Cláudio Buffara and William McWorter
Tue, 25 Mar 2003
The Math Lotto
The "Math Lotto" betting card is a 6×6 table. The gambler is to mark 6 of the 36 slots in the card. The official result is published with 6 slots chosen as the "LOSING SLOTS". The gambler wins if he doesn't pick any losing slot.
 Prove that it is possible to fill out 9 betting cards so that at least one of them is a winner. Describe the markings on the 9 cards.
 Prove that 8 cards are not sufficient to ensure a win.
Solution

Assume each betting card consists of a 6element subset of
{1, 2, ..., 36}. Fill out 9 cards as follows:
C_{1} = {1, 2, 3, 4, 5, 6}
C_{2} = {4, 5, 6, 7, 8, 9}
C_{3} = {1, 2, 3, 7, 8, 9}
C_{4} = {10, 11, 12, 13, 14, 15}
C_{5} = {16, 17, 18, 19, 20, 21}
C_{6} = {22, 23, 24, 25, 26, 27}
C_{7} = {25, 26, 27, 28, 29, 30}
C_{8} = {22, 23, 24, 28, 29, 30}
C_{9} = {31, 32, 33, 34, 35, 36}
Form the sets:

A = C_{1} ∪ C_{2} ∪ C_{3} ∪ C_{4} ∪ C_{5}
B = C_{6} ∪ C_{7} ∪ C_{8}Let T be any 6subset of {1, 2, ..., 36}.
If T meets A in at most 3 elements or B in at most 1 element, then one of the cards in A or B is disjoint from T. If T meets A in 4 or more elements and B in 2 or more elements, then T meets A in exactly 4 and B in exactly 2 elements. Hence T is disjoint from C_{9}.
Conclusion: one of these 9 cards is a winner.

Let A_{1}, ..., A_{8} be any 8 6subsets. If an element x appears in three of the A_{i}, then choose x and one element each from the remaining five sets, making a 6element set meeting each A_{i}. Thus none of the A_{i} is a guaranteed winner. Otherwise, no element appears in more than two A_{i}. We count the ordered pairs
(A_{i}, x) , x in{1, ..., 36} in two ways:
sum of number of elements contained in each A_{i} = 8×6 = 48, and
sum of number of A_{i} containing each x = d_{1} + ... + d_{36}, d_{i} being the number of A_{j} containing the i^{th} element.We know each d_{i} is less or equal 2 from above. Hence there must be at least 12 d_{i}'s equal to 2. Let x be one of those contained in two A_{i}, say A_{i} and A_{j}. Then
A_{i} ∪ A_{j} ≤ 11; whence there is a y outsideA_{i} ∪ A_{j} contained in two A_{i}'s, say A_{k} and A_{l}, with all four of these sets distinct. Hence x, y, and one element each from the remaining four A_{i}'s makes a 6subset which meets all 8 of the A_{i}. Thus none of the A_{i} is a guaranteed winner.Contact Front page Contents Up
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