# Euro 2008 Soccer Seedings

• Nov 21st 2007, 09:18 PM
chopet
Euro 2008 Soccer Seedings
Euro 2008 seedings for draw on Dec. 2 in Vienna:

Pot 1: Switzerland (hosts), Austria (hosts), Greece (holders), Netherlands

Pot 2: Croatia, Italy, Czech Republic, Sweden

Pot 3: Germany, Romania, Portugal, Spain

Pot 4: Poland, France, Turkey, Russia

4 groups will be created, with each Pot contributing 1 member to each group.
So Group A may have Switzerland, Croatia, Germany and Poland.

I was wondering about the probability for Netherlands, Italy, Germany and France to be drawn into the same group. Is it:
${1\over(4!)^4}$
• Nov 23rd 2007, 10:14 AM
TKHunny
Whoa!!! That is WAY too small. There are only $4^{4}$ ways to pick teams. You have selected one such way. There are only four ways to achieve the one way you have selected, that is, they must be selected together in one of the four selection rounds.

Pr(Selected in the first round) = (1/4)^4

The problem with the next step is that not all paths lead to the desired result. If any of the four teams is selected in the first round, they all must be. If any of the four teams is selected in the first round, but any of the others is not, the process ends. The ONLY way to get to round two is for NONE of the teams to be selected in round 1. There are only a few ways to do that.

Pr(None selected in round 1) = (3/4)^4

Thus, Pr(Selected in round 2) = (3/4)^4*(1/3)^4, since round 2 doesn't exist without its dependence on the outcome of round 1.

Using the same logic:

Pr(Selected in round 3) = (3/4)^4*(2/3)^4*(1/2)^4

If we get to round four with hope, obviosuly they will be selected together.

Pr(selected in round 4) = (3/4)^4*(2/3)^4*(1/2)^4*(1)^4

Thus, (3/4)^4 + (3/4)^4*(1/3)^4 + (3/4)^4*(2/3)^4*(1/2)^4 + (3/4)^4*(2/3)^4*(1/2)^4*(1)^4 = 4*(1/4)^4 = 0.015625
• Nov 23rd 2007, 11:20 AM
Plato
I missed this question when it was first posted.
There is another way to see the answer is $\frac{1}{{\left( {4!} \right)^3 }}$.
Think of the first row of host teams as urns: S, A, G, N.
Then arrange those four letters on each of the three lines to provide one possible selection.
• Nov 26th 2007, 04:27 AM
chopet
Hi Plato, I was thinking along the same line as you but I missed out fixing the top line.

But the answer is too small. and TKHunny's one is more intuitively correct.
Then I realise we should fix even more than we should.

Let's give each country a label, with Pot 1 labelled A1,A2,A3,A4, and so forth, and we can create a table with the rows representing the pots and the columns representing the groups:

1A 1B 1C 1D
2A 2B 2C 2D
3A 3B 3C 3D
4A 4B 4C 4D

Each row can be arranged in 4! way. There are 4 rows giving: $4!^4$ ways.
But I am interested only in:

x x 1C x
x x 2C x
x x 3C x
x x 4C x

I don't care what happens to all other groups, so I have to divide by $3!^4$ ways.
And the group {1C,2C,3C,4C} can exist in column 1, 2, 3 or 4, I divide by 4.

Finally, the probability is: $4{ {3!^4}\over{4!^4}}={1\over 4^3}=0.015625$

Thanks guys!
• Nov 28th 2007, 12:10 AM
PHANTOM
Just like Plato said
I agree with the answer Plata wrote.

every POT has 4 members and the way we scatter these 4 members into different groups is : 4!

We have 4 pots, so the number of ways for all combinations is: 4! x 4! x 4! x 4!
that's the number of elements of the sample space.

now lets figure out how many elements of the sample space account for " the 4 specific teams being together". There are 4! ways to arrange the 4 groups.

so P= 4!/(4!)^4= 1/(4!)^3 same as Plato's answer.
• Nov 30th 2007, 05:05 AM
chopet
Your method doesn't discount the fact that while your 4 countries are in the same group, any 4 OTHER countries can make up the OTHER groups.

e.g.
{France,Germany,Italy,Holland} in one group is what we want.

{Austria,Croatia,Spain,Poland} in another group is actually INDISTINGUISHABLE from
{Switzerland,Croatia,Spain,Poland}
or
{Switzerland, Sweden,Spain,Poland}
or
{Sweden, Czech, Portugal,Poland}
...

I can go on, but I know you have $3!^4$ groups to discount.