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Subject: Chaining Outbreaks rss

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Todd Woodward
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Bowling Green
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I wasn't sure whether this should be in Rules or Sessions. In a just completed game we drew St. Petersburg from the infection deck which caused an outbreak. We thought we had lost due to all the blue cubes being placed. However, a blue cube wasn't placed in Moscow due to having 3 black cubes already in it. Did we really just win because 1 blue cube was left, or is the Moscow outbreak considered blue since blue caused the original outbreak? Has anyone else had this happen?
 
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Ax Bits
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Ottawa
Ontario
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Regardless of the number of non-blue cubes in Moscow, an outbreak in St Petersburg would require placing a blue cube in Moscow.

And you don't lose when all the cubes of a colour are placed. You lose when you need to place a cube and you can't because all cubes of that colour have been placed.
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DC
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Grand Rapids
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Nope, you can have up to three of each color on each city. So yes, you would have had to place a blue cube on Moscow.

Any time you have to put a cube on a city which already has 3 cubes of that color, you must then outbreak that city (and put blue cubes on every city adjacent to it) -- unless it has already outbroke in that chain, in which case you wouldn't outbreak again. So even if Moscow had had 3 blue cubes in it, you wouldn't be out of the woods yet!
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Isidoros Sarantinos
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We have 3 cities of the same color (black). Let's say A, B, C. They form a triangle on the map and have 3 cubes each. During the infector phase, one of them is drawn. A chain reaction is under way. I say 3 outbreaks occur. Am I right?

My questions : How many outbreaks can happen in one city during this phase? What if A and B were drawn?
 
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Caleb Martin

Rhode Island
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Paixtaras wrote:
We have 3 cities of the same color (black). Let's say A, B, C. They form a triangle on the map and have 3 cubes each. During the infector phase, one of them is drawn. A chain reaction is under way. I say 3 outbreaks occur. Am I right?

My questions : How many outbreaks can happen in one city during this phase? What if A and B were drawn?


The rule is that a city can't outbreak twice in the same chain reaction, not twice in the same turn. If you draw A as the infector, then A, B, and C would all outbreak, so put one black cube on every city adjacent to that triangle and move the outbreak track up by 3. If you draw B as your second infection card for the turn, then yes, those three cities all have a second outbreak, since this is a new chain reaction. Your turn would end with all cities adjacent to A, B, and C getting 2 new black cubes, and 6 outbreaks total. Good luck...

Edit: so to answer your question explicitly, the worst-case scenario would be having the infection rate all the way up at 4, and having multiple adjacent cities all with 3 cubes of the same color. If you draw 4 of those cities as your 4 infection cards, every single one of them has four outbreaks, and you more or less lose immediately (there probably wouldn't even be enough cubes to resolve the latter outbreaks).
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Isidoros Sarantinos
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Iranon wrote:
The rule is that a city can't outbreak twice in the same chain reaction, not twice in the same turn. If you draw A as the infector, then A, B, and C would all outbreak, so put one black cube on every city adjacent to that triangle and move the outbreak track up by 3.

And what if the adjacent cities to that triangle are connected to both outbreaking cities? Then they get 2 cubes instead of one, don't they? Please correct me if I am wrong.
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Carl Brousseau
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Paixtaras wrote:
And what if the adjacent cities to that triangle are connected to both outbreaking cities? Then they get 2 cubes instead of one, don't they? Please correct me if I am wrong.


You're right.
 
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tim thorson
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and adjacent cities are connected with red lines correct?
 
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Robert Sheets
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Lebanon
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Quote:
and adjacent cities are connected with red lines correct?


you are correct.
 
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