I know there is quite a few of these, but for some reason I have so much trouble understanding probability properly, so was hoping to get some help.
Basically there are 8 horses racing, assuming they are all equal, I would assumed they have a 1/8 chance of placing in any particular value (first, second, third etc).
My first question is, assuming that I want to figure out the chance of placing within the top 4. Can I do (1/8)+(1/8)+(1/8)+(1/8) = 4/8 or 50 percent chance of placing on the top 4?
Second question is if I had two horses, what would be there chance of placing on the top 4? This I'm a little confused what to do, partly because I know if one horse places, that means the second horse only has a 1/7 chance of placing the other numbers? So I would guess you have 4/8 for horse 1 and also (1/7)+(1/7)+(1/7)+(1/7) = 4/7 for horse 2?
So that would get a (4/8)*(4/7) = 28.6 percent chance of both horses getting on the top four?
Any help would be amazing
Best Answer
Assuming there aren't any hidden horse dependencies, your approach to the first question is right. Any given horse has a $4/8 = 1/2$ chance of placing in the top half.
As far as the second question is concerned, it's easiest to consider the probability of neither horse finishing in the top half. Horse A finishes in the bottom half with probability $1/2$; horse B then finishes in one of the three remaining bottom half spots with probability $3/7$. Therefore, both finish in the bottom half with probability $1/2 \times 3/7 = 3/14$, so at least one of them finishes in the top half with probability $1-3/14 = 11/14$.
There is a more systematic way to approach these questions, using the binomial coefficients. We write $\binom{n}{k}$ (read "$n$ choose $k$"), to denote the number of ways to choose $k$ objects (order unimportant) out of $n$. So if we have two horses, and we want the probability that they both finish in the bottom half, we find the number of ways for them to finish in two of the bottom four spots, and divide that by the number of ways for them to finish in any spots:
$$ \frac{\binom{4}{2}}{\binom{8}{2}} $$
The expansion for the binomial coefficients is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n! = 1 \times 2 \times 3 \times \cdots \times n$. That gives us $\binom{4}{2} = 6$ and $\binom{8}{2} = 28$, so the probability of them both finishing in the bottom half is $6/28 = 3/14$, as before.
ETA: If the desired probability is that both horses fall in the top half, that's equal to the probability that both of them fall in the bottom half—the halves being indistinguishable from that perspective—or, again, $3/14$.