Probability of at least one random number out of 3 being greater than 3 other random numbers? $$\{?,?,?\mid?,?,?\}$$
There are 6 random numbers (drawn from arbitrarily large pool). What is the probability that biggest number lies in second half? Answer is $1/2$ but I tried to solve it with combination. Can you show me where I went wrong?
I thought, for one particular number in second half, the probability that it is greater than all three points in first half is $1/8$ and likewise probability that it is not greater than all three points in first half is $7/8$, so
$${3 \choose 1}  \left(\frac18\right)^1 \left(\frac78\right)^2$$ is probability that only of them is bigger than first half
$${3 \choose 2} \left(\frac18\right)^2 \left(\frac78\right)^1$$ is probability that two of them are bigger than first half
$${3 \choose 3} \left(\frac18\right)^3 \left(\frac78\right)^0$$ is probability that all three of them are bigger than first half
Total is $\dfrac{169}{512}$ not $\dfrac{256}{512}$. So what is my mistake?
Edit: I think I figured out, biggest number, rank 6 must be in second half; so I can choose other two numbers 
(5 choose 2) different ways * 3! configuration for right side * also 3! configuration for left side; 
divided by 6! equals to 1/2. now i can generalize it to solve secretary problem, : )

If none of the numbers in the pool are equal, then the fact that six numbers are drawn from the pool is irrelevant.  You are just dealing with six numbers, none of which are equal to any of the others.  Since we are dealing with a set of six (an even number) numbers, then each number is either above the median or below it.  So no need to do combinations or anything else fancy.  The question you are asking is the same as asking what the probability of picking a red ball is from a set of 3 red balls and 3 blue balls.
