George Jaros
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Pirates Enjoy Math (Don't Say Arrrrrrithmetic) is a micro card game for 2-4 players in which players use math and the Order of Operations to try to become the pirate with the most Loot.

The game consists of two rounds per hand and as many hands as players would like. The first phase is a card drafting phase where players try to build a Crew that will help them get the Loot. The second phase is where players use their Crew to create a mathematical formula that will get them closest to the Loot Value. The winner gains the Loot, plus each player has a chance to earn Bonus Loot based on the cards they play.

PnP rules can be found here:
http://georgejaros.com/Files/Pirates%20Enjoy%20Math%20(Don%2...

PnP Cards can be found here (2 pages, 18 cards):
http://georgejaros.com/Files/Pirates%20Enjoy%20Math%20(Don%2...

Note: Artwork is just clipart at this point. The card backs will have condensed rules on them (in the form of a parchment with the Pirate Code), but the PnP files do not currently have backs.

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Martin Windischer
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Poor players who have to calculate 9^3
Just one question: Why didn't player 3 choose 9+5+4 in your example?
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George Jaros
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MartinWin wrote:
Just one question: Why didn't player 3 choose 9+5+4 in your example?


Two reasons...

1) Because when I originally set up the sample cards I had player 3 with a hand of 1, 4, 5, 5. At some point last night I changed his hand to 1,4,5,9, I think so that player 4 didn't have a hand of 6,7,8,9. And I never went back to check if swapping the cards gave another possible solution.

2) Player 3 isn't good at math.

Thanks for pointing that out though, I'll tweak the cards to make it work better. But that points out one potential issue I'm having with the game, it's too easy to get one of the target values. I'm thinking of having different card counts for different numbers instead of just 2 of each and see how that works...
 
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George Jaros
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MartinWin wrote:
Poor players who have to calculate 9^3


I'm considering having a cheat-sheet for the cubes... But maybe that's too much help =) Pirates wouldn't help each other, but they would cheat... It's a tough decision.
 
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George Jaros
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OK, I've fixed the example in the rules so that player 3 now doesn't have a correct solution.
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Martin Windischer
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gjjaros wrote:
2) Player 3 isn't good at math.


This explains it

And I have another rule question: Let's say the cards 3/ and 9- are in the loot. Does that mean that there are 4 possible Loot Values?
3-9=-6
9-3=6
3/9=1/3 rounded to 0
9/3=3

And are divisions always immediately rounded down?

So can I make (7/3)*6 to get 12?
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George Jaros
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Loot has to be a positive (or 0) integer, so in your example 6 and 3 and 0 would be possible Loot Values.

Divisions are only rounded down in the Loot Value, but rounded up in the points you lose when scoring if you don't have an exact match. So in your example you'd have a Crew Value of 14. But if you did (7/6)*3 your Crew Value would be 3.5 and if the Loot Value was 4 you'd have a difference of .5 and lose 1 Gold. I can add examples of that in the instructions. I'm just wondering how I'm going to condense the rules onto one card. I'm definitely going to have to link to a page with more in-depth rules and examples, but that's been the case for all my games so far...

So if you had a 3,6,7 (and something else) in your hand you could have Crew Values of:

(7-6)^3 = 1
6-7+3 = 2
3(7-6) = 3
(7*3)/6 = 3.5 (although this wouldn't be needed since you can get both 3 and 4 with other methods)
7-6+3 = 4
7-(6/3) = 5
7-(3/6) = 6.5
7+(3/6) = 7.5
6/3 + 7 = 9
6-3+7 = 10
6*3-7 = 11
7(6/3) = 14
3*7-6 = 15
3+6+7 = 16
7(6-3) = 21
6(7-3) = 24
6*3+7 = 25
7*3+6 = 27
(6^3)/7 = 30.86...
3(7+6) = 39
7*6+3 = 45
(7^3)/6 = 57.167...
6(7+3) = 60
7(6+3) = 63
7*6*3 = 126

Etc. I'm sure I missed a bunch. Also, some numbers can be reached multiple ways, e.g. 7*6-3 = 39
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George Jaros
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Thinking more about the division I think I'm going to limit the Divide cards to 2 and 3 (on the flip sides of the Exponent cards). I think that'll give better variation of numbers since so many divides would have a target Loot Value of 0, and anything with a Divide is going to have a low value anyway. I'll add a new PDF shortly.

Thanks for forcing me to think through this more Martin!
 
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Martin Windischer
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To avoid the many Loot Values of 0 I think you should force the lower card to be the smaller number the divisor (similar to the forcing of 2 or 3 be the exponent).
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George Jaros
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Yes, I thought of that as I was editing the rules. Updated PDFs now with 2 and 3 the only Divide cards (and the only Exponent cards) and a caveat that cards with a Divide suit use that number as the denominator. That way the only time you should get 0 for a target is if the other number is 1 or you have a 3 of Divide and a 2.

Thanks again for your help!
 
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George Jaros
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Images done for two of the cards. My friend Neal is doing the characters, the layout and background is by me. Likely won't have all done by tomorrow, but if the game does well we'll get the rest of the artwork done soon!


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George Jaros
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Logo done:

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George Jaros
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Background artwork is complete. Just waiting on more characters from my friend Neal, however I don't think he'll have them all done by tonight for the end of the contest. =(

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