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Subject: Who won this game of Knizia's Samurai? rss

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Kevin O'Brien
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Who won this game of Knizia's Samurai?

The Set-Up:

There are four players - A, B, C, and D. The 13th Buddha has just been scored, ending the game. Here are the figure totals for the four players:

Player A - 4 Buddhas, 1 High Helmet, 3 Rice Fields (8 Total)

Player B - 4 Buddhas, 2 High Helmets, 4 Rice Fields (10 Total)

Player C - 2 Buddhas, 3 High Helmets, 1 Rice Field (6 Total)

Player D - 3 Buddhas, 3 High Helmets, 4 Rice Fields (10 Total)

Here are the rules for determining the winner, taken from the PDF file on BGG - http://www.bggfiles.com/viewfile.php3?fileid=9765

Scoring and determining the winner
When the game ends, all players remove their screens and count the number of figures of each type.
If any player has captured the most figures of 2 or 3 types, he is the winner!
If there is no immediate winner, all players who have captured the most (ties for the most do not count) of one type of figure are eligible to win; players with no "most" cannot win.
Players with the most of a figure set aside those figures and count the number of other figures they captured. The player with the most other figures is the winner!
If there is still a tie, the players count all their captured figures; the player with the most is the winner!
If there is still a tie, the tying players share the victory.

And here's the dilemma:

If "ties for the most do not count" and "players with no 'most' cannot win" then it seems that no player can win! Did all four players lose? Should we just say "To heck with the rules" and declare Players B and D the tied winners?

Comments are welcome.

Cheers,

Kevin
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Steve Hope
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You all lose (or you all place equally, to avoid pedantic arguments about what it means for everyone to "lose").

The only thing I don't like about Samurai is its slightly wonky scoring system. Well, that and the fact that the player after me might be a bozo ensuring that the player after THAT will win.
 
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Chapel
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Player B and D share the win... Gotta love Knizia! .... cool


EDIT: Hmmm...maybe wrong. four way tie for last. But wouldn't that also be a four way tie for first? This is a scenario that I've never encountered thank god!
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Richard Pardoe
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My first thoughts were the same as Steve's - everyone places equally.

I also believe this is consistent with the theme of the game - trying to seek plurality control of the major factions.

Therefore, (as mentioned in the first winning condition) if a player gets the plurality in 2 or more types, that person is the winner.

If no player gets the plurality in 2 or more types, then the players with clear pluralities in a type look at secondary control to see who has the most influence.

In the game presented above, no player has the plurality in any of the types, so there is no winner.

And as a result, Japan enters in the Sengoku Jidai (Warring States Period) until a clear winner ala Tokugawa Ieyasu plays the game.
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Stan Mamula
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We've ended up with so many ties in this game it's crazy. I love the gameplay, but ties occur so often that it starts to feel futile to even play.

What happens if you do away with "ties for the most do not count" and consider ties as shared for the most? The scoring becomes a bit easier. With this rules tweak, in your example above, Player B and D would share the victory with the most influence in two factions. This rules tweak could cause other problems, but I haven't any done any playtesting to see. Plus, it still resulted in a tie. Only 2 players instead of 4, but still a tie. shake
 
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Glenn Ironhat
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Personally, I think you are taking the term "the most" too literally. No where does it say that you need more than all the other players to have the most.

I think player B wins with 4 of two kinds of figures. No other player has 4 of two kinds.

The comment about "ties for the most do not count" kind of shoots down the whole "most" definition. How can someone have the most and still be tied? Also that refers to a second condition if no one has one yet with the first condition.
 
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Adam Smiles
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gironha wrote:
Personally, I think you are taking the term "the most" too literally. No where does it say that you need more than all the other players to have the most.


No where? Have you read the rules? They specifically state that if you are tied with another player, you don't have the most.

By the rules, no player has the most, so no player may win the game.
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Un Streetfighter avec un doctorat
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This translation (http://www.bggfiles.com/viewfile.php3?fileid=9451) has an additionnal winning condition, at the end: "If no player has the 'most' of a figure, the player with the most captured figures is the winner". The French version (Descartes, which I own) has that same rule. So B and D would win.

"The most" has to be taken literaly. It means A > B in a figure type, not A >= B, otherwise the system makes no sense as written.

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George Kinney
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cyberkev63 wrote:

Player A - 4 Buddhas, 1 High Helmet, 3 Rice Fields (8 Total)

Player B - 4 Buddhas, 2 High Helmets, 4 Rice Fields (10 Total)

Player C - 2 Buddhas, 3 High Helmets, 1 Rice Field (6 Total)

Player D - 3 Buddhas, 3 High Helmets, 4 Rice Fields (10 Total)


Only players A, B and D are eligible to win, since C doesn't have a 'most' of anything. (or potential 'most', bear with me a minute...)

It doesn't matter what they have a 'most' in, just that it is a 'most'. (Or a tie for a 'most') You don't compare each group of figures individually among the players, you just compare the totals in decending order, irregardless of which piece that total represents.

