Monday, September 22, 2008

Quite A Stretch -- Up/Down Card Game

Here is a card game for two players. (Although it could be easily
modified for more players.)
This game seems slightly familiar to me. Are major aspects of it taken
from pre-existing card games?
All that matters in this game as far as the cards are concerned is
each card's numerical value (with Ace = 1, Jack = 11, Queen = 12, King
= 13).
Start by shuffling a deck of playing cards. (No jokers.)
Divide up the cards evenly between players. A player is not allowed to
see his/her opponent's cards until the cards are played. (The cards
belonging to each player that have not yet been played I will call a
player's "hidden hand".)
All cards once played in this game are placed face-up.
Each player starts their pile of cards by placing any one card they
choose (face-up, of course) down between the players. (So we have two
piles, one for each player.)
On each round (of two moves each) players take turns being the
offensive player and the defensive player.
The offensive player puts a card down (face-up) next to his pile. Say
that the top card (ie, the last card played by that player in the
previous round) in the offensive player's pile has a numerical value n
and the card he/she just put down next to the pile has a value m. And
say the card on top of the defensive player's pile has a value k. Then
the defensive player must, if he/she can, place a card (any card he/
she chooses from his/her hidden hand) down on top of his/her pile that
differs from k by less than or equal to the absolute value of (n-m)
AND is in the same numerical direction from k as m is from n. So, in
other words, if the card the defensive player plays has a value j,
then (m-n) has the same sign (+, -, or is zero) as (j-k).
And |m-n| is >= |j-k|. (|j-k|, the absolute value of the difference between the
card the defensive player last played and the card the defensive
player is now playing, must be any value from 0 to |m-n| {ie, from 0
to the absolute difference between the card that the offensive player
played in the previous round and the card the offensive player has
currently played}.)
(See example.)
If the defensive player cannot move, then he/she skips his defensive
move, and therefore does not remove a card from his/her hidden hand in
that round.
The players then move the cards that are next to their piles onto the
top of each pile (face-up).
Then, for the next round, the players switch who is offense and who is
defense, and the formerly defensive player then plays offense.
(So players play cards like this {if both players can move}: {player
1, player 2}, {player 2, player 1}, {player 1, player 2}, {player 2,
player 1}, etc.)
(Even if the defensive player did not play a card in the previous
round, the defensive player then immediately becomes the offensive
player of the next round and plays a card then anyway, whatever the
outcome of the previous round.)
The first player to run out of cards in his/her hidden hand wins.
Here is the beginning of a sample game:
Start: Player 1 puts down a 5, and player 2 puts down a 6.
Round 1:
Player 1 puts down a 10. (So player 2 must put down a 6,7,8,9,10, or
Player 2 puts down a 7.
Round 2:
Player 2 is now offense, and he puts down a 3. (So player 1 must put
down a 6,7,8,9, or 10. -- Since the card put down in round 1 by player
2 was a 7, and last card put down by player 1 was a 10.)
Player 1 places down an 8.
Round 3:
Player 1 then places down another 8. (So player 2 must put down a 3.)
Player 2 does not have a 3, so player 2 skips defensive move.
Round 4:
Player 2 places down a 7. (So player 1 must put down a 8,9,10,11, or
12. -- Note: Last card played by player 2 was a 3, from round 2.)
Player 1 places down a 12.
What would be a good strategy for this game?
Leroy Quet

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