Problem of the Week 1184
World Cup Groups
Solution Notes
Well, the World Cup has been dramatic so far.... If you would like another appropriate soccer-related problem, check out Who Won? It has a penalty kick problem I thought of 13 years ago.
I received solutions to 1184 from several of you: Jerrold Grossman, Al Zimmerman, Stephen Meskin, James Brawner, John Watkins, Dan DiTursi.
As James Brawner wrote:
There are 40 distinct vectors (a, b, c, d) that can arise:
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(9, 6, 3, 0)
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(7, 7, 1, 1)
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(7, 4, 3, 3)
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(6, 5, 4, 1)
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(5, 5, 2, 2)
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(9, 6, 1, 1)
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(7, 6, 4, 0)
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(7, 4, 3, 2)
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(6, 5, 2, 2)
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(5, 4, 4, 3)
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(9, 4, 4, 0)
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(7, 6, 3, 1)
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(7, 4, 3, 1)
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(6, 4, 4, 3)
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(5, 4, 4, 2)
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(9, 4, 3, 1)
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(7, 6, 2, 1)
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(7, 4, 2, 2)
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(6, 4, 4, 2)
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(5, 4, 3, 2)
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(9, 4, 2, 1)
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(7, 5, 4, 0)
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(7, 3, 2, 2)
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(5, 5, 5, 0)
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(5, 3, 3, 2)
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(9, 3, 3, 3)
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(7, 5, 3, 1)
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(6, 6, 6, 0)
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(5, 5, 4, 1)
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(4, 4, 4, 4)
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(9, 2, 2, 2)
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(7, 5, 2, 1)
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(6, 6, 4, 1)
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(5, 5, 3, 2)
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(4, 4, 4, 3)
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(7, 7, 3, 0)
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(7, 4, 4, 1)
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(6, 6, 3, 3)
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(5, 5, 3, 1)
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(3, 3, 3, 3)
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The most likely vector depends on the value of p, the probability of a tie.
If p < 1/4, then the most likely vectors are (9, 6, 3, 0) and (6, 6, 3, 3), each with a probability of (3/8)(1 − p)6.
If 1/4 < p < 1/3, then (6, 4, 4, 3) is most likely, with a probability of
(9/8)p(1 − p)5.
If 1/3 < p < (2 √6 − 3) / 5 ~ 0.38, then (7, 4, 4, 1) is most likely, with a probability of (9/4)(p2)(1 − p)4.
If (2 √6 − 3) / 5 < p < 1/2, then (5, 4, 3, 2) is most likely, with a probability of 6(p4)(1 − p)2.
If 1/2 < p < 6/7, then (5, 3, 3, 2) is most likely, with a probability of
6(p5)(1 − p).
If p > 6/7, then (3, 3, 3, 3) is most likely, with a probability of p6.
The least likely vector also depends on p.
If p < √2 − 1 ~ 0.414, then (3, 3, 3, 3) is least likely, with a probability of p6.
If p > √2 − 1, then the least likely vectors are (9, 3, 3, 3) and (6, 6, 0, 0), each with a probability of (1/8)(1 − p)6.
In the 1982 and 1990 World Cups, there was a group with 5 of the 6 round
robin matches ending in a tie. (Back then, only two points were awarded for
a win.)
The problem has been studied before. See http://www.actuaries.org/FIFA.pdf. But there are errors there.
Here is my self-contained Mathematica code that gets the 40 possibilities by
brute force checking all possible game results.
games = Subsets[Range[4], {2}];
tups= T uples[{smallWin, largeWin, draw}, 6];
vector[tup_] := (tt=Flatten[Table[{games[[i]], tup[[i]]}, {i,6}]/. {
{{i_,j_}, smallWin} :>{ {i,3}},
{{i_,j_}, largeWin} :> {{j,3}},
{{i_,j_}, draw} :> {{i,1}, {j,1}}},1];
Reverse@Sort@ Table[Total[Last/@ Select[tt, #[[1]]==i&]], {i,4}]);
possibles = DeleteDuplicates[Sort/@ vector/@ tups]
Length[possibles]
{{0,3,6,9},{1,1,6,9},{3,3,3,9},{1,3,4,9},{0,4,4,9},{1,2,4,9},{2,2,2,9},
{3,3,6,6},{0,6,6,6},{1,4,6,6},{1,3,6,7},{3,4,4,6},{0,4,6,7},{1,4,5,6},
{2,4,4,6},{1,2,6,7},{2,2,5,6},{3,3,4,7},{0,3,7,7},{1,3,5,7},{2,3,4,7},
{0,4,5,7},{1,3,4,7},{1,4,4,7},{2,2,4,7},{1,1,7,7},{1,2,5,7},{2,2,3,7},
{3,4,4,5},{2,3,5,5},{3,4,4,4},{4,4,4,4},{2,4,4,5},{1,4,5,5},{2,3,4,5},
{0,5,5,5},{1,3,5,5},{2,2,5,5},{2,3,3,5},{3,3,3,3}}
40
[Back to Problem 1184]
© Copyright 2014 Stan Wagon. Reproduced with permission.
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