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Subject: Quantum 6 - a no-cards variant rss

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Richard Walter
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This is a very nice 2-player card game.

However, my son & I have come up with a no-card variant of this that we're calling "Quantum 6" that merges this game with another one (whose name I've forgotten that was referenced here on the geek sometime in the last few months.)

The idea is that you don't use physical cards, you just pretend to have them.

So, each player pretends to have been dealt his hand of 6 cards from the deck of 16.

A player has the same options as normal: Play a card to the field, discard a 6, take a card from the field, etc... When a player plays a card that was previously undefined, he may decide at that point which card it is.

However, a declaration of a card must maintain logical consistency with the deck. ie: only two 1's exist in the deck, so once two cards have been declared as a 1, then no more 1's may be declared.

When one player declares the hand over, there still may be undefined cards in the player's hands. In this case the player who did *not* declare the hand over defines 1 card in his hand; the hand-ending player then declares 2 undefined cards in his hand, and the players alternate declaring 2 undefined cards until each hand is completely known and the winner determined.

A player can only the declare the hand over if there exists a possible hand of cards that he could have that is less than the field. However, in this game, there is a chance (due to the final declarations of cards in hand) that both players' hands end up larger than the field. In this case, the player who did not declare the hand over is the winner.


Example of play:
. Both players have 6 undefined cards in their hands and the field is empty.
. Player A plays a 2 to the field. He now has 5 undefined cards in his hand.
. Player B discards a 6. He now also has 5 undefined cards in his hand. The field total is 2.
. Player A discards a 6. He now has 4 undefined cards in his hand.
. Player B plays a 2 to the field.
. Player A plays a 4 to the field.

At this point, the field contains 2,2,4.
There are two 6's discarded out of the game.
Player A has 3 undefined cards in his hand.
Player B has 4 undefined cards in his hand.
But, none of the undefined cards can be 2's, because both of them have already been played.

. Player B discards a 6. He now has 3 undefined cards in his hand.
. Player A declares the hand over.

Each player still has 3 undefined cards in hand, so they now reveal them:
Player B declares a 1.
Player A declares a 1,3 (A can't declare another 2 because both have been played, and can't take two 1's because player B already declared one, leaving only 1.)
Player B declares a 4,3
Player A declares a 4. (A can't declare 1, 2 or 3, because they're all gone, and the smallest one left is a 4.)

Both players have a total of 8, and the field has 8, so player A wins since she was the player who declared the hand over.

As a variant on the variant, we've thought about allowing a new type of action where a player can define a card in the 4 that aren't used in the hand to further constrain the possible cards in play without affecting the value of the field.

-Richard
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