % win
:2: 30% win
:3: 20% win
:4: 10% win
Now the :3: horse scratches What are the new chances? Is :1: chances now 50% gotten by taking :1: 's chances divided by the total of chances left or 40/80? :2: 30/80 or 37.5% :4: 10/80 or 12.5%. Or do you look at it by prorating the gain in win % by each horse's chances of not winning (1-Win%). In :1: 40%+[20%*(60/220)] or 45.4% :2: 30% +[20%*(70/220)] or 36.4% :4: 10% +[20%*(90/220)] or 18.2% The 220 is the sum of the :1:'s ,:2:'s, and :4:'s (1-win%) numbers. Which way is the correct math way of looking at. I have been pondering this for years and am leaning toward the later method.

Neither approach is correct nor is either approach incorrect under some narrow circumstances. I believe JP and Cratos have stated the correct approach.

In just words, I would say how you arrived at the initial percentages has to be re-evaluated. You have navigated a path to the percentages in some logical framework. If a horse scratches, the path has "changed". You now have to determine a "new" path" using the same framework. What the new path is, is a function of the particular framework employed. Scaling in most cases isn't correct.

Let's look at the problem in this way. Let's assume you have a super metric that encompasses all you will ever need to correctly predict the percentages. There will exist a joint probability density function for that metric for all the horses in the race. If you were to eliminate (scratch) a horse, the shape of the probability density function will change. The extent or manner in which it changes is a function of the horse that scratched and the remaining horses. Scaling in most cases would be incorrect. In other words new information has been introduced (scratched horse) and is a case of conditioning. The "new" solution is conditional on new information.

Mike (Dr Beav)

[HUSKER55]

QUESTION:

if horse A has a 40% chance of winning in a field of 4 which is 25% and you eliminate horseC then your horse still has a 40% chance except the field is now33%.

so isn't it ((.4+.25)/.25) or about 1.6

now it is ((.4+.33)/.33) or about 1.2

your chances on horse A improved?

[TrifectaMike]

I have a three horse field, Horse A (100%), Horse B (0%), Horse C (0%). Horse A is scratched. What are the new percentages assigned to Horse B and Horse C?

Mike (Dr Beav)

[thaskalos]

I knew I had a problem explainlng it. Leaving aside such things as pace scenarios let look at it this way. a 4 horse race
:1: 40 % win
:2: 30% win
:3: 20% win
:4: 10% win
Now the :3: horse scratches What are the new chances? Is :1: chances now 50% gotten by taking :1: 's chances divided by the total of chances left or 40/80? :2: 30/80 or 37.5% :4: 10/80 or 12.5%. Or do you look at it by prorating the gain in win % by each horse's chances of not winning (1-Win%). In :1: 40%+[20%*(60/220)] or 45.4% :2: 30% +[20%*(70/220)] or 36.4% :4: 10% +[20%*(90/220)] or 18.2% The 220 is the sum of the :1:'s ,:2:'s, and :4:'s (1-win%) numbers. Which way is the correct math way of looking at. I have been pondering this for years and am leaning toward the later method.
It is your ORIGINAL premise that is faulty, IMO.

You cannot "leave aside" an important concept such as pace...and still claim to possess a realistic view of the probable win percentages of the field.

[HUSKER55]

MIKE,

In your example, would that race be a pass as you would not have any edge?


Roll the dice would not be unreasonable in that scenerio, would it?

[Tom]

I think you need to recalculate each horse's probability of winning the new race. The original numbers were their probability of winning the match up with the 3. It is a new race now.

[thaskalos]

MIKE,

In your example, would that race be a pass as you would not have any edge?


Roll the dice would not be unreasonable in that scenerio, would it?
Ahhh...but there IS an edge...ESPECIALLY in 2-horse fields.

It has been determined that, in "match race" situations, the horse which goes to the lead wins about 80% of the time...regardless of distance.

[Ocala Mike]

I have a three horse field, Horse A (100%), Horse B (0%), Horse C (0%). Horse A is scratched. What are the new percentages assigned to Horse B and Horse C?

Mike (Dr Beav)


You would need to refer this special case to quantum physicists who would take you to task for assigning 0% to begin with, which is impossible in their world.


Ocala Mike

[Robert Goren]

Since I want answers not connected to what happens in the horse race lets move it move it from a horse race to say a golf tournament or a car race. Now which way is correct?

[HUSKER55]

OK thaskalos, I stand corrected. thank you:)

[Cratos]

You would need to refer this special case to quantum physicists who would take you to task for assigning 0% to begin with, which is impossible in their world.


Ocala Mike

Yes, in a world of subjectivity I will agree that 0% probability and 100% are impossible, but Mike raises a very good theoretical point and that is if Horse A has a theoretical probability of 100% to win and is scratched leaving Horses B and C in the race and each had a theoretical 0% chance of winning before the Horse A was scratched it would be reasonable to assumed that their probabilities each would randomly be 50% each to win. If conditional information came forward then the probabilities of each horse might change.

The point here is that Horses B and C had equal probabilities before Horse A was scratched and since the probability set must add to 100% then the two remaining horses would share equally in the chance to win unless there was some conditional information made available after Horse A was scratched.

[stu]

Since I want answers not connected to what happens in the horse race lets move it move it from a horse race to say a golf tournament or a car race. Now which way is correct?


Robert, wouldn't there be an interdepence among contestants regardless of the event?

Tony Stewart plans to keep wrecking drivers who block him, regardless of consequences (http://www.washingtonpost.com/sports/othersports/autoracing/tony-stewart-plans-to-keep-wrecking-drivers-who-block-him-regardless-of-consequences/2011/07/01/AGqEN0tH_story.html) e.g. Brian Vickers has little to no chance of winning if Tony Stewart is in the race but has a finite chance of winning if Tony Stewart is out.

