Tourney Seeds
8 top seeds enter the quarter-finals of a tourney, seeded 1 to 8, Number 1 seed being the top seed, and so on. From these, 4 will enter the semi-finals, and finally 2 players will play the final.
- Pairings are made at complete random in both the quarter-finals and the semi-finals.
- A higher seeded player will always defeat a lower seeded player in any match.
The probability that Number 4 seed enters the final is given by y x , where x and y are coprime positive integers. Find the product x × y .
The answer is 140.
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2 solutions
I added some LaTeX for you. Also, it's more typical to add the coprime integers; is there some specific reason why are you multiplying?
I think the question is worded incorrectly.
It states that "Pairings are made at complete random in both the quarter-finals and the semi-finals", which implies that the seeds are re-randomized for the semis.
That is, it is not
((1v2) v (3v5)) and (4v6) v (7v8) but rather (1v2) and (3v5)) and (4v6) and (7v8)
and the next round could be 1v3 and 4v7 or 1v4 and 3v7 or 1v7 and 3v4.
I don't think this produces the same answer but I'm not sure
I thought higher seeded defeat lower seeded means 4 defeats 3.
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Higher seed means higher place in the ranking so 1 beats 2,3,4,5,6,7,8.
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Thank you Peter, but well, asker should have make it clear. Not all of us have English as mother tongue. I only know as much as I need to get by. I know seed as the egg of trees.
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@Saya Suka – I get what you mean. I answered wrong btw. I used the fact that there are 2 groups and used a vase model: suppose seed 4 has 1 colour, seeds 5 to 8 a 2nd colour and 1 to 3 a third colour. The scheme of matches can be divided in 2 sides with 4 matches. 4 has to be on 1 side with 5 to 8, so P (4 gets to the finals) = 1 over 1 times 4 over 3 divided by 8 over 4. This gets you the answer 2/35. I made the mistake that you can make 2 schemes: one with this configuration to the right and one to the left. So the answer is 4/35.
To achieve the semi-finals, the seed 4 must face the seed 5, 6, 7 or 8. Additionally, it's important to note that if 1, 2 and 3 each face a team with lower seed than them, they will be the only teams 4 can face, and 4 will win none of them in the semi-final.
So one of the following match-ups must take place: 4-5, 4-6, 4-7 or 4-8 (4 choices). Additionally, one of the following match-ups must take place (two of them can't happen the same time): 1-2, 1-3 or 2-3(3 choices), so team 4 has a chance to win the semi-finals. Of the remaining four teams, there are always 3 ways to arrange match-ups between them.
Let's calculate the possible quarter-final matching-ups where 4 goes to semi-final and has a chance of achieving the final:
\(4\times3\times3=36)
The total ways of matching up eight teams in the quarter finals can be calculated by finding out how many ways there are to arrange the teams to an order(1st and 2nd are a match-up, 3rd and 4th are a match-up and so on). However, as the teams can be in 2 orders in a single match-up(in each 4 of them), and there are \(4!\) possible orders between the match-ups, we get the total number of ways to arrange the teams into match-ups is:
2 4 × 4 ! 8 ! = 1 0 5
And the probability for the conditions above to be fulfilled is 1 0 5 3 6 = 3 5 1 2 .
With those conditions, there is the team number 4, two higher seeded teams and one lower seeded team is the semi-finals. Team 4 has the equal chance of facing any of them, so the probability they will go to the final at this situation is 3 1 .
The probability of achieving finals from the starting situation is 3 5 1 2 × 3 1 = 3 5 4
4 × 3 5 = 1 4 0 .
There are 2 pools of players, one pool defeated directly/ indirectly by number 1 seed and one pool by the runner-up. 4 players in each pool. The total number of ways to select 3 other players in the seed number 4's pool will be ( 3 7 ) = 3 5 .
In how many situations, does number 4 seed go to the finals? Only if the 3 other players in the pool are from 5-8 seeding does 4 go through. The number of ways to select 3 players from 4 is ( 3 4 ) = 4 .
Answer: 3 5 4 .