Tournament problem Sixteen players $S_1,S_2......S_{16}$ 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 exactly one of s1,s2  are present in the  winners.
My attempt : 
There are two cases possible
1) when S_1 and S_2 are not paired with each other
 In this case they can be paired in $14C1*13C1$ ways. now we want exactly one from S_1 and S_2 as winner..
This can be possible when S_1 wins and S_2 loses or S_1 loses and S_2 wins so the total number of ways comes out to be $14C1*13C1*2$!
2)when S1 and and S2 plays against each other..there are 2 cases possible and both are favourable..as we want exactly one from S1 and S2 as a winner
Sample space= 14C1*13C1*(both win,both loose,one wins one loose)*4+2
Probability
=14C1*13C1*2+2/(14C1*13C1*4+2 )
I know the question has already answered before...but I don't what is wrong with this approach
 A: Consider $S_1$'s opponent. Either it is $S_2$ at $\frac{1}{15}$ chance or another player at $\frac{14}{15}$ chance.
In the first case the desired condition will be fulfilled (as you observed). 
In the second case, there's a $\frac 12$ chance each that $S_1$ or $S_2$ will win and $\frac 12 $ chance each they will lose. So $\frac 14$ chance of each of ($S_1$ win, $S_2$ win), ($S_1$ loss, $S_2$ win), ($S_1$ win, $S_2$ loss), ($S_1$ loss, $S_2$ loss). This gives $\frac 14 + \frac 14 =\frac 12$ chance of ending up with exactly one of $S_1,S_2$ as winners. 
The probability follows easily.

In your analysis you need to weight the two different cases appropriately, which is perhaps not so easy. But once $S_1$ is not playing $S_2$ it's certain that $S_2$ is not playing $S_1$, of course, so $14C1$ is enough to ensure they aren't playing each other. Then because there are 4 outcomes of interest when $S_1$ is not playing $S_2$, but only 2 when $S_1$ is playing $S_2$, we need to double the weight of the $S_1$ v $S_2$ option:
$(14\times 2+1\times 2\times 2)/(14\times 4 + 1\times 2\times 2) = 32 / 60 = 8/15$
Perhaps more generally it is better to weight the successful options down by the size of the option space for each case:
$(14\times 2 / 4 +1\times 2/ 2)/(14 + 1) = 8/15$
Which starts converging to the previous analysis.
A: METHOD 1:
Pair =>  (x and y)
P(pair) => 1/15
P(exactly one of them will will the match) = 1/15 *1 = 1/15
Next Condition : NOT PAIRED.  i.e. x fights with others(except y ). y fights with others(except x).
P(Not Pair) = (1 - 1/15) = 14/15
If NOT PAIRED , then 4 conditions will come :

*

*x win and y win

*x win and y loose

*x loose and y win

*x loose and y loose

According to the problem, then we are interested in conditions 2 and 3.
That's why ,
(1- 1/15)[x loose y win OR x win y loose]
=> (1- 1/15)[ 1/4 + 1/4 ]
Therefore,
P(when paired) + P(not paired) with conditions [x loose y win OR x win y loose]
=  1/15 + (1- 1/15)[ 1/4 + 1/4 ] = 8/15
...................................................................................
METHOD 2(By a renowned IIT Maths Professor KD JOSHI) :
(A OR B) will be among the 8 winners i.e. P(A U B ) =  P(A) + P(B) - P(A ⋂ B)
P(A ⋂ B)  Explanation:
This means that A and B both win the Final. This cannot happen if Paired.
If A is paired with others as well as  B  is paired with others, Then 2 things can happen.
If A fights with others(except B ) A WIN.
If B fights with others(except A ) B WIN.
Probability(A and B paired) = 1/15
Then, Probability(A and B Not Paired) = [1 - 1/15]
P(A) and P(B ) are independent events.
So, P(A ⋂ B) = P(A) * P(B)
Therefore The Total value of P(A ⋂ B) = P(A and B Not Paired) * P(A) * P(B)
As all men are of equal strength. So, P(A) = P(B) = 1/2
P(A U B ) =  P(A) + P(B) - P(A ⋂ B)
=> P(A U B ) =  P(A) + P(B) - P(A and B Not Paired) * P(A) * P(B)
=>  P(A U B ) =  1/2 + 1/2 - [1 - 1/15] * 1/2 * 1/2
=>  P(A U B ) =  (1 - 7/30) = 23/30
.......................................................................
Method 3 :
P(exactly one of A or B to happen)
= P(ABcomplement) + P(BAcomplement) = P(A) + P(B) – 2P(A ⋂ B)
P(A and B paired) + P(A and B not paired) * P(exactly one of A or B to win)
= P(A and B paired) + P(A and B not paired) *  [ P(A) + P(B) – 2P(A ⋂ B)]
= 1/15 + (1 - 1/15) * [1/2 + 1/2 - 2 * 1/2 * 1/2] = 8/15
...........................................................................
Method 4(Best) :
let, A is the winner and B is the looser.
16 men =>  8 pairs
#Winners = 8
#Losers = 8
if A is winner then (16 - 1 ) = 15 empty slots are to be occupied by B and others.
#Empty slots = 15
Then Probability( Exactly one of the two players is among the 8 winners)
= ( #Losers / #Empty slots) = 8/15
