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, : )

Your problem is that the probability that the fourth number is greater than the preceding three is not $\frac 18$, it is $\frac 14$.  Your calculation of $\frac 18$ assumes that the fourth number being greater than the first and the fourth number being greater than the second are independent.  They are not.  If the fourth number is large, it has a good chance of being greater than any of the first three.
