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Math Help - UEFA Champions League

  1. #1
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    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
    Last edited by steel214; March 19th 2009 at 01:40 PM.
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  2. #2
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    Quote Originally Posted by steel214 View Post
    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.
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  3. #3
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    Quote Originally Posted by steel214 View Post
    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}}.
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  4. #4
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    Quote Originally Posted by Plato View Post
    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
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