Odds of two players meetings in an eight person single elimination tournament There is an $8$ person tournament. The odds of winning are $50\%$ for each player.
What is the probability of any $2$ players meetings at any point in the tournament? 
I understand that there are $7$ matches, and there are are a total of $\binom 82$ ways to choose $2$ people out of $8$. So the probability is $\dfrac{7}{\frac{8!}{2!6!}} = 0.25$.
What I am stuck at is another way of approaching the problem. The question is asking what is the probability of $A$ and $B$ would meet. There are $3$ rounds of matches. In the first round, P($A$ meeting $B$) is $1/7$. In the second round, given that $A$ hasn't met $B$ in round $1 (p=6/7)$, and that both $A$ and $B$ proceeded into the second round $(p=0.5*0.5 = 1/4), A$ has a probability of $1/3$ to meet $B$. So the probability that they meet in round $2\;\; is\;\; 6/7*1/4*1/3 = 1/14$. 
To make $1/7+1/14+P(A, B$ meet in round $3) = 0.25$, $P(A,B$ meet in round $3) = 1/28.$ But I could not figure out why the probability that $A$ and $B$ meet in round $3$ is $1/28.$ Any ideas?
 A: There are $\binom{8}{2}$ ways to choose $2$ players from $8$. Any set of two players is equally likely to be the set of finalists.
We can also do it in a more complicated way. The probability they are in different pools of $4$ is $\frac{4}{7}$. Then they each have to win two games, probability $\frac{1}{4}\cdot \frac{1}{4}$.
A: Here's another way of approaching the problem.
A will meet 1, 2 or 3 opponents, with probabilities $\frac{1}{2}$ (he gets knocked out in round 1 [R1]), $\frac{1}{4}$ (he wins in R1 and loses in R2), and $\frac{1}{4}$ (he wins in R1 and R2).
If he meets 1 opponent, there's $\frac{1}{7}$ chance it's B.
If he meets 2, there's $\frac{2}{7}$ chance that one of them is B.
If he meets 3, there's $\frac{3}{7}$ chance that one of them is B.
So the probability he meets B is $(\frac{1}{2}*\frac{1}{7})+(\frac{1}{4}*\frac{2}{7})+(\frac{1}{4}*\frac{3}{7})=\frac{1}{4}$.
And if he meets 3 opponents and one of them is B, then the probability that he meets B in any particular round is $\frac{1}{3}$. So the overall probability that he meets B in the 3rd round is 
$(\frac{1}{4}*\frac{3}{7}) * \frac{1}{3} = \frac{1}{28}$.
A: Here's another approach.  
Let's say that the two players of interest are A and B.  Player A has a probability of $1$ of playing a game in round one; $\frac12$ of playing a game in round two; and $\frac14$ of playing a game in round three. 
We have no reason to prefer any particular opponent for A in any particular round.  So the chance of A
playing B in round one, two, or three respectively is $1\cdot\frac17$; $\frac12\cdot \frac17$; and $\frac14\cdot \frac17$ 
A: I think the stage wise probability break-up that you are looking for
while concluding your question is:
P(don't meet in 1st round and both win) $\times$ P(don't meet in 2nd round & both win)
= $\dfrac67\cdot\dfrac14 \times \dfrac23\cdot\dfrac14 = \dfrac1{28}$
Added
For the first part of the question P( A and B meet), OP's way of solving seems simplest.
There are $\binom82$ possible pairs, $7$ matches, and a pair can play only one match, so $\dfrac7{\binom82}= \dfrac14$
Similarly, on the same lines,
the simplest way to compute P(A and B meet in the final) $=\dfrac{1}{\binom82} = \dfrac1{28}$,
bur OP wanted a method that continued in the way that was started. 
A: Work in progress
Chance of the players meeting in the first round:
Since there are 8 players, the chance of meeting is $1/7$ (the chance that they are together).
Chance of the players meeting in the second round (given that they haven't met in the first round, which is $6/7$):
Since there are 4 players left, the chance of meeting is 6/7 (refer to above) * 1/2 (the chance of person 1 making it through) * 1/2 (the chance of person 2 making it through) * 1/3 (the chance that they are together), which is $1/14$.
Chance of the players meeting in the third round (given that they haven't met in the first two rounds, which is $13/14$):
Since there are 2 players left, the chance of meeting is 13/14 * 1/2 * 1/2 * 1/1, which is $13/56$.
$1/7 + 1/14 + 13/56$ is the answer.
A: More generally, assuming n rounds of a knockout tournament ($2^n$ players) and each player has even odds in any single game, then the probability they meet is:
$$\sum \limits_{r=1}^{n} \frac{1}{2^{r-1}(2^n-1)}$$
A: The probability is 1/4.
The chance of the two players playing each other in the first round is 1/7. 
The chance of any player making the second round is 1/2. The chance of both of them making the second round is then 1/4. In an arrangement of four players, there is a 1/3 chance that two players will play each other once multiplied by 6/7 (the chance they haven't played each other in the first round). This means the chance of them playing each other in the second round is 1/4*1/3*6/7=1/14. The chance of a person making the last round is 1/4. This means the chance of both of them being in the last round is 1/16. When this is multiplied by 4/7 (the chance they haven't met before) we get 1/28. If they both are in the last game they have a 100% chance of playing each other. This means the total chance of them playing each other is 1/7+1/14+1/28=1/4. Thanks fellas hope you liked this answer.
