unJon
unJon
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April 16th, 2022 at 10:39:06 AM permalink
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 donít 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?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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April 16th, 2022 at 1:53:28 PM permalink
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.
It's not whether you win or lose; it's whether or not you had a good bet.
jedijon
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April 16th, 2022 at 3:26:58 PM permalink
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
DeMango
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April 16th, 2022 at 3:49:00 PM permalink
Interesting thread! Thanks. All we need now is Chump Changes view on the subject!
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
Dieter
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April 16th, 2022 at 4:39:41 PM permalink
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.
May the cards fall in your favor.
ThatDonGuy
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April 16th, 2022 at 4:48:16 PM permalink
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.
jedijon
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April 16th, 2022 at 11:28:59 PM permalink
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
jedijon
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April 16th, 2022 at 11:52:14 PM permalink
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.
Last edited by: jedijon on Apr 17, 2022
ThatDonGuy
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April 17th, 2022 at 7:29:12 AM permalink
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.
jedijon
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April 17th, 2022 at 12:07:57 PM permalink
Ah ok, I understand. Thanks for your help with this.

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