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Machi Koro» Forums » Variants

Subject: The no purchase variant/s rss

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Jeff Pearce
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Hey,

So I've played this a few times and it can be a different spin on the game. We usually use a meeple for this, but you can use just a regular coin from the game, so you don't need to add anything for this variant.

We call it the no purchase variant.

I usually play base and harbour, but this will work with any expansion. Once any player has exceeded 10 coins in the game, this variant is activated. From then on, every player before they roll, must roll both dice to determine the no purchase card.

We set the game up in three rows of four, and the two dice will determine which card from the top left to the bottom right will remain unavailable for purchase for that player's turn.

This is repeated for each player and the game otherwise plays as normal. A variation on this would be to do it per round.

Some other variations on this could be that rather than a roll that determines non-purchase, you could just move the die up or down through each stack per player or per round.

This obviously fixes the problem that the top left card would never be unavailable for purchase.

Other variants could be six rows of two, and you roll one dice per player/round, and that row would be unavailable for purchase that player/round.

I decided on 10 coins, because by that time you've generally unlocked at least one more monument and have a few cards in front of you, so you should already be making money and have enough options.

Anyway - it's a different spin. Not something I'd play with each time - but simple enough to give it a try. Tell me what you think?
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Simon Lundström
Sweden
Täby
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Now who are these five?
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2 dice to determine a random number between 2 and 12 skewes the result heavily towards 7. Is there a reason why you want the 7th card to be unavailable more often?

Otherwise, use a 12-sided die, or emulate one with one die and a coin flip (head=6, tails =0, then add the die roll). Or you could emulate the coin flip with the other die (1–3=0, 4–6=6, then add the other die roll).
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Andy Burgess
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Or how about restricting to 9 stacks in a 3x3 grid - roll both dice with one indicating the column (1,2 = first column, 3,4 = second column, 5,6 = third column) and the other indicating the row in a similar fashion. Then every stack would have an equal chance of being blocked.
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Jeff Pearce
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I chose using the two dice because then there is no additional requirements. I usually play with twelve stacks so this suits me. Obviously, you can play around with this variant for what suits you. I felt that the element of chance with one stack being unavailable made it a little more eventful.
 
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