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, : )

The probability is $1/2$. Here is a possible explanation. 
Let's suppose we have two groups containing three random numbers each. 
Group 1: $\{x_1,x_2,x_3\}$
Group 2: $\{x_4,x_5,x_6\}$
Assume that these numbers are uniformly distributed and in [0,1]. All the random variables are also independent of each other.
\begin{equation}
p_x(X) = \left\{\begin{array}{rl}
1, & \textrm{if $X \in [0,1]$}\\
0, & \textrm{otherwise}
\end{array}\right.
\end{equation}
Defining $z_1 = \max(x_1,x_2,x_3)$ and $z_2 = \max(x_4,x_5,x_6)$, we want to obtain $P(z_2 > z_1)$.
This probability can be obtained by integrating the joint density $p_{z_1z_2}(Z_1,Z_2)$ of the random variables $z_1,z_2$, over the region where $z_2 > z_1$. Since $z_1$ and $z_2$ are independent because all the $x$ are independent, we have that $p_{z_1z_2}(Z_1,Z_2)=p_{z_1}(Z_1)p_{z_2}(Z_2)$.
The density $p_{z_1}(Z_1)$ is obtained by deriving the cdf $F_{z_1}(Z_1)$:
\begin{eqnarray*}
p_{z_1}(Z_1) &=& \frac{d}{dZ_1}\,F_{z_1}(Z_1) = \frac{d}{dZ_1}\,P(z_1 \le Z_1) \\ \\
&=& \frac{d}{dZ_1}\,P(\max(x_1,x_2,x_3) \le Z_1) \\
&=& \frac{d}{dZ_1}\,P(x_1 \le Z_1, x_2 \le Z_1, x_3 \le Z_1) \\
&=& \frac{d}{dZ_1}\,P(x_1 \le Z_1)P(x_2 \le Z_1)P(x_3 \le Z_1) \\
&=& \frac{d}{dZ_1}\,Z_1^3 = 3Z_1^2, \quad \textrm{for $Z_1 \in [0,1]$}.
\end{eqnarray*}
The same procedure is used for $z_2$ and we obtain $p_{z_2}(Z_2) = 3Z_2^2$, for $Z_2 \in [0,1]$.
The joint density is:
\begin{equation}
p_{z_1z_2}(Z_1,Z_2) = \left\{\begin{array}{rl}
9Z_1^2Z_2^2, & \textrm{if $(Z_1,Z_2) \in [0,1] \times [0,1]$}\\
0, & \textrm{otherwise}
\end{array}\right.
\end{equation}
We now can obtain the desired probability $P(z_2>z_1)$:
\begin{eqnarray*}
P(z_2 > z_1) &=& \int_{0}^{1}\int_{0}^{Z_2}p_{z_1z_2}(Z_1,Z_2)\,dZ_1dZ_2 \\
 &=& 9\int_{0}^{1} Z_2^2\int_{0}^{Z_2}Z_1^2\,dZ_1dZ_2 \\
 &=& 3\int_{0}^{1} Z_2^5dZ_2 \\
 &=& \frac{1}{2}\,Z_2^6\bigg|_0^1 = 1/2.
\end{eqnarray*}
