Sunday, May 22, 2011

How do you bid in Spades?

Anybody who knows me knows that I enjoy playing games, particularly card games. And when it comes to card game, I'm particularly fond of playing spades.

Spades is a four player game where players pair up on teams and collect "books", competing to reach some predetermined score. A book is a set of four cards given to the team who plays the highest card under the rules of the game, which I will explain later. Teams collect points by bidding [or trying to predict the number of books they will collect]. If a team does not collect the number of books they bid, they will lose ten times their bid. If a team does collect this sum of books, then they gain ten times their bid (I'm assuming a simple version of Spades here with no sandbags to simplify the explanation). 

So we can see the importance of precise bidding. What then should be the standards of good bidding? 

We can start by mathematically describing the rules of the game. 

1) Trumps / Spades beat allother cards (hence the name of the game). 
for all x \in Spades, y \notin Spades x > y
(because there are only four suits, the statement y \notin Spades could be replaced by y \in Hearts \cup Clubs \cup Diamonds). 

2) In every suite, the cards 3, 4, ..., K, A follow their natural order, i,e,

3D < 4D < 5D < 6D < 7D < 8D < 9D < 10D < JD < QD < KD < AD
3H < 4H < 5H < 6H < 7H < 8H < 9H < 10H < JH < QH < KH < AH
3C < 4C < 5C < 6C < 7C < 8C < 9C < 10C < JC < QC < KC < AC
3S < 4S < 5S < 6S < 7S < 8S < 9S < 10S < JS < QS < KS < AS

(There are many ways of playing this game. I'm going to assume the deck is set up the way I originally learned it. This is sometimes called "Joker, Joker, deuce, deuce", which refers to the top 4 Spades - the big joker, the little joker, the high deuce, and the low deuce. Under this setting, there is no 2C and no 2H, and generally the 2D is the high deuce and 2S is the low deuce, but this generally depends on where the game is being played and who else is playing). 

And by 1) we have that AD < 3S, AH < 3S and AC < 3S. This establishes a partial ordering amongst the suits and cards. Note that this is not a total ordering because ther is no comparison between a JH and a 3C. This is handled by the next rule. 

3) All players must follow suit unless a player has no cards of that suit in their hand. This means that the suit of the first card thrown determines the type of cards that all other players must choose from. If a player does not have that suit in their hand, then they are free to play anything, but the only suit that can beat the original suit is a spade (per the partial order). 

4) If a plater does not follow suit (and is caught), it is called reniging and costs that team 3 books. 

5) The first player to play a card in the first book is the player to the dealer's right. The first player to play a card in all successive books is the player who won the last book. 

From this we can see that the only cards that beat Aces are the Spades. One strategy is to play aces first, in which case the only way an ace doesn't bring in a book is if a player on the opposing team has no cards of that suite. What is the probability that an Ace walks then? 

We can set this up as the following fraction: 

number of ways that another player can be dealt a hand without receiving a card of this suite
total number of ways that another player can be dealt a hand. 

How many cards are left that are in this suit? Suppose that I have x cards of this suit in my hand. There are 12 cards total in a (non-spade) suit. So this leaves 12 - x cards in this suit that are not in my hand. 

How many cards are left that are not of this suit? Suppose again that I have x cards of this suit in my hand. There are 52 cards total, 13 of which are in my hand (13 = 52 / 4), leaving 52 - 13 = 39 cards remaining. Each (non-spade) suit has 12 cards.  So there are 39 - (12 - x) = 39 - 12 + x = 27 + x cards that are not in this suit and not in my hand. 

If a player has no cards in this suit, then all their cards come from this set of size 27 + x. 

So this equation can be represented as 

27 + x

This simplifies to
(27 + x)! 26!
(14 + x)! 39!

I've taken a basic approach to this game and there are many further questions to be answered about the game of spades. I could continue and find the probabilities for other non-spades, as well as the probabilities for spades. But my time has run out and this has been a fun exercise to play around with one of my favorite games and one of my favorite languages.

    1 comment:

    ashaki said...

    I love your enthusiasm for the game, but all that math may have just deterred me from ever playing. But you have fun. :)

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