You should look at it as:
Player A: 4 3 1
Player B: 4 4 2
Player C: 3 2 1
Player D: 4 3 3

So none of them have a 'most' in column 1, since they all three tie, but B has one in column 2, so B wins. Only if two of them tied in column two would the 'total remaining' or 'overall total' matter as far as scoring goes.

I think this is where a lot of people get confused, and although I think the examples they provide make it pretty clear, they could have added one more line to make it more explicit.

Either A or D could have won with one more Buddha, and D could have won with one more High Helmet as well.

FWIW, if you do actually have a lot of games ending in ties, then you can stop that cold by forcing ties on individual pieces to remove them from the game which guarantees that someone will have a majority.
 
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Seth Ben-Ezra
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vialiy wrote:
This translation (http://www.bggfiles.com/viewfile.php3?fileid=9451) has an additionnal winning condition, at the end: "If no player has the 'most' of a figure, the player with the most captured figures is the winner". The French version (Descartes, which I own) has that same rule. So B and D would win.


The Rio Grande rules have this rule as well. Players B and D win.

Seth Ben-Ezra
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Seth Ben-Ezra
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Gecko23 wrote:

It doesn't matter what they have a 'most' in, just that it is a 'most'. (Or a tie for a 'most')


A tie for a 'most' doesn't count.

From the rules:

Quote:

If there is no immediate winner, all players who have captured the most (ties for the most do not count) of one type of figure are eligible to win.


Seth Ben-Ezra
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Adam Smiles
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George, I'd have another look at the rule book if I were you.

I'll quote from the actual rules:

Quote:
- When the games ends, all players from remove their screens and count the number of figures of each type.
- If any player has captured the most figures of 2 or 3 types, he is the winner!
-If there is no immediate winner, all players who have captued the most (ties for the most do not count) of one type of figure are eligible to win; players withe no "most" cannot win.
-Players with the most of a figure set aside those figures and count the number of other figures they captured. The player with the most other figures is the winner!
-If there is still a tie, players count all their captured figures; the player with the most is the winner!
-If there is still a tie, the tying players share the victory.
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George Kinney
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I've looked at the rulebook plenty, but I'll be sticking with what I posted. I don't think Dr. Knizia is going to show up at my door and rough me up for it.

We had a couple of games end up with ambiguous outcomes, or more importantly, unsatisfying outcomes like the one posted here. After reading through the scoring rules for the third time after a game one night, I proposed this 'interpretation', and thats what we've used since.

Everyone's happy, games end cleanly, life goes on.








 
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Robert Jones
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George, are you saying that you use different scoring rules or that your interpretation of the actual is correct? I've never heard of anyone interpreting the rules in the way you demonstrated.

Bob
 
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Jim Cote
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Gecko23 wrote:
We had a couple of games end up with ambiguous outcomes, or more importantly, unsatisfying outcomes like the one posted here. After reading through the scoring rules for the third time after a game one night, I proposed this 'interpretation', and thats what we've used since.


Then you should call it a house rule, and not answer the original question so 'authoritatively'.
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George Kinney
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ekted wrote:
Then you should call it a house rule, and not answer the original question so 'authoritatively'.


Yep, my bad. Get a little sugar in me and off I go.

I'll shame myself when I get home.
 
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Tim Kilgore
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The scoring works like this, with cases presented in order of likihood...

Case 1. If you win the support of 2 or 3 castes, you win. If you are the only person to win the support of caste (the others are tied), you win. Period. You are done. No need to continue. Otherwise...

Case 2. If you win a caste but no player has won the support of 2 or 3 castes, the players that have won castes remove all of the figures used to win the caste (and any excess) from your pile. Players winning castes compare the remaining pieces - he who has the most wins and if the numbers are the same, a tie happens.

Case 3. In the event no one wins a caste (what we are discussing now), all of the players total all of their caste support (i.e. total the pieces won for each of the 3 castes). The player with the most overall support wins. Ties are possible.


So in your game, players B & D tied with 10 points of overall support. Players A & C lost.

I hope this helps clear up the funky scoring rules.

Tim



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What Tim said. cool
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Matthew Wills
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GreatWolf wrote:
vialiy wrote:
This translation (http://www.bggfiles.com/viewfile.php3?fileid=9451) has an additionnal winning condition, at the end: "If no player has the 'most' of a figure, the player with the most captured figures is the winner". The French version (Descartes, which I own) has that same rule. So B and D would win.


The Rio Grande rules have this rule as well. Players B and D win.

Seth Ben-Ezra
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Weird - I'll have to check that at home. I thought the rules at http://freespace.virgin.net/chris.lawson/rk/samurai/rules.ht... were the same as the RGG ones - and that rule isn't mentioned there...
 
