<s>[INST] 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?
 [/INST] 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 
$(\frac{1}{4}*\frac{3}{7}) * \frac{1}{3} = \frac{1}{28}$.
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