Greyhounds & Bayes?

I think this should be an easy Bayes Theorem calculation but I'm struggling...

1. Over the last three years the probability that Trap 1 wins is 17.40% (6 traps) based on actual results.
2. Over the last three years the probability that that the favourite wins is 32.48% based on actual results.

What is the probability that Trap 1 wins when it's the favourite?

Thanks in advance

I hope you can get an answer soon.
I think there are only about 5 tracks still active in the US, and one of them is scheduled to close in about 5 weeks.

I'm in the UK! About 20 active tracks with over a 100 races a day on average!

My bad! I thought this was a question about the bus company.

I probably should have guessed UK from 6 traps.
I think I still saw 8 traps recently in the US.

(And after May 15, it looks like we'll be down to two tracks in West Virginia. No idea if Southland is already closed.)

If I understand two-condition Bayes correctly:
P(C given both A and B) = P(A given both B and C) x P(C given B) / P (A given B)
In this case, A is "the dog is in Trap 1," B is "the dog is the favorite," and C is "the dog wins"
I also assume that, in dog racing, traps are assigned randomly, so P(A) = 1/6 under all conditions.
Therefore:
P(dog wins, given Trap 1) = P(dog is in Trap 1, given it is the favorite and wins) x P(dog wins, given it is the favorite) / P(dog is in Trap 1, given it is the favorite) = 1/6 x 32.48% / (1/6) = 32.48%

Something about this doesn't pass the smell test - mainly, the fact that Trap 1 wins 17.4% of the time isn't used; I am substituting 1/6 for this. Since the probability that the favorite is in Trap 1 is not given, I don't think you can solve it "as is."

Isn't trap 1 closest to the inside rail or the hare or something?
It might have a slight positional advantage, making the number not 1/6.

Quote: ThatDonGuy

If I understand two-condition Bayes correctly:
P(C given both A and B) = P(A given both B and C) x P(C given B) / P (A given B)
In this case, A is "the dog is in Trap 1," B is "the dog is the favorite," and C is "the dog wins"
I also assume that, in dog racing, traps are assigned randomly, so P(A) = 1/6 under all conditions.
Therefore:
P(dog wins, given Trap 1) = P(dog is in Trap 1, given it is the favorite and wins) x P(dog wins, given it is the favorite) / P(dog is in Trap 1, given it is the favorite) = 1/6 x 32.48% / (1/6) = 32.48%

Something about this doesn't pass the smell test - mainly, the fact that Trap 1 wins 17.4% of the time isn't used; I am substituting 1/6 for this. Since the probability that the favorite is in Trap 1 is not given, I don't think you can solve it "as is."
link to original post

Thanks, looking at the stats - the probability that Trap 1 is favourite is 22.1%...

Quote: Dieter

Isn't trap 1 closest to the inside rail or the hare or something?
It might have a slight positional advantage, making the number not 1/6.
link to original post

True - it is kind of like saying, "If you run a 400m race, but lane 8 only has to run 300m, and the lanes are assigned randomly, the probability of someone being in lane 8 given that they won is 1/8."

In any case, it seems that any solution requires that we know the percentage of the time the favorite is in Trap 1.
There appear to be two ways to solve this:

P(dog wins, given it is the favorite and in Trap 1) = P(dog is in Trap 1, given it is the favorite and wins) x P(dog wins, given it is the favorite) / P(dog is in Trap 1, given it is the favorite)

or:

P(dog wins, given it is the favorite and in Trap 1) = P(dog is the favorite, given it is in Trap 1 and wins) x P(dog wins, given it is in Trap 1) / P(dog is the favorite, given it is in Trap 1)

In either case, there is not enough information to solve it. You might as well just check every race and count how many times the favorite was in Trap 1 and won the race.

Thanks so much, I was trying to apply simple Bayes, not multiple condition Bayes.

Just checking you saw my post - The probability of Trap 1 being the favourite is 22.1%

Quote: jedijon

Thanks so much, I was trying to apply simple Bayes, not multiple condition Bayes.

Just checking you saw my post - The probability of Trap 1 being the favourite is 22.1%
link to original post

The problem is you don’t have two independent variables. Being in trap 1 is advantageous. Being the favorite is a mix of trap position and dog speed/acceleration/stamina/etc.

22.1% is greater than 1/6 illustrates that.

Sounds like you have the data to get the calculation you want based on actuals anyway?

Quote: jedijon

I think this should be an easy Bayes Theorem calculation but I'm struggling...

1. Over the last three years the probability that Trap 1 wins is 17.40% (6 traps) based on actual results.
2. Over the last three years the probability that that the favourite wins is 32.48% based on actual results.

What is the probability that Trap 1 wins when it's the favourite?

Thanks in advance
link to original post

I don't think there is enough information here to answer the question.

Apologies, I'm getting my data from the below link and am compiling both Track & Favourite stats for 2019 + 2020 +2021

Using 'Central Park' as an example:

Trap 1 has won 1449 races out of 8329 (17.4%)
2649 favourites have won out of a total number of 7565 favourites (35%)
(I assume 7565 < 8329 is due to joint favourites)
Trap 1 has been favourite 1526 times (20.17%)
When favourite, Trap 1 has won 501 times (32.83%)

I want to know, do we have enough information and is it possible to calculate the "expected probability" of Trap 1 winning when it's favourite, to compare to the actual result of 32.83%?