Or in golf, Tiger Woods has often intimidated whomever he is paired with on day 4. If Tiger withdrawls, everybody's chances go up but not as much as guy who who would have been paired with tiger.

In short, all of the variables are interdependent and can't be easily be deconvoluted if one participant is withdrawn.

[openhorse]

thought experiment:
Lets say your betting the Pick 4 and they scratch a horse 3 races ahead and you want to project what will happen to the odds. I think this tool solves that problem - ? (please see pdfs attached : before/after scratch )

Let's assume you have a super metric that encompasses all you will ever need to correctly predict the percentages.

For these reports, call this metric the 'Handicap_Index'.
And... you have high confidence because you ran a betting simulation. results: 53% wins, +10% roi. so you can rely on your odds and have faith.


There will exist a joint probability density function for that metric for all the horses in the race.

a Joint distribution maybe, 5x5: spd,race,pace,trk-profile,connections x Clm, Mdn, Alw, Hcp, Stk



The extent or manner in which it changes is a function of the horse that scratched and the remaining horses.

I convert odds back to probability to see the change easier.
In this test case the sort order of the horses doesnt change - but i think it is possible based on conditions such as IV's

Right click and open in a new window to compare side-by-side.
We scratch the 3 horse to see how it skews the odds and probabilities.

[Robert Goren]

I am about to pull my hair out. This is a math question. Move it to slot machines or what ever you want where the dynamic is not effected by the withdrawal of entrant. Please, I do not want lectures on pace scenarios or whatever but a simple math answer.

[stu]

http://wizardofodds.com/slots

In the linked odds table, blank-blank-blank happens 12.5% of the time (or one-eighth of the time).

If you could guarantee that blank-blank-blank would not happen, then the probability of any other outcome is increased proportionately from 1 out of x to 1 out of seven-eighths of x.

So the new odds are 8/7ths of the original odds.

[Red Knave]

Or do you look at it by prorating the gain in win % by each horse's chances of not winning (1-Win%). In :1: 40%+[20%*(60/220)] or 45.4% :2: 30% +[20%*(70/220)] or 36.4% :4: 10% +[20%*(90/220)] or 18.2% The 220 is the sum of the :1:'s ,:2:'s, and :4:'s (1-win%) numbers. Which way is the correct math way of looking at. I have been pondering this for years and am leaning toward the later method.
I think that your 'not winning' calculation is incorrect because you are not using the correct win% any more. If the 3 horses left are the only remaining horses then the win% must still be recalculated to 100%. You are using their win% from a 4 horse race which will not be run. In essence those numbers, (40%, 30%, 10%) are just random numbers now.
FWIW, I think the 'simple' answer is your first solution; 50%, 37.5%, 12.5%. All the other answers are more correct with regards to a horse race, though.

[ALL CIRCUITS GO]

the way I see it:

Most state lotteries have a daily number drawing where a random number between 000 and 999 is generated. The odds of any individual 3 digit number coming out is 1000-1.

A state uses ping pong balls numbered 0-9 in 3 separate containers, drawing a ball from each container to get the number. What if a ball number '0' is removed from the last container? The odds of getting any individual 3 digit number not containing '0' as its final digit is now 900-1 whereas the odds of getting any 3 digit number where the final digit is '0' is now impossible.

If there are 10 horses in a race, then the random chance of any horse winning the race is 1 in 10 or a 10% chance of winning. Removing 1 horse from the race, each horse now has a 1 in 9 chance of winning or an 11% chance of winning. If there are only 5 horses, then each has a 1 in 5 chance of winning, or 20%. Taken to the extreme, what if the field were narrowed to only 2 entries? Then each has a 1 in 2, or 50% chance of winning.

So I think that removing an entrant from a race increases the chances of any other entrant winning, evenly for each.

What if we went to the track and were told there were 10 horses ready to run, but we didn't know their pps, nor could we view the horses until the race started. We would have no information to assign our odds value other than randomness. If all bettors had the same limited information, would the market efficiency of the betting public be evident from the tote board? Would all entrants have odds of 10-1? What if we saw odds of 8-1 on an entrant? Would we assume smart money was in play?

[Robert Goren]

What it is not a state lottery, but a weighted lottery like the NBA draft lottery after one team has been eliminated? No conspiracy theories please.

[pondman]

I knew I had a problem explainlng it. Leaving aside such things as pace scenarios let look at it this way. a 4 horse race
:1: 40 % win
:2: 30% win
:3: 20% win
:4: 10% win
Now the :3: horse scratches What are the new chances? Is :1: chances now 50% gotten by taking :1: 's chances divided by the total of chances left or 40/80? :2: 30/80 or 37.5% :4: 10/80 or 12.5%. Or do you look at it by prorating the gain in win % by each horse's chances of not winning (1-Win%). In :1: 40%+[20%*(60/220)] or 45.4% :2: 30% +[20%*(70/220)] or 36.4% :4: 10% +[20%*(90/220)] or 18.2% The 220 is the sum of the :1:'s ,:2:'s, and :4:'s (1-win%) numbers. Which way is the correct math way of looking at. I have been pondering this for years and am leaning toward the later method.

Your initial percentages are based on experience over time-- the grand daddy of all perfect knowledge. They themselves are an average over time, and I would think, be skewed and oddly distributed, within each of these % clusters (this type of horse, winning this type of race.) In other words, you've group thses horses for a reason--which in itself might include 2 or 3 variables. Therefore, the 40% cluster would have a different curve (distribution) and characteristics than the 10% cluster (I think.)


You aren't going to get neat and clean calculations out of this.


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