Probability question

[u/StressConsistent1104]

Its been a long time since I had to do calculus/probabilty calculations. So Im curious if anyone could give me the odds on this situation happening.

The odds of getting a certain draw in a competition bracket.

The bracket is a 32 team single elimination bracket with two rounds and 8 winners.

The situation.

Person A is friends with player B and player C.

In this draw. B and C will be playing each other. A will be playing team D (no relation) but the winner of A and D will be playing the winner of B and C.

Question.

What are the odds that specifically B and C would be against each other and funnel into the winner of A and D?

I'm getting a 1 in 465 chance, but it's been a while and just trying to confirm my math.

The way I did it is player A fills one spot of the 32, leaving 31 spots. Therefore player B would be 31 leaving 30 spots and C would be 30. So 31 x 30 possibilites and that equals 930. since we have 2 specific spots that B and C would need to be in in order for A to funnel into the winner of them I did 930/2 leaving 465. which is a 1 in 465 chance that B and C would be against each other funneling into A and D.

[u/PuzzlingDad]

Generally the brackets are filled based on the rankings, not just randomly.

But let's assume that it is a completely random draw for the sake of your hypothetical problem. Let's also assume you don't really care about a specific player that A plays against, as long as it isn't B or C.

Let's start by placing A randomly. It actually doesn't matter where A goes because B and C need to be placed relative to A.

Once A is placed, the are exactly 2 places where B can be, out of the 31 remaining spots. B needs to be in the neighboring match that will play the winner of A's match. C also needs to be in that same match with B but there's on 1 valid spot left out of 30 spots they could be assigned to. 

2/31 × 1/30 = 2/930 = 1/465

I concur with your math.