Simple Card Game Of Ups and Downs
Leroy Quet
Start with a deck of 2n cards labeled 1 through 2n, one distinct
number per card. (n is some number, such as 10.)
Shuffle the deck, and deal n cards to each player.
Each player secretly arranges their n cards in any order they choose.
The players then, without further arranging their cards, take turns
alternately placing cards, in order, face-up onto a common pile
between the players.
If a card placed in the central pile is greater in numerical value
than the last card placed down by the previous player, then player 1
gets a point.
If a card placed in the central pile is smaller in numerical value
than the last card placed down by the previous player, then player 2
gets a point.
Continue until all 2n cards have been placed in the central pile.
The player with the greatest number of points wins.
-
Variation: Play with a standard deck of 52 cards. (No jokers.)
(Therefore, in this case, n = 26. Ace =1, Jack = 11, Queen = 12, King
= 13.)
Same rules as before, but if a card is of the same value as the
previous card, then player 1 gets a point if the newly placed card is
of a red suit (hearts or diamonds), and player 2 gets a point if the
newly placed card is of a black suit (ace or clubs).
Is this game unoriginal? It sounds familiar, maybe just because it is
so simple.
Thanks,
Leroy Quet
Leroy Quet
pile, they MUST place them in the order they originally ordered them,
from left to right in their hand (or from top to bottom in their own
face-down card pile).
Leroy Quet
psychological game -- trying to guess in what order your opponent will
play their cards?
By the way. More of my games (many of which are more interesting than
this game) at:
http://gamesconceived.blogspot.com/
Thanks,
Leroy Quet
scattered
Interesting game. In effect, each deal specifies a two player zero sum
game which in principle has an optimal strategy (though perhaps
mixed). Some deals make the game blatantly unfair (with the extreme
case of one player getting the top half of the deck).
Suppose that player 1 recieves cards 1,3,5, ..., 2n-1 and player 2
recieves cards 2,4,...,2n. It is obvious that any pure strategy that
player 1 adopts is defeated by one of player 2's (the one in which
player 2 happens to always play the successor of the card that player
1 plays - this assures player 2 will get at least n of the 2n-1 total
points. It is irrelevant that player 2 is unlikely to know what this
strategy actually is - its existence demonstrates that player 1 has no
optimal pure strategy). Player 1 can almost do something similar to
player 2 - but not quite since their 1 always loses and player 2's 2n
always wins when placed down. Thus most pure strategies of player 2
are defeated by some strategy of player 1. I suspect that all of
player 2's are defeated by at least 1 of player 1's (at least for n
sufficiently large) - but I haven't bothered to prove it. I suspect
that this deal is (slightly) unfair to player 1. Does *any* deal lead
to a fair game?
The game reminds me of a modification to the card game war called
Napolean's War that I read about here: http://www.pagat.com/invented/war_vars.html
-scattered
scattered
Math before enough coffee doesn't mix. I misread how the game was
scored, thinking that either player got a point if the card they
played was larger than the previous card. Reading the game more
carefully, in the case where player 1 gets 1,...,n and player 2 gets n
+1,...,2n, player 1 wins in all cases by a score of n to n-1, in the
other case I mentioned it seems that it is player1 with the advantage
since they could (given psychic ability) always play the predecessor
of the card player 2 is about to play.
An interesting variation on the game is to start with the same
scenario but determine the winner in a different way. The two players
jointly generate a permutation of 1,2,...,2n. Declare player 1 the
winner if it is an odd permutation and player 2 the winner if it is
even.
I think that this game is fair given any deal with simply shuffling
your deck randomly an optimal mixed stratgey, but I haven't proved it
rigorously.
-scattered
Ilmari Karonen
>
> An interesting variation on the game is to start with the same
> scenario but determine the winner in a different way. The two players
> jointly generate a permutation of 1,2,...,2n. Declare player 1 the
> winner if it is an odd permutation and player 2 the winner if it is
> even.
> I think that this game is fair given any deal with simply shuffling
> your deck randomly an optimal mixed stratgey, but I haven't proved it
> rigorously.
Should be, yes. Either player can always flip the outcome of the game
by swapping any two of their own cards: thus this game is essentially
equivalent to the well known "matching pennies" game.
--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
scattered
> On 2009-12-14, scattered <still.scatte...@gmail.com> wrote:
>
>
>
> > An interesting variation on the game is to start with the same
> > scenario but determine the winner in a different way. The two players
> > jointly generate a permutation of 1,2,...,2n. Declare player 1 the
> > winner if it is an odd permutation and player 2 the winner if it is
> > even.
> > I think that this game is fair given any deal with simply shuffling
> > your deck randomly an optimal mixed stratgey, but I haven't proved it
> > rigorously.
>
> Should be, yes. Either player can always flip the outcome of the game
> by swapping any two of their own cards: thus this game is essentially
> equivalent to the well known "matching pennies" game.
That is roughly what I was thinking. A variation of the game would be
to have the two players play a card of their choosing at each stage
(rather than sequence the cards in advance). In this case it is easy
to see that player 2 has a winning strategy since their second to last
play is the last play in which either player has any choice and one of
their two choices will result in an even permutation. Trivial as a
game, but proving that player 2 has a winning strategy might make for
an interesting homework problem in an introductory abstract algebra
course.
nate
> Is this game unoriginal? It sounds familiar, maybe just because it is
> so simple.
>
> Thanks,
> Leroy Quet
this was cross-posted to rec.games.abstract and rga
is a group studying abstract games with no element of chance.
while interesting in principle, this was off-topic over here.
cheers.
- nate
marks...@gmail.com
>
> this was cross-posted to rec.games.abstract and rga
> is a group studying abstract games with no element of chance.
>
Normally, yes, but Leroy has posted a lot of abstract games here. It''s
interesting, at least to me, what occasional, original, luck based games he
might produce. It's not like we can't spare the bandwidth. Amputating Bill
freed up a sizeable chunk.
I wouldn't want to open the flood gates on luck based games. Quite the
opposite. But, that being said, an abstract game is so awash with luck
already that a tumbling pair of dice only adds a little more. Every (turn
based) abstract game begins with a coin toss. And no, the pie rule doesn't
cancel that out. Even if perfect play were known the pie rule might not
cancel out the luck of the coin toss. The possibility of invoking the pie
rule might only transfer, though greatly diminish, the advantage from Player
1 to Player 2. Assuming that the game's perfect play is *not* known by the
players, since otherwise it wouldn't be a game for the players, the players
would only be guessing about which moves to call pie rule on, said guesswork
being subject to the laws of chance (i.e. luck). Add that luck to the luck
of the coin toss. It's cummulative. Moving on to the gameplay... When
pitting your guesswork against the guesswork of your opponent there's only
one deciding factor - luck. You don't even have to be equally matched. You
might get lucky and beat a more skilled player. A couple of clever moves
are made - obvious moves that suddenly and surprisingly became available to
you at precisely the right moment, definitely not planned in advance.
-Mark
Mark Steere Games
http://marksteeregames.com