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Six teams play a single round-robin tournament, in which each team plays every other team exactly once, there are no ties, and for each team, the probability of winning any given game is 1/2. What is the probability that at least two different teams finish the tournament with the same number of wins? Express your answer as a common fraction.

 

Thanks in advance!

 May 14, 2024
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Interestingly, the probability that NO teams finish with the same number of wins is much easier to calculate than the probability of at least two teams having the same number.

 

Here's why:

 

Total Order of Wins: In a round-robin tournament with no ties, there must be a total order for the teams based on their wins. There will be one team with the most wins (let's call it W), another team with the second-most wins (W-1), and so on, down to a team with the least wins (L).

 

Number of Wins: Since each team plays every other team exactly once, the total number of wins and losses for all teams must be equal: (Number of teams) * (Wins per team) = (Number of teams) * (Losses per team) --> 6 * Wins = 6 * Losses.

 

Therefore, each team must have either 3 wins and 3 losses (middle ground) or 2 wins and 4 losses (bottom half) or 4 wins and 2 losses (top half). There are 3 possible wins for each team (2, 3, or 4).

 

Ordering the Wins: Once we know the number of wins for each team, we just need to order them. There are 6! (6 factorial) ways to order 6 distinct elements, but in this case, 3 of the elements have the same value ("3 wins"), so we've overcounted.

 

There are 3! ways to order the elements with 3 wins (W-1, W, W+1), so we need to divide 6! by 3! to account for this. This gives us the total number of ways for a complete ordering of wins (no ties) as: (6!)/3! = 120.

 

Probability of No Ties: Since there are no ties, all the games must be decided with a win or loss, and the probability of winning any game is 1/2, the total number of possible outcomes for the entire tournament is 2 raised to the power of the total number of games (each game has 2 outcomes).

 

In a round-robin tournament with n teams, there are n*(n-1)/2 games played. So, the total number of outcomes is: 2^(n*(n-1)/2). For n=6 teams, this is: 2^(6*5)/2 = 2^15.

 

Therefore, the probability that there are NO ties and a complete ordering of wins (no teams with the same number of wins) is: (Number of favorable outcomes) / (Total number of outcomes) = (120) / (2^15)

 

This is the easiest outcome to calculate. The actual question asks for the probability of AT LEAST two teams having the same number of wins.

Since the total probability is 1, the probability of at least two teams having the same number of wins is the complement of the probability of no ties:

 

Probability (At least two teams with same wins) = 1 - Probability (No ties)

= 1 - (120) / (2^15)

 

This is the answer, but it's a bit cumbersome. We can simplify it further.

 

Notice that the 2 in the denominator (2^15) can be factored out as 2 * (2^14).

 

In the numerator, we can rewrite 120 as (2^3 * 3 * 5).

 

So the probability becomes:

 

1 - ( 2^3 * 3 * 5) / (2 * 2^14)

 

Rewrite to group the 2's:

 

1 - ( 2^2 * 3 * 5) / (2^15)

 

Take out a common factor of 2:

 

1 - (2 * 3 * 5) / (2^14)

 

Since 2 is in both the numerator and denominator, we can divide them, leaving:

 

1 - (3 * 5) / (2^13)

 

This is the most simplified form of the answer: 1 - (15/8192) = 8177/8192

 May 14, 2024

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