Assume that all the position in the tournament ladder is randomly selected, and 16 teams in the tournament ladder are evenly matched, and they include Brazil and England. What is the probability that both team will meet in a match during the tournament?

Given that the probability to win the match is 0.5

The probability cannot be calculated without being biased. The results are biased because of the way a team preforms. England and Brazil probably have a better chance of meeting because they are better teams and will progress. The USA (no offence) and Japan (no offence) would have a less likly outcome of facing each other because they are less likly to progress. What you also have to take into account is that each starting group there is a dominant team and 3 weaker teams (so there is a big game in the end and so a lesser team doesn't end up in second place.) That said, someone in the group with Brazil has a slim chance of facing England.

worthbeads, obviously you did not read the question close enough.

The question tells you to assume that ALL the teams are evenly matched.

So that England has the same chance and calibre of players as Japan.

You should fix his problem, not mine

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Originally Posted by worthbeads

you should fix his problem, not mine

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If you are going to give incorrect information, then I am going to fix YOUR problem.

As for the OP's question, it's been a while since I have done probability calculations

Assuming the teams are already picked, the probability can be between 100% (play each other right away) through 1 out of 256 (playing each other at the final). That would be 0.4%. I believe my calculations are right, but if anyone else has another answer, please tell me.

Worthbeads, you again are incorrect.

The question states the following:

The tournament ladder is RANDOMLY selected…
16 teams, 4 rounds
Each time has a 50% chance to win their match

England's chance of playing Brazil in Round 1 is: 1 in 15

England's chance of playing Brazil in Round 2 is:
England's chance of playing Brazil in Round 3 is:
England's chance of playing Brazil in Round 4 is:

I do not remember off hand how to calculate this, but I can assure you 1 in 256 is NOT correct.

That's what I was saying. I just did it more vaguely. The probabilities are are and between 100% and 1/256. It is not every percentage between, but between is the probabilities of the others. I got 100% if the two teams play their first game together, assuming the teams are already placed.

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Originally Posted by worthbeads

Assuming the teams are already picked, the probability can be between 100% (play each other right away) through 1 out of 256 (playing each other at the final)

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The probability of 1/256 was found this way:

The probability of the first team, team#1, to win their first game is 1/2
The probability of them winning their second game is also 1/2
Therefor, the probability of them winning both games is 1/4 (1/2*1/2)
Then the third, 1/2 (1/4*1/2=1/8)
And semifinal, 1/2 (1/8*1/2=1/16)

so there is a 1/16 chance team#1 makes it to the final (if it is necessary to play the other selected team)

That means the other selected team also has a 1/16 chance of going to the finals. You multiply 1/16*1/16 to get the probability of them both going to the finals! 1/16*1/16=1/256

By the way


Team#1 meeting right away (assuming teams have already been picked):1/1 (100%)
Team#1 meeting team#2 in second round:1/4 (25%)
Team#1 meeting team#2 in third round: 1/16 6.25%
Team#1 meeting team#2 in finals: 1/256 (0.39%)

For England and Brazil to meet in the finals, three things must happen:

1. England must win three games
2. Brazil must win three games
3. England and Brazil must start on different sides of the tournament tree (otherwise, they'd meet before the finals)

Multiply these three probabilities together to get the final answer. The first two are easy. Let's consider the third:

a) how many different ways are there to rearrange the teams?
b) how many of those ways result in England in the first 8, Brazil in the second?
c) how many result in Brazil in the first 8, England in the second?

a) Simple combinatorics, there are 16! Ways to arrange the teams (16 factorial)

b) break this one down:
I) how many ways to get England in the first eight teams?
ii) how many ways to get Brazil in the second eight teams?

ii) is easy - you have 8 teams to arrange (one is Brazil), so 8!
I) tougher - England in position 1-8, plus seven other teams(of 14) in any of the other seven spots:
8 x 14 x 13 x 12 x 11 x 10 x 9 x 8

So the final # of ways to get England in the first 8 and Brazil in the second 8 would be 8 x 8 x 14!

c) exactly the same as b): 8 x 8 x 14!

So, you should be able to figure out the answer for 3, then get the answer to your initial question.

Oops! I didn't read the question well enough, apparently!

Anyway, my method can be used to find the probability of England and Brazil meeting in the final. You should also be able to adapt it to find the probability of them meeting in round 3, round 2, and round 1. Add all four probabilities together (because they are exclusive outcomes so they don't interfere), and you'll get your answer.