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I think this boils down to how you interpret this sequence in the RGG rules:

Quote:
* If there is no immediate winner, all players who have captured the most (ties for the most do not count) of one type of figure are eligible to win; players with no "most" cannot win.

* Players with the most of a figure set aside those figures and count the number of other figures they captured. The player with the most other figures is the winner!

* If there is still a tie, the players count all their captured figures; the player with the most is the winner!


The contentious bit seems to be players with no "most" cannot win. Either it applies only to the first quoted bullet point, meaning players with no most cannot win by the bulleted method, or it's, uh, globally scoped, and the affected players are precluded entirely from winning.

My take on it, which I think is sensible but not literal, is that if the winner cannot be otherwise determined, the player(s) with the most pieces win.

I'd be interested to hear from a native German speaker as to how the original German rules are phrased.
 
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Glenn Ironhat
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asmiles wrote:
gironha wrote:
Personally, I think you are taking the term "the most" too literally. No where does it say that you need more than all the other players to have the most.


No where? Have you read the rules? They specifically state that if you are tied with another player, you don't have the most.

By the rules, no player has the most, so no player may win the game.


Yes but that is on the second line. It may only apply "If there is no immediate winner".
 
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Glenn Ironhat
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vialiy wrote:
This translation (http://www.bggfiles.com/viewfile.php3?fileid=9451) has an additionnal winning condition, at the end: "If no player has the 'most' of a figure, the player with the most captured figures is the winner". The French version (Descartes, which I own) has that same rule. So B and D would win.

"The most" has to be taken literaly. It means A > B in a figure type, not A >= B, otherwise the system makes no sense as written.



First you say the "the most" has to be taken literally. Then the rule says that the player with the most captured figures is the winner. Well no one has the most of that either. How can there be a tie?
 
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gironha wrote:
vialiy wrote:
This translation (http://www.bggfiles.com/viewfile.php3?fileid=9451) has an additionnal winning condition, at the end: "If no player has the 'most' of a figure, the player with the most captured figures is the winner". The French version (Descartes, which I own) has that same rule. So B and D would win.

"The most" has to be taken literaly. It means A > B in a figure type, not A >= B, otherwise the system makes no sense as written.



First you say the "the most" has to be taken literally. Then the rule says that the player with the most captured figures is the winner. Well no one has the most of that either. How can there be a tie?


Heh. Good catch! The French rules distinguishes between "majority" and "most", and "majority" should be taken literaly. But then, I feel I'm veering towards interpretation. The last rule is ambiguous enough that you may be right. Anyone has access to the original German rules?

The PC game renders this more explicitly, supporting my view:
"If no player wins Caste Support, then either the player with the most Total Support wins or the game is a tie between the players who are tied for Total Support"
 
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Steve Hope
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sbszine wrote:
I think this boils down to how you interpret this sequence in the RGG rules:

Quote:
* If there is no immediate winner, all players who have captured the most (ties for the most do not count) of one type of figure are eligible to win; players with no "most" cannot win.

* Players with the most of a figure set aside those figures and count the number of other figures they captured. The player with the most other figures is the winner!

* If there is still a tie, the players count all their captured figures; the player with the most is the winner!


The contentious bit seems to be players with no "most" cannot win. Either it applies only to the first quoted bullet point, meaning players with no most cannot win by the bulleted method, or it's, uh, globally scoped, and the affected players are precluded entirely from winning.

My take on it, which I think is sensible but not literal, is that if the winner cannot be otherwise determined, the player(s) with the most pieces win.

I'd be interested to hear from a native German speaker as to how the original German rules are phrased.


Now I'm curious too. My presumption would be that you COULDN'T win unless you had a "most"--otherwise you might have this curious situation:

A: 5/4/2
B: 2/4/5
C: 4/4/4

Where player C is discarded for consideration in step 1 but then, surprise! He comes back to win in step 3. From a logical standpoint, I don't like situations where a player wins but if another player had done WORSE, the winner would have lost. In this example, if player A or B had captured 1 less in the middle column, the other would have won. But since they both captured 4, player C wins! So player C wins by Phil's interpretation, but would have lost to player A if player B had done slightly worse.

In the "must have a most to win" interpretation, you have the unsatisfactory result of occasionally having ties like this. But I believe that at least you avoid the even more unsatisfactory result of "if YOU had only done a little better, then I would have won!"
 
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I agree with Tim. Ties are not a majority, so you don't control the caste. My version (only six months old) clearly states that if the first two critera all end in ties you go to overall piece count, which has B & D tied.

For fun though, you could then eliminate A & C, and reapply the criteria to just B & D, in which case, B has majority in Buddhas, D has majority in Helmets-- a tie at 1 caste each, so compare non-majority pieces, B has 6, D has 7, so D wins. This would just be for fun though, since it aint really the rules.

BTW, to annoy your opponents, it's fun to hide your face behind your screen, peeking out at them during there turn and occasionally popping up and saying, "Mmmmmmmmbuddha!"

-BoB
 
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