sixteen players $s_1, s_2, s_3, \ldots, s_{16}$ playing a tournament are divided into eight pairs at random Sixteen players $s_1, s_2, s_3, \ldots, s_{16}$ playing a tournament are divided into eight pairs at random.  From each pair, a winner is decided on the basis of a game played between the two players of the pair.  Assume all players are of equal strength.  


*

*Find the probability that $s_1$ is among the eight winners. 

*Find the probability that exactly one of the players $s_1$ and $s_2$ are present in the $8$ winners. 

 A: Another way to see this is to note that there are going to be $8$ winners and $8$ losers.  Whatever $s_1$ does, we are interested in $s_2$ doing the opposite.  But $s_1$ fills one of the $8$ slots of one type (W or L).  So of the 15 remaining slots, $8$ are of the opposite type.  So there is an $\frac8{15}$ chance that $s_2$ is of the opposite type.
A: There's nothing wrong with it, but since $s_1$ has to be paired with someone, and  all players are of equal strength, $P(s_1$ is among the winners) is simply $\dfrac12$
Part 2
Let $s_1$ be anywhere, there are $15$ spots for $s_2$ of which $1$ is with $s_1$, with $Pr=\dfrac1{15}$
If they are together, only one of them will win,
else we need $s_1W-s_2L$, or vice-versa.
hence $Pr = \dfrac1{15}\cdot1 + \dfrac{14}{15}\cdot2\cdot\dfrac12\dfrac12 = \dfrac8{15}$ 
A: For the first part, player $s_1$ must be in one of the eight matches.  By assumption, each player is equally likely to win a match against one of the other players in the tournament.  Thus, player $s_1$ has probability $\frac{1}{2}$ of winning her match, so she has probability $\frac{1}{2}$ of being one of the eight winners.
As true blue anil pointed out to me, your answer for the first part is correct for the reasons he stated in his comment.
For the second part, there are two possibilities.

*

*Player $s_1$ plays player $s_2$.

*Player $s_1$ plays a player other than $s_2$.

Case 1:  There are $15$ possible opponents for player $s_1$, each of which is equally likely to be assigned to her.  Hence, the probability that she plays player $s_2$ is $\frac{1}{15}$.  The probability that exactly one of them will win the match is $1$.  Hence, in this case, the probability that exactly one of the players $s_1$ and $s_2$ will win the match is $$\frac{1}{15} \cdot 1 = \frac{1}{15}$$
Case 2:  The probability that player $s_1$ does not play player $s_2$ is
$$1 - \frac{1}{15} = \frac{14}{15}$$
There are four equally likely possible outcomes of their two matches.

*

*Both player $s_1$ and player $s_2$ win their matches.


*Player $s_1$ wins her match, while player $s_2$ loses hers.


*Player $s_1$ loses her match, while player $s_1$ wins hers.


*Both player $s_1$ and player $s_2$ lose their matches.
Since these four events are equally likely, they each occur with probability $\frac{1}{4}$.  We are only interested in the second and third possibilities.  Hence, the probability that exactly one of the players $s_1$ and $s_2$ wins her match if they do not play each other is
$$\frac{14}{15}\left(\frac{1}{4} + \frac{1}{4}\right) = \frac{7}{15}$$
Since the two cases outlined above are mutually exclusive, the probability that exactly one of the two players $s_1$ and $s_2$ wins her match is $$\frac{1}{15} + \frac{7}{15} = \frac{8}{15}$$
A: The Hypergeometric distribution Concept is useful here.
16 players. First-round => 16 players. Second-round => 8 players and These 8 men are the Winners of 1st Round. We can select 1 player out of 8 players in $\binom{8}{1} $ ways. We also select 1 player from  remaining = (Total players - players in second round i.e. 8 ) = 16 - 8 = 8 players.These 8 players are the losers of 1st Round.This can be done in $\binom{8}{1} $ ways. Also, we can select these 2 players let's say A and B in $\binom{16}{2}$ ways. Total probability = $\frac{\binom{\text{Winners Of First Round} }{1}\binom{\text{Losers Of Second Round} }{1}}{\binom{\text{Total Players}}{2}}$ = $\frac{\binom{8}{1}\binom{ 8 }{1}}{\binom{16}{2}}$ = $\frac{8}{15}$(Ans)
