Knockout Stage

Consider the "two sides" of the knockout stage table, being the left side of the table, and the right side of the table. As you can see in the figure, teams from opposite sides of the final can only meet in the final. Therefore, we need to meet two conditions:

• Team #3 needs to be on the opposite side of the table compared to Team #1, otherwise they won't meet in the final.
• Team #2 needs to be on the same side as Team #1, such that it will be eliminated agains Team #1 and it won't face (and eliminate) Team #3.

So, the probability that Team #1 and Team #3 meet in the final is equal to \begin{equation} 1 * \frac{16}{31} * \frac{15}{30}  = \frac{8}{31}  \end{equation}

• "1" because Team #1 can start at either the left side or the right side.
• $ \frac{16}{31}$ because Team #3 needs to be at the opposite side from Team #1, where it can be at any of the 16 of the remaining 31 spots.
• $\frac{15}{30} $ because team #2 needs to be at the same side as Team #1, where it can be at any of the 15 of the remaining 30 spots. The reason it's 15 and not 16 is because Team #1 already has one spot at that side of the tree.

Alternative method - Combinatorial Analysis

It would be possible to solve this using combinatorial analysis as well, however, it's more intensive to do so.

• In order to construct a half of the three, there is a total of $\binom{32}{16} = 601.080.390$ ways to do so.
• Considering the positions of Teams #1 (first half), #2 (first half), and #3 (second half) are fixed, the number of successful ways to order the teams in the first half is $\binom{32-3}{16-2} = 77.558.760$.

This is only done for if Teams #1 and #2 are in the first half of the tree, and Team #3 in the other half. It can also happen the other way around. Therefore, the probability that Team #1 and Team #3 end up in the final, is equal to \begin{equation} 2 * \frac{77.558.760}{601.080.390} = \frac{155.117.520}{601.080.390} = \frac{8}{31} \end{equation} Simulation
If you want to simulate this with Python, try the following:

import random



def simulate_tournament():

    teams = list(range(1, 33))

    random.shuffle(teams)

    

    half1 = teams[:16]

    half2 = teams[16:]

    

    if 1 in half1 and 3 in half2 and 2 in half1:

        return True

    elif 1 in half2 and 3 in half1 and 2 in half2:

        return True

    else:

        return False



def estimate_probability(num_simulations=100000):

    count = 0

    for _ in range(num_simulations):

        if simulate_tournament():

            count += 1

    return count / num_simulations



probability = estimate_probability()

probability