Game of tickets 
In a game of tickets, $8$ tickets numbered $1,2,3,\dots,8$ are grouped in the pairs at random and observed pair wise, the lower numbered tickets of pairs are kept and others are thrown and this process is repeated once more. Find the probability that out of remaining two tickets, one is numbered four.  

 A: Hint: divide the tickets into two groups of four, and then into pairs, so that the surviving ticket from each pair in the first round is put with the surviving ticket from the other pair in the same group in the second round. Then actually the pairs within the groups doesn't matter for determining which two tickets survive, which are always the lowest number from each group. 
Now $4$ survives if and only if $1,2$ and $3$ are all in the opposite group. How many different ways are there to divide into groups? How many of these have a group containing $1,2$ and $3$ but not $4$?
A: I propose an alternative interpretation of the problem which makes it easier to solve.
We have to evaluate the probability that in a tennis tournament among players $1,2,3,4,5,6,7,8$, where in a match the lower number wins, player $4$ arrives at the final. So player $4$ should not meet any player in $\{1,2,3\}$ before the final. That is, the side of the board of player $4$ has to contain $3$ players chosen in $\{5,6,7,8\}$. Hence the probability is
$$\frac{\binom{4}{3}}{\binom{7}{3}}=\frac{4}{35}.$$
