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June 4th, 2017, 11:35 AM   #1
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Help Un-Fry 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.
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June 5th, 2017, 05:18 AM   #2
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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

$\dfrac{1}{56} + \dfrac{1}{56} + \dfrac{6}{56} + \dfrac{6}{56} = \dfrac{14}{56} = 0.25.$
Thanks from billnc

Last edited by JeffM1; June 5th, 2017 at 05:20 AM.
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June 5th, 2017, 04:28 PM   #3
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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.
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