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Question about World Cup probability and random selection

  1. Jun 27, 2014 #1
    Of the 32 teams that qualify for the world cup (8 groups with 4 teams each), what percentage would a roster of 16 teams-to-advance-to-the-2nd-round (2 teams from each of the 8 groups) should be correct if the teams were chosen at random?

    Some background: A group of us at work filled out brackets before the start of the world cup. Of the 16 teams that have now advanced to the elimination round, the best of us made 11 accurate picks (disregarding whether the picks finished first or second in their respective groups). It made me wonder how much better that would have been than most completely random picks.

    I do not know enough statistics to do the requisite math, except to understand that it is a little more complicated than just saying 16 out of 32 = 50% on average.

    I got this far:
    Each group has 4 teams. 2 of those teams advance.
    The probability of correctly choosing both teams for a single group is 1/6
    The probability of correctly choosing just one team is 4/6
    the probability of not choosing either team is 1/6
    As such, the probability of correctly choosing at least one or two of the correct teams in each group is 5/6.
    So, for one group, most people making choices (two thirds of them) will have a 50% accuracy in predictions - assuming that all choices are made randomly ad infinitum.

    What do you guys think?
     
  2. jcsd
  3. Jun 27, 2014 #2

    AlephZero

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    That's correct, for one group.

    To extend it to two groups (say A and B), you can count up the possibilities like this:

    To get 4 correct, you have to get 2 correct in each group. You have 1 chance out of 6 in each group, so the total chance is (1/6)(1/6) = 1/36.

    To get 3 correct, you have to get 2 in group A and 1 in group B, or 2 in B and 1 in A. The chance of getting 2 in A and 1 in B is (1/6)(4/6) = 4/36. The chance of 1 in B and 2 in A is the same, so the total chance is 8/36

    To get 2 correct, if you count up all the possibilities you get 18/36.

    The arithmetic for 1 correct is the same as for 3, and 0 correct is the same as for 4. So the answer is

    4 correct: 1/36
    3 : 8/36
    2: 18/36
    1: 8/36
    0: 1/36

    You can then "double up" from 2 to 4 groups, and from 4 to 8. For 4 groups, the answer is
    0 or 8 correct: 1/1296
    1 or 7: 16/1296
    2 or 6: 100/1296
    3 or 5: 305/1296
    4: 454/1296.
     
  4. Jun 27, 2014 #3

    D H

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    That looks like a typo. I suspect you meant 304/1296 rather than 305/1296 for getting 3 or 5 teams correct in 4 groups.
     
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