Answer

Since we're not given any details about the two players A and B, they are just any two arbitrary players. So question is essentially asking, what's the probability of any given pair playing together. Notice there's nothing making one pair more likely to play than another, so all are equally likely. Therefore, we can find the probability as the number of pairs that end up playing divided by the number of all possible pairs we can form. You can think of this as: there's X possible pairs we could form amongst the 32 players, but only Y pairs will actually get to play (given the structure of the game), so chance of any of the X possible pairs actually becoming a pair in the game is $\frac{Y}{X}$

So how many pairs can we form amongst the 32 players? By definition this is $\binom{32}{2}=\frac{32!}{30!2!}=\frac{32\cdot 31}{2}$. And how many pairs will get to play in the game? Well each game eliminates one player, and to find the best player we must eliminate all others, of which there are 31. So there's 31 games in total. Therefore the probability of any two players meeting during the turnament is: $$\frac{31}{\frac{32\cdot 31}{2}}=\frac{1}{16}$$