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 February 20th, 2013, 04:21 AM #1 Newbie   Joined: Feb 2013 Posts: 2 Thanks: 0 probability Hello, I would like that you check my solution is it ok. There are dogs racing - 6 dogs numbered [1, 2 ... 6] What is the probability of first dog being in 1 or 2 place? If the probabilities were equal - then it is: P(dog is first) + P (dog is second) = 1/6 + 1/6 = 1/3 If the probabilities are different? We have run results already of 400+ runs. Let's say that that we know from database that first dog finishes at first place 1/5 of the time and finishes in 2nd place 1/7 of the time. So then by the same it is 1/7 + 1/5 = 0.32 Is that correct thinking?
 February 21st, 2013, 12:45 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 You realize that 1/7 + 1/5 doesn't equal 0.32? What determines what number a dog has? Is there any reason why having the number 1 would be advantageous?
February 21st, 2013, 09:40 PM   #3
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Re: probability

Quote:
 You realize that 1/7 + 1/5 doesn't equal 0.32?
yeah, its 0.34 I now recalculated. Maybe i did roundings before addition or something like that.

Quote:
 What determines what number a dog has? Is there any reason why having the number 1 would be advantageous?
I don't know what determines. The reason he has advantage is by database of results. The fist dog lets say finished in first place more than other dogs. Its just simple results.

I mean there are few hundred dog race results. WE pick a random dog race from database. The task is to solve - what is the chance that 1st dog will be in first or 2nd place.

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