Wednesday, March 08, 2006

Not according to Hoyle

We've invented a new card game, for those who love mental math. Or who have calculators handy.

We call it Pemberley, because it's our card game and we can call it what we want to. It's for two or more players--we're not sure how many more, because we haven't tried it with more than three.

The objective is to win.

One wins by scoring the most points.

One scores points by getting the cumulative total of cards played so far to land on certain selected numbers.

If you're still reading, here are the rules:

Dealer deals seven cards to each player. Player to right of dealer cuts the deck, and dealer turns over the exposed card. This card is the starter.

Cards are treated as numbers 1-13, Kings being 13, naturally. Each suit is assigned an operation, as follows:


The players set down cards alternately (keeping their piles separate), calculating the new total based on applying the operation and number indicated by the card last played. That sounds confusing, but it's quite simple, as you'll see when I get to a sample round.

All totals must be whole numbers, and they cannot be greater than 500 or less than -500.

Points are granted as follows:

  • Landing on 1 is worth 1 point the first time, then 1 more point every time it is reached in the game.
  • Landing on 12 or a multiple of 12 is worth 1 point.
  • Landing on a prime number is worth 1 point. (List of primes.)
  • Landing on 0 is worth 2 points
  • Landing on 7 or a multiple of 7 is worth 2 points.
  • Landing on 13 or a multiple of 13 is worth 3 points.

Because points are acquired as play proceeds, it's easiest to use a cribbage board to keep score. If you don't have one, perhaps hash marks would work.

Any cards that cannot be played will deduct one point from the player's final score.

Once all cards have been played, the players gather up their own cards and arrange them in sets which, again working off the starter card, make numbers that score points. Any card that cannot be used in one of these sets is another card to deduct from the final score. These points are to be added to the ones scored in the first part of the game, extra cards are then subtracted, and the winner is the one with the most points.

OK, here's a sample game to make it clearer: ♥=Add ♠=Subtract ♦=Multiply ♣=Divide

Starter card is 3♣ (Disregard the suit on the starter card), so starting value is 3.

P1: Q♦, 3*12 is 36. Divisible by 12, 1 point.
P2: Q♥ 36+12 is 48. Divisible by 12, 1 point.
P1: 8♣ 48 / 8 is 6.
P2: 2♣ 6/ 2 is 3. Prime, 1 point.
P1: 8♠ 3-8 is -5. Prime, 1 point.
P2: 5♣ -5 / 5 is -1.
P1: 8♥ -1 +8 is 7. Divisible by 7, 2 points.
P2: 4♦ 7*4 is 28. Divisible by 7, 2 points.
P1: 7♣ 28 / 7 is 4.
P2: 3♠ 4-3 is 1. First hit of 1, 1 point.
P1: K♦ 1*13 is 13. Divisible by 13, 3 points.
P2: A♥ 13+1 is 14. Divisible by 7, 2 points.
P1: 7♠ 14-7 is 7. Divisible by 7, 2 points.
P2: Has a 10♣ unused, which is set aside.

At this point, P1 has 9 points and P2 has 7 points. Now they take up their hands and lay them out to make the maximum points they can find:

P1: (8♥, 7♠, Q♦, 8♣, 8♠), i.e. ((3 (starter) +8-7)*12)/8)-8=-2, prime, for 1 point.
(K♦) 3*13=69, divisible by 13, for 3 points.

P1's score is now 13, but he was unable to use his 7♣ this time, so one point is deducted, and his final score is 12.

P2: (Q♥, 5♣) i.e., (3+12)/5=3, prime, for 1 point.
(3♠) 3-3 =0, for 2 points.
(4♦, 2♣, A♥) 3*4/2+1=7, for 2 points.

P2's score is now 12, but the 10♣ must now be deducted, so his final score is 11, and P1 wins. If you want to do multiple rounds, you can set a winning point value and keep a running tally.

We welcome thoughts and suggestions (except for "You guys are hopeless nerds.")


the Joneses said...

No, I wouldn't say necessarily hopeless. I'm sure there's a gleam of hope...somewhere...hiding behind the cloud of nerdiness. :)

Actually, it sounds pretty fun to me. My lovely non-math-loving wife, however, would probably consider it equivalent to the seventh circle of hell.


The Duke said...

Two notes: the one who cuts the deck is the one who plays first. Also, the points for hitting #1 increase only in that round, not through the entire game. In the next round, landing on one for the first time is again only one point.

Play continues through the end of the round in which one player tops 30 points.

DJ said...

It appears from the sample game that when you fulfill multiple scoring criteria, you have to pick one of them. (I.e., if you got 91 you'd have to pick either the 2 points for multiple-of-seven or the three points for multiple-of-thirteen.) Is this what you intended?

Rachel's Jeremy said...

Good question. I wondered that myself.

Personally, I would think you'd get to score both!

Queen of Carrots said...

No, that's not correct; you would get 5 points for landing on 91. But you don't get double points for landing on squares; so 144 is only 1 point, not 2.

DJ said...

In the sample transcript when player 1 makes a total of 7, you listed it as being worth 2 points (divisible-by-seven). Shouldn't there also be a point because 7 is prime, for a total of 3 points? Or am I misunderstanding the rules?

Queen of Carrots said...

Ah. No, the primes and the multiples don't overlap like that. Or, if you like, the prime point is included in the 2 points for hitting 7 (or the 3 points for hitting 13). But the different multiples do.

Juliana M said...

Does the one point for hitting 1 for the first time only apply to 1 and not -1?

P.S. Having played this game last weekend with QOC and DOB, I can verify that it is lots of fun. The real points stack up when you get to numbers that get you double-points like 91 (divisible by both 7 and 13 gets you 5 points).

P.P.S. Each round can take a considerable amount of time, especially for first-time players. In case you're interested, I tied for first place with DOB. We didn't have time for another round.

The Duke said...

Correct: only 1 counts, not -1.

Uncle Steve said...

Aunt Bettie and I played a hand of Pemberly over the weekend. In addition to placing us into the top 20 Pemberly players in the world, it raised a few questions.

1) The initial seven cards in each player's hand - are those known to the player, or flipped in sequence as they are stacked face down (like the game of war)?

2) The rule that says "All totals must be whole numbers" - does that mean "Non-integer results of operations force a card to be declared unused"? Or is rounding acceptable? Is 8 divided by 10 equal to 1, or is that not a valid operation?

3) What happens when the operation bumps against the limit of 500 (or -500)?

4) When is the deck shuffled? At the end of each round, or when no more rounds can be played?

Inquiring Pemberly players want to know.


Queen of Carrots said...

1) The cards are known; otherwise their play might violate the other rules.

2)Rounding is unacceptable; thus if a division would result in a non-integer, the card cannot be played.

3)If playing a card would cause you to go over 500 or under -500, then you may not play the card.

4)We shuffle at the end of this round.

Anonymous said...

we used to play a math-camp card game that was a version of "war", I guess--or maybe a game like set.

you use the numbered cards, aces as 1 (or 11, I guess, if you want).

you lay out 4 cards, face up.
The goal is to find a way to string operations and parentheses and the numbers on the cards to create a true math equation. The first to shout out a true math equation wins the cards. The allowed operations were addition, subtraction, multiplication, division, exponentiation, and anything else you wanted--logs, mods, etc.

for example:
4 hearts
2 clubs
10 clubs
3 diamonds

(10 - 4) / 2 = 3

or (equivalently)
3 *2 + 4 = 10

maybe you will enjoy this game too.

The Duke said...

Who is anonymous?