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09-03-2017 , 05:17 AM

ThePressure

adept

Join Date: Jul 2008 Posts: 988

Hi,

I was wondering if someone could help me, I'm trying to work out how I do the calculations and probabilities for the below scenarios, any help would be great.

In a 6 dog race, what is the probability any given dog would finish in the top 2?
6 dog race, what is the probability with 2 selections, one will finish in top 2, and both finish in top 2?

same with 3 and 4 selections, chances of picking the winner, and chances of picking 1st and 2nd?

thanks, can't find any information about this online, I thought it would be easy to find.

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09-03-2017 , 06:46 AM

md46135

grinder

Join Date: Mar 2015 Posts: 420

assuming each dog has 1/6 probability of winning and 1/6 for any other place:

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In a 6 dog race, what is the probability any given dog would finish in the top 2?

2/6 = 33.33%

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6 dog race, what is the probability with 2 selections, one will finish in top 2, and both finish in top 2?

only one in top 2: 2C1*4C1/6C2 = 8/15 = 53.33%
both in top 2: 2C2/6C2 = 1/15 = 6.67%
both outside top2 : 4C2/6C2 = 2/5 = 40%

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same with 3 and 4 selections, chances of picking the winner, and chances of picking 1st and 2nd?

if you choose 3 selections:
you have both dogs in top 2: 2C2*4C1/6C3 = 1/5 = 20%
only one dog in top 2: 2C1*4C2/6C3 = 3/5 = 60%
no dogs in top 2: 4C3/6C3 = 1/5 = 20%

with 4 selections:
top 2 covered: 2C2*4C2/6C4 = 2/5 = 40%
only 1 dog in top 2: 2C1*4C3/6C4 = 8/15 = 53.33%
no dogs in top 2: 4C4/6C4 = 1/15 = 6.67%

Last edited by md46135; 09-03-2017 at 07:08 AM.

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09-03-2017 , 11:19 AM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

For those who rather see direct probability calculations: Assume pick is A and B

One in top 2: P(A in top 2)*P(B in bottom 4 given A in top 2) * 2 = 2/6*4/5*2=16/30 = 0.5333
(2 multiplier for the case B in top 2)

Both in top 2: P(A in top 2)*P(B in top 2 given A in top 2) = 2/6*1/5 = 2/30 = 0.0667

Neither in top 2: P(A in bottom 4)*P(B in bottom 4 given A in bottom 4) = 4/6*3/5 = 12/30 =0.400

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09-03-2017 , 11:50 AM

ThePressure

adept

Join Date: Jul 2008 Posts: 988

both excellent responses thanks! what does all the 2C2 stuff mean?

also being greedy, what are the odds of picking a winner if you back 4 selections, 4/6 ? 66%? but thats more likely than having the top 2 covered from 4 selections? that doesn't make sense?

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09-03-2017 , 12:33 PM

Didace

Carpal \'Tunnel

Join Date: Nov 2009 Posts: 19,878

Quote:

Originally Posted by md46135

assuming each dog has 1/6 probability of winning and 1/6 for any other place

Don't overlook this. It is not true in the real world.

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09-03-2017 , 02:27 PM

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

Quote:

Originally Posted by ThePressure

Why?

Requiring 2 of 4 to be in certain positions is more restrictive (=generally less likely) than requiring only 1 of 4.

2C2 is math shorthand for the number of combinations (ways) of selecting 2 things out of 2 things. In general

nCr = C(n,r) = n Choose r (Google nomenclature) = n! /(r!*(n-r)!)

where x! = x*(x-1)*(x-2)* … *3*2*1

C(5,2) =5!/(3!*2!)= 5*4/(2*1)= 10.

Search for math combinations on the internet for more detail.

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09-06-2017 , 01:11 AM

akkopower1

centurion

Join Date: Feb 2017 Posts: 198

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Originally Posted by ThePressure

2C2 measn how many ways can you choose two objects from two objects, here the order is nit relevant.

If there are two drinks in the fridge there is only one way you can choose two drinks. Take them both!!.

3C2, means if there are 3 drinks in the fridge (drinks A,B and C), how many ways can you choose 2.
You can take; AB, AC, BC, that is three different ways

8C3 means if there are 8 drinks how many ways can i choose 3, it would be 56 different combinations

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09-07-2017 , 09:13 AM

davmcg

veteran

Join Date: Jan 2003 Posts: 2,535

Quote:

Originally Posted by Didace

Don't overlook this. It is not true in the real world.

This.

Because the odds are all different, you would be better designing a spreadsheet (that gives the odds of all 30 1st/2nd permutations - dog A first, dog B first of the remaining 5 runners, dog A first, dog C first of the remaining 5 etc) and inputting actual true odds.

The information is also basically useless for dogs as in many cases the probability of two dogs running well can be related ie they can interfere with each other (or not). Also the odds of the bookmakers on dog races are very poor value (although pooled bets can always offer opportunities).

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