# Thread: UEFA Champions League

1. ## UEFA Champions League

Hi,
I have a discussion with my friends about the next games.
As you know for the next step there are 4 British teams, 2 Spanish teams, 1 Germany and 1 Portuguese.

The problem is the following.

What is the probability of at least to have one game between British teams?
The problem can be reduced to 4 White balls + 4 Black balls in a bag?

Answers we have discussed without agreement:
1) 50%
2) 21,4% ( 6/28)
3) 77%

Thanks for your colaboration

2. Originally Posted by steel214
Hi,
I have a discussion with my friends about the next games.
As you know for the next step there are 4 British teams, 2 Spanish teams, 1 Germany and 1 Portuguese.

The problem is the following.

What is the probability of at least to have one game between British teams?
The problem can be reduced to 4 White balls + 4 Black balls in a bag?

Answers we have discussed without agreement:
1) 50%
2) 21,4% ( 6/28)
3) 77%

Thanks for your colaboration
Hi steel214,

It's easier to find the probability that there will be no game between British teams, then subtract that probability from 1 to find the probability that there will be at least one game between Brits.

There are $\binom{8}{2 \; 2\; 2\; 2} = \frac{8!}{(2!)^4} = 2520$ possible ways to pair the teams into games without any restrictions (a multinomial coefficient). All of these arrangements are equally likely.

How many of the pairings result in no British-British pair? Well, there are 4 possible ways to match the first British team with a non-Brit. Once this selection is made, there are 3 remaining non-Brit teams to pair with the second British team. Then there are 2 possible pairings for the third British team, and finally only 1 choice for the fourth British team. So there are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways to pair the teams with no British-British game.

Therefore the probability of no British-British game is p = 24 / 2520 = 0.009524, and the probability of at least one British-British game is 1 - p = 0.9905, approximately.

3. Originally Posted by steel214
The problem is the following.
What is the probability of at least to have one game between British teams?
Answers we have discussed without agreement:
1) 50%
2) 21,4% ( 6/28)
3) 77%
I agree with answer #3.
To separate eight individuals into four groups of two each can be done in $\frac{{8!}}{{\left( 2 \right)^4 \left( {4!} \right)}} = 105$ ways.
These are unordered partitions. Thus we avoid an over count.
Now there are 24 ways for the four British teams not to play one another.
So we get $1 - \frac{{{\text{24}}}}{{{\text{105}}}}{\text{ = 0}}{\text{.771428}}$.

4. Originally Posted by Plato
I agree with answer #3.
To separate eight individuals into four groups of two each can be done in $\frac{{8!}}{{\left( 2 \right)^4 \left( {4!} \right)}} = 105$ ways.
These are unordered partitions. Thus we avoid an over count.
Now there are 24 ways for the four British teams not to play one another.
So we get $1 - \frac{{{\text{24}}}}{{{\text{105}}}}{\text{ = 0}}{\text{.771428}}$.
Yikes! Plato is correct, I neglected to take the ordering (or the lack thereof) of the games into account.

-Awkward