
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 4th, 2017, 11:35 AM  #1 
Newbie Joined: Jun 2017 From: North Carolina Posts: 2 Thanks: 0  Help UnFry My Brain  Horse Probability
Hi! Newbie here. I am trying to figure out a pretty basic natural probability question regarding a horse race and I would greatly appreciate anyone's help. So here goes: Imagine a horse race with 8 entries. Throw away all advantages and imagine that all 8 horses are equal. Now the natural probability of the #1 horse to come in in FIRST OR SECOND is .25, right? .125 chance for 1st .125 chance for 2nd What would be the natural probability for the #2 horse to then come in FIRST OR SECOND? Is it .125 or is it .1428? If we are going to say that the #1 horse WILL come in 1st or 2nd, then there are only 7 spots left. So, I'm thinking the correct answer is .1428. So the odds of the #1 horse coming in first or second AND the #2 horse coming in first or second should be .25 x .1428 = .0357 NOT .25 x .125 = .03125 Is this right? Thanks, in advance, for your input. Last edited by skipjack; June 5th, 2017 at 01:00 AM. 
June 5th, 2017, 05:18 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,028 Thanks: 420 
No, it is obviously wrong. If the horses are all alike, they have equal chances of coming in first or second. Probability that the second horse will come in second if first horse comes in first = $\dfrac{1}{7} * \dfrac{1}{8} = \dfrac{1}{56}.$ Probability that the second horse will come in first if first horse comes in second = $\dfrac{1}{7} * \dfrac{1}{8} = \dfrac{1}{56}.$ Probability that the second horse will come in first if the first horse does not come in first or second = $\dfrac{1}{7} * \dfrac{6}{8} = \dfrac{6}{56}.$ Probability that the second horse will come in second if the first horse does not come in first or second = $\dfrac{1}{7} * \dfrac{6}{8} = \dfrac{6}{56}.$ Probability that second horse will come in first or second $\dfrac{1}{56} + \dfrac{1}{56} + \dfrac{6}{56} + \dfrac{6}{56} = \dfrac{14}{56} = 0.25.$ Last edited by JeffM1; June 5th, 2017 at 05:20 AM. 
June 5th, 2017, 04:28 PM  #3 
Newbie Joined: Jun 2017 From: North Carolina Posts: 2 Thanks: 0 
So the probability that the #2 horse will come in first or second, IF the #1 horse comes is first or second, is 1 /56 + 1 / 56 which is 2/56 which is .0357142, which is where I had correctly landed, save a rounding error on my part. I wrote: "So the odds of the #1 horse coming in first or second AND the #2 horse coming in first or second should be .25 x .1428 = .0357" Thanks so much! Last edited by billnc; June 5th, 2017 at 04:41 PM. 

Tags 
brain, horse, probability, unfry 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
angles on a saw horse  mickthetree  Geometry  2  July 12th, 2014 10:01 AM 
the horse  Haley  New Users  6  November 8th, 2012 05:55 AM 
Probability brain teaser  Demo_Vers.  Advanced Statistics  3  December 18th, 2009 04:22 PM 
Horse Combinations  M. Einstein  Math Events  0  July 26th, 2009 09:14 AM 
Probability of a horse racing game  Zomis  Advanced Statistics  3  June 29th, 2007 08:03 AM 