8 teams have advanced to the quarter-finals of UEFA Champions League. With 4 of them coming from England (Manchester United, Liverpool, Tottenham and Manchester City), what is the probability that no two English teams are drawn together? If that happens, it's possible to have an all-English semi-finals. There are no rules in this draw, each team can be drawn against any other opposition.
Step 1: Total number of possible drawings
There are two ways to calculate this. First, without loss of generality, pick any team say Manchester United to be the first team in the draw, meaning the first ball coming up in the draw is for Manchester United. There are 7 possible opponents, all equally likely. One of them will be drawn to face Manchester United. Next, there are 6 teams remaining and the process continues. The next team drawn has 5 possible opponents, and so on. Hence the total number of possible drawings is 7*5*3*1=105.
Alternatively, this can be calculated with combinations. Initially there are 8 teams in the draw, and 2 teams are drawn to play against each other. The number of ways to do so is 8C2. Next there are 6 teams left, the number of ways to find 2 teams is 6C2. Thus this can be the product of 8C2*6C2*4C2*2C2 = 8*7*6*5*4*3/2/2/2 = 7*6*5*4*3. To account for repetitions from ordering, for example ABCD is same as ACBD, this should be divided by 4!. Hence the result is 7*6*5*4*3/4! = 7*5*3 = 105, same as above.
Step 2: Total number of favorable outcomes, i.e. no 2 English teams drawn together
Using similar argument as above, suppose Manchester United is the first team drawn, there are 4 possible non-English opponents (Juventus, Ajax, Porto or Barcelona). The second English team have 3 possible non-English opponents and so on. Thus the total number is 4*3*2*1=24. There are 24 ways to draw where every English team is facing a foreign team in the quarter finals.
The probability that no two teams from England play together is 24/105 = 8/35, which is about 22.86%. Thus there is over 77% chance that two of them will play each other.
What about the 4 teams from England all drawn against each other? The number of such possible pairings are only 3*3=9! The probability is 9/105 = 3/35 = 8.57%.
The probability that exactly 1 pair of English team playing together is 1-24/105-9/105=72/105 = 24/35, which is over 68.57%. This is the most likely scenario and it exactly occurred in 2009 when Manchester United defeated Porto, Chelsea defeated Liverpool and Arsenal defeated Villarreal in the quarter finals.
Overall, each team has 1 in 7 chance to face against any other team. Who will Manchester United face after beating PSG by away goals? Is history going to repeat to have exactly 2 English teams facing each other at 68.57%, denying a possibility of all 4 semi-finalists from Premier League?
Step 1: Total number of possible drawings
There are two ways to calculate this. First, without loss of generality, pick any team say Manchester United to be the first team in the draw, meaning the first ball coming up in the draw is for Manchester United. There are 7 possible opponents, all equally likely. One of them will be drawn to face Manchester United. Next, there are 6 teams remaining and the process continues. The next team drawn has 5 possible opponents, and so on. Hence the total number of possible drawings is 7*5*3*1=105.
Alternatively, this can be calculated with combinations. Initially there are 8 teams in the draw, and 2 teams are drawn to play against each other. The number of ways to do so is 8C2. Next there are 6 teams left, the number of ways to find 2 teams is 6C2. Thus this can be the product of 8C2*6C2*4C2*2C2 = 8*7*6*5*4*3/2/2/2 = 7*6*5*4*3. To account for repetitions from ordering, for example ABCD is same as ACBD, this should be divided by 4!. Hence the result is 7*6*5*4*3/4! = 7*5*3 = 105, same as above.
Step 2: Total number of favorable outcomes, i.e. no 2 English teams drawn together
Using similar argument as above, suppose Manchester United is the first team drawn, there are 4 possible non-English opponents (Juventus, Ajax, Porto or Barcelona). The second English team have 3 possible non-English opponents and so on. Thus the total number is 4*3*2*1=24. There are 24 ways to draw where every English team is facing a foreign team in the quarter finals.
The probability that no two teams from England play together is 24/105 = 8/35, which is about 22.86%. Thus there is over 77% chance that two of them will play each other.
What about the 4 teams from England all drawn against each other? The number of such possible pairings are only 3*3=9! The probability is 9/105 = 3/35 = 8.57%.
The probability that exactly 1 pair of English team playing together is 1-24/105-9/105=72/105 = 24/35, which is over 68.57%. This is the most likely scenario and it exactly occurred in 2009 when Manchester United defeated Porto, Chelsea defeated Liverpool and Arsenal defeated Villarreal in the quarter finals.
Overall, each team has 1 in 7 chance to face against any other team. Who will Manchester United face after beating PSG by away goals? Is history going to repeat to have exactly 2 English teams facing each other at 68.57%, denying a possibility of all 4 semi-finalists from Premier League?
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