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The Oracle of Delphi» Forums » Rules

Subject: Monster setup rss

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Jon Ben
Canada
Vancouver
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Of course I've been up all night! Not because of caffeine, it was insomnia. I couldn't stop thinking about coffee.
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I just want to clarify something about the monster setup.

Quote:
Take as many Monsters of each color as players
participating in the game. Place any 2 different Monsters
on the 3 marked Monster Islands. Distribute the remaining
Monsters evenly among the remaining 6 Monster
Islands, so that no color occurs twice on any island.


So there are 3 special monster spaces that are marked, and then 6 unmarked monster spaces. The marked spaces each get two monsters of different colours. Then the rest of the monsters get randomly divided as evenly as possible between the other 6 spaces but not allowing the same colour more than once.

This means that in a 2p game every unmarked space will have exactly 1 monster. In a 3p game every unmarked space will have exactly 2 monsters. In a 4p game every unmarked space will have exactly 3 monsters.

Is all of that correct? Why bother with the marked spaces, are they on the ring tiles?
 
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Jay Cat Five
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Monterey
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Non-authoritative: I supposed it made it easier to systematically distribute 12, 18, and 24 tiles among nine tiles.

I hope that was marginally helpful.
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Jon Ben
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Of course I've been up all night! Not because of caffeine, it was insomnia. I couldn't stop thinking about coffee.
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ngajoe wrote:
Non-authoritative: I supposed it made it easier to systematically distribute 12, 18, and 24 tiles among nine tiles.

I hope that was marginally helpful.


So you think it's just there so the maths work out nicely. That's certainly a component of it, but I was curious if the location of them on the tiles is special in someway. Do you know on which tiles the marked spaces appear?

 
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Ralph Bruhn
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ngajoe wrote:
Non-authoritative: I supposed it made it easier to systematically distribute 12, 18, and 24 tiles among nine tiles.

I hope that was marginally helpful.
Official answer: Exactly this is the only reason!


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Matthias M
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I suppose it is just the math. There are 12/18/24 tiles to distribute in a 2/3/4 player game, as ngajoe already said. The greatest common divisor of these numbers is 6, so I guess that publisher and author (who is a math teacher, BTW) tested 6 monster island first and X monsters for each island with X = number of players.

What if playtesting showed that there should be more monster islands? The solution is straightforwand: take out a group of six and distribute them evenly among 1/2/3/6 secondary monster islands (with 6/3/2/1 monsters each), and then put X-1 monsters on each regular monster island. If the number of monsters on the secondary islands should not deviate much from the number of monsters on the regular islands, you get on optimization problem which has exactly the solution you can actually find in the game: 2 monsters on each of the three secondary islands and 1/2/3 monsters on the six regular islands, depending on player count.

It's the logical solution if 6 monster island turned out to be too few in playtesting.
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