Funny card game
I have a special deck of cards. Each card has four numbers, one on each corner. Each number is either 1, 2, or 3. Every combination of numbers is in the deck, so there are 81 cards in total.
I deal out these cards face up, and you pick up groups of three cards that satisfy the rule: If two cards have the same number in the same corner, then the third card must also have that number in that corner.
So you could pick up three cards that all say 3 in the upper left corner, or you could pick up three cards which respectively have 1, 2, and 3 in their upper left corners--but you can't pick up two cards that say 1 and another that says 2, this would violate the rule.
The hard part is, this rule must be satisfied for all four corners simultaneously.
So this is a valid set, if we write out the numbers on each card:
2, 1, 3, 2
2, 3, 3, 1
2, 2, 3, 3
And so is this:
1, 3, 2, 2
2, 2, 3, 1
3, 1, 1, 3
But this is not:
1, 2, 2, 2
2, 2, 3, 1
3, 2, 3, 3
(Oops- the third column does not satisfy the rule, there are two "3's and one "2")
If this is confusing there is an actual game you can play with, with shapes and colors instead of numbers.
Some questions to think about:
1. Not counting different orderings of the three cards, how many distinct groups of three cards that meet the rules are there?
2. We can keep picking up groups of three cards that meet the rules. Eventually we will be left with some number of cards left over that cannot be made into a set. How likely is it that three cards will be left over?
3. How many cards could you possibly lay out at once and still have no matching set? Can you come up with an upper bound at least?
Monday, April 5, 2004
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