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Samurai» Forums » General

Subject: Reformulation of the victory point description rss

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Henrik Johansson
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The rules are precise on how to calculate the (or several) winner, I'm not complaining about that. But the formulations are very mathematical and possibly hard to understand without several readings in my opinion. My idea is to describe it more like the "Longest road" thing in "The Settlers".
There should be 3 victory certificates, "Majority Helmets", "Majority Buddhas" and "Majority Rice", where Majority means strict majority. On the victory certificates there should be the following information:
"Strict majority Helmets 10000 VP. Rice and Buddhas: 100 VP each. Helmets: 1 VP each.".
"Strict majority Buddhas 10000 VP. Rice and Helmets: 100 VP each. Buddhas: 1 VP each.".
"Strict majority Rice 10000 VP. Buddhas and Helmets: 100 VP each. Rice: 1 VP each.".
When a player obtain a strict majority, he takes the appropriate certificate and the accompanying VP formulas. When contested and the majority is no longer strict, he looses it, and reverts to the default 1 VP a piece.
So a player with a 2 rice majority, 2 buddhas and 2 helmets would immediately calculate his score to 2+10000+200+200 = 10402, and evaluate an extra rice to 1 VP, and an extra Buddha or Helmet to 100 VP, unless also getting a second majority.
This suggestion will not change anything in the victory determination process, and given a little language brush-up it would make the game rules much simpler in the victory department.
 
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  • Last edited Sun Jun 19, 2011 10:52 am (Total Number of Edits: 1)
  • Posted Mon Jun 6, 2011 10:57 am
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W M Shubert
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I always describe it this way, and it goes over well:

Most helmets is one point. Tie means no points to anybody.
Most buddhas is one point. Tie means no points to anybody.
Most rice cakes is one point. Tie means no points to anybody.

So there are three points in the game. Most points is the winner.

If there is a tie, take away all the figures you got points for. Most figures left wins the tiebreaker.

If there is still a tie, then just count ALL your figures. Most figures wins the 2nd tiebreaker.

If there is still a tie, then it's a tie.

I think I prefer this to your system because all the numbers are bothersome. But yeah, I agree that the original rules are hard to understand.
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Henrik Johansson
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wmshub wrote:
If there is a tie, take away all the figures you got points for. Most figures left wins the tiebreaker.

If there is still a tie, then just count ALL your figures. Most figures wins the 2nd tiebreaker.

That is one of the weak points of the original rules and your rewrite, but not in my original posting: Figures are taken away, and then reappearing for counting again. A little adjustment here, and I will happily accept your rewrite attempt.
 
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Indiana Jones
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This is what we use:

____________________


At game end, tokens are awarded to players with the most of each figure.

A token is not awarded if there is a tie for most.

The winner is the player with the most tokens.

If tokens are tied, the winner is the player with the most figures not matching their token.

If figures are tied, the winner is the player with the most total figures.

If total figures are tied, then it's a tie.


 
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