1993 Paper 2 Q1

Year: 1993
Paper: 2
Question Number: 1

Course: LFM Pure
Section: Proof

Difficulty: 1600.0 Banger: 1500.0

Problem

In the game of ``Colonel Blotto'' there are two players, Adam and Betty. First Adam chooses three non-negative integers \(a_{1},a_{2}\) and \(a_{3},\) such that \(a_{1}+a_{2}+a_{3}=9,\) and then Betty chooses non-negative integers \(b_{1},b_{2}\) and \(b_{3}\), such that \(b_{1}+b_{2}+b_{3}=9.\) If \(a_{1} > b_{1}\) then Adam scores one point; if \(a_{1} < b_{1}\) then Betty scores one point; and if \(a_{1}=b_{1}\) no points are scored. Similarly for \(a_{2},b_{2}\) and \(a_{3},b_{3}.\) The winner is the player who scores the greater number of points: if the socres are equal then the game is drawn. Show that, if Betty knows the numbers \(a_{1},a_{2}\) and \(a_{3},\) she can always choose her numbers so that she wins. Show that Adam can choose \(a_{1},a_{2}\) and \(a_{3}\) in such a way that he will never win no matter what Betty does. Now suppose that Adam is allowed to write down two triples of numbers and that Adam wins unless Betty can find one triple that beats both of Adam's choices (knowing what they are). Confirm that Adam wins by writing down \((5,3,1)\) and \((3,1,5).\)

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

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Problem source
In the game of ``Colonel Blotto'' there are two players, Adam and
Betty. First Adam chooses three non-negative integers $a_{1},a_{2}$
and $a_{3},$ such that $a_{1}+a_{2}+a_{3}=9,$ and then Betty chooses
non-negative integers $b_{1},b_{2}$ and $b_{3}$, such that $b_{1}+b_{2}+b_{3}=9.$
If $a_{1} > b_{1}$ then Adam scores one point; if $a_{1} < b_{1}$ then
Betty scores one point; and if $a_{1}=b_{1}$ no points are scored.
Similarly for $a_{2},b_{2}$ and $a_{3},b_{3}.$ The winner is the
player who scores the greater number of points: if the socres are
equal then the game is drawn. Show that, if Betty knows the numbers
$a_{1},a_{2}$ and $a_{3},$ she can always choose her numbers so
that she wins. Show that Adam can choose $a_{1},a_{2}$ and $a_{3}$
in such a way that he will never win no matter what Betty does. 
Now suppose that Adam is allowed to write down two triples of numbers
and that Adam wins unless Betty can find one triple that beats both
of Adam's choices (knowing what they are). Confirm that Adam wins
by writing down $(5,3,1)$ and $(3,1,5).$