# word smithing skills needed

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emxibus
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Joined: 10/24/2008

I'm trying to word the following mechanic, so that it fits on half of a poker sized card. Fewer words the better :)

There are 2 dice, call maneuver dice, that are rolled. There is a card in your hand that has a die (1 - 6) on it. You can play the card if you can figure out a way to add and subtract, in any order, the two maneuver dice and the die on the card so that the total is divisible by 3. You have to use all three dice.

for example: you maneuver dice are 5 and 2. the die on the card is a 1. To play the card, you could use the following math:
5 - 1 + 2 = 6. 6 is divisible by 3.

Here's my current attempt:

Play this card when adding and subtracting, in any order, the two maneuve dice and this card results in the total being divisible by three.

Cogentesque
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Joined: 08/17/2011

Play this card when adding and subtracting. Add or subtract this card and the maneuver dice in any order. The result must be divisible by three.

?

akanucho
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Joined: 11/10/2009
word smithing skills needed

Short: You may play this card if adding and subtracting the dice value of this card and the maneuver dice produces a multiple of three.

Shorterer: Play if adding and subtracting this card and the maneuver dice produces a multiple of three.

If exactly two out of three dice are either 3 or 6, the three cannot be added or subtracted to make a multiple of three. In every other case, they can. Therefore:

Shortest: Play if exactly two dice are not 3 or 6.

But this phrasing is awkward. Instead, I would prefer one of the following:

Clearer: Play unless only one die is 1, 2, 4, or 5.

However, if your intention is to involve multiples of three in many aspects of the game, this completely misses the point.

SLiV
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Joined: 10/21/2011
Mathematical remarks

The word forging apart, two things I'd like to add:

1: If you're only adding and subtracting, you can leave out "in any order" all together; the order in which you add and subtract is completely irrelevant (so a+b-c = a-c+b = -c+a+b = a-(c-b) etc).

2: Akanucho remarked correctly that the only way you cannot play the card, is if exactly two dice are 3 or 6. The chance of that happening is 3! / 2! × 1/3 × 1/3 × 2/3 = 6 / 27 ≈ 22%. Unless of course your game revolves around multiples of the number 3, this mechanic seems a bit gimmicky for what is almost the same as a 5-sided die (or a six-sided die where 6 is reroll).

Saying "play if exactly two dice are not 3 or 6," is incorrect, though, since {5, 1, 2} (i.e. three dice are not 3 or 6) and {3, 3, 6} (i.e. zero dice are not 3 or 6) are also solutions; I think you mean "play unless exactly two dice are 3 or 6".

emxibus
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Joined: 10/24/2008
Thanks!

Thanks for the input.

Some background:

This is for my gladiator game. I've been trying to come up with a way to entertain the players not currently fighting (fights are only between two players). I thought I would try this little mini game. The two players fighting have a maneuver die. The players watching would draw 3 observation cards (with the die on it), and watch the fight. During the fight the maneuver dice are maniplated by different moves the gladiators can perform. The maneuver dice usually increase in value (six being the max). The players watching would play an observation card (one per round. There are multiple rounds per fight) when the condition on the card were met. The goal would be to make 5 observation before the fight was over. If the goal was met they would get a bonus for their gladiator's next fight.

Now that I know that the cards can be "played unless exactly two dice are 3 and 6", the solution to the mini game doesn't take much thought and is not as interesting to me. O'well, back to the drawing board. I'm glad we have some mathematicians here, I don't think I would have figure that out on my own.