I am not concerned about trap bias or odds for now, just purely on the stats.

Thanks all for your help and input on this.

Can't post the link until i get to 20 posts... it's greyhoundstats dot co dot uk

Interesting thread! Thanks. All we need now is Chump Changes view on the subject!

Quote: jedijon

Can't post the link until i get to 20 posts... it's greyhoundstats dot co dot uk
link to original post

https://greyhoundstats.co.uk/

Looks OK to me.

Quote: jedijon

I want to know, do we have enough information and is it possible to calculate the "expected probability" of Trap 1 winning when it's favourite, to compare to the actual result of 32.83%?

I am not concerned about trap bias or odds for now, just purely on the stats.

Thanks all for your help and input on this.

The website has all of the necessary information. However, when you boil it down with Bayes, you get this:

P(win, given favorite and in Trap 1) = the number of times the winner was both the favorite and in Trap 1 / the number of times the favorite was in Trap 1.

In other words, most of the numbers cancel each other out, and you end up with the actual result.

Quote: Dieter

Quote: jedijon

Can't post the link until i get to 20 posts... it's greyhoundstats dot co dot uk
link to original post

/

Looks OK to me.
link to original post

Thanks, it's because this website prevents me from posting links as I am yet to make 20 posts across all forums

P(win, given favorite and in Trap 1) = the number of times the winner was both the favorite and in Trap 1 / the number of times the favorite was in Trap 1.

In other words, most of the numbers cancel each other out, and you end up with the actual result.
link to original post

Thanks I understand the 'actual' result - very simple calculation and easy to get from the data.

I just wanted to know, and maybe it's not possible, but assuming you didn't have the 'the number of times the winner was both the favorite and in Trap 1' information (501), is it possible to calculate the expected result?

I applied the (second) formula you mentioned in an earlier post:

P(dog wins, given it is the favorite and in Trap 1) = P(dog is the favorite, given it is in Trap 1 and wins) x P(dog wins, given it is in Trap 1) / P(dog is the favorite, given it is in Trap 1):

Using:

Trap 1 has won 1449 races out of 8329 (17.4%)
2649 favourites have won out of a total number of 7565 favourites (35%)
(I assume 7565 < 8329 is due to joint favourites)
Trap 1 has been favourite 1526 times (20.17%)
When favourite, Trap 1 has won 501 times (32.83%)

(501 / 1449) * (1449 / 8329) / (1526 / 7565) = 29.82%

I guess the problem here is I'm using 501 which I'm .not supposed to know' given my objective.

Is the first part of the equation a Bayes in itself to solve? (P(dog is the favorite, given it is in Trap 1 and wins)

Thanks once again, really appreciate your time/input on this.

Quote: jedijon

I applied the (second) formula you mentioned in an earlier post:

P(dog wins, given it is the favorite and in Trap 1) = P(dog is the favorite, given it is in Trap 1 and wins) x P(dog wins, given it is in Trap 1) / P(dog is the favorite, given it is in Trap 1):

Using:

Trap 1 has won 1449 races out of 8329 (17.4%)
2649 favourites have won out of a total number of 7565 favourites (35%)
(I assume 7565 < 8329 is due to joint favourites)
Trap 1 has been favourite 1526 times (20.17%)
When favourite, Trap 1 has won 501 times (32.83%)

(501 / 1449) * (1449 / 8329) / (1526 / 7565) = 29.82%

You're close, but you are using 7565 instead of 8939 for the number of times a dog was in Trap 1 in the third part; it is the number of races thta had a dog in Trap 1, not the number of times a race had a single favorite.
(501 / 1449) * (1449 / 8329) / (1526 / 8639) = 501 / 1526.
Again, the numbers cancel each other out, and you end up with "the number of races a dog in Trap 1 was the favorite and won" divided by "the number of races where there was a dog in Trap 1".

To use Bayes properly in this instance, you are going to have to know in advance how many times a dog that was in Trap 1 and the favorite won, which is what you are trying to figure out in the first place.

Ah ok, I understand. Thanks for your help with this.

Quote: ThatDonGuy

(snip)
To use Bayes properly in this instance, you are going to have to know in advance how many times a dog that was in Trap 1 and the favorite won, which is what you are trying to figure out in the first place.
link to original post

If I read this correctly, then I think the quotes below may be an example of the type of information needed ?

" . 54% of all winners from box 1 were the favorities (sic) "
and possibly (?)
" .38% of all Wentworth Park TAB favourites over the past 3 years won their races..."
". Of the eight boxes allocated for each greyhound race*, Box One (red) generated the most winners with 18.5%...."

Note 1: Box = Trap = Bay
Note 2: This is just for one track (an Australian track) and there are up to 8 dogs per race (instead of 6 in the UK?).
Note 3: Data is "based on 3650 races between June 2011 - August 2014".
Note 4: Scroll down to "Statistics" in the link below to see the full quotes/ stat info

http://www.wentworthpark.com.au/racing/how-to-bet-on-greyhounds