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Subject: Mismatch strategy? rss

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Ulrich Roth
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Here's a question I have after reading the rules and seeing a few videos about the game (which I have yet to play):

Wouldn't it be an unbeatable (if not winning) strategy to ALWAYS go for the mismatch, and try to win with four whites?

Once ahead in whites, I don't see how the opponent could stop that, other than eternally mismatching himself, resulting in a frustrating draw.

I would be delighted to be told what I am missing here.
 
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Nushura
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If your opponents matches and you don't....then HE gets the matching token
 
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Ulrich Roth
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Nushura wrote:
If your opponents matches and you don't....then HE gets the matching token


Sure, but don't I get a white token in this case regardless?

As I have understood it, the only way my opponent has to deny me that white token is to play a mismatch himself, in which case nobody gets anything.

If he doesn't do that and goes for the "one of each kind" approach (5 tokens in all), then my primitive "all white" strategy is likely to be faster, since it only requires 4 tokens to win.
 
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David Harding
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You also don't get a white if your opponent plays nothing against you. This will happen.

Also, your opponent can't always match, even if they want to, so the double unmatch happens often.

Play it and see.

The four whites can be an easy win occasionally against a new player but IMO opponents quickly learn to work to not let it happen.
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Ulrich Roth
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huffa2 wrote:
You also don't get a white if your opponent plays nothing against you. This will happen.

Also, your opponent can't always match, even if they want to, so the double unmatch happens often.


Hmm - but in both these instances the opponent fails to get ahead of the "always systematically mismatch"-player.

I guess you are right: I will have to try it out in practice, and I sure hope that my worries are unfounded.

Thanks for responding!
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Ulrich Roth
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huffa2 wrote:
Play it and see.

We have now done so and found that - fortunately! - there is indeed at least one sure-fire refutation of the dull "Always Mismatch"-strategy:

Counter that by systematically mismatching yourself, EXCEPT one (any) colour.
(Preferably a high-ranking one, in case the opponent switches strategy, but if he really only ever mismatches, it doesn't matter which colour it is.)

Whenever you can win a card of that one colour, do so. DO NOT go for cards in other colours, because that will put you behind in the race.

Typically the game will then end in a tie of 4 tokens each in one colour (white for the opponent, the chosen colour for you), with the tiebreaker in your favour.
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Chris
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ludopath wrote:
Typically the game will then end in a tie of 4 tokens each in one colour (white for the opponent, the chosen colour for you), with the tiebreaker in your favour.


It's an interesting strategy, and a cool development that you played it through to the point where you found a decent counter-move. I have to confess, when I read this thread last week, I couldn't immediately come up with a solid rebuttal and was quarter-worried that you'd semi-broken the game (inasmuch as something like a house rule might be required to prevent N consecutive chasen / tea whisks). I've only played it for one session, though, and enjoyed it very much, so had made a note to test the 'always mismatch' the next time we played. I'm glad you got there first.

However. There is a slight issue, although I'd consider it rules-lawyer-ish rather than terminal and it is as follows. The rules actually state:

Quote:
It is possible for both players to have met one of the winning conditions at the end of the same round. In the case of a tie, the winner is the player with the most chadogu tokens of the higher suit - whoever has the most green matcha tokens wins. If this is a tie, then the player with the most red tea bowl tokens wins, and so on down the suit rankings. If the game is still tied, further round(s) should be
played until one player has more of a higher ranking chadogu token.


The issue is that I can't find anything in the rules that suggests that the chasen (whisk) tokens are worth less than the 'suited' ones (I may well be over-looking something and would be happy to be corrected). The ranking chart only references the suited chadogu, not the chasen, and there's nothing in the rules to indicate where the chasen actually rank.

Therefore, unless I've missed something, there's nothing official to say that a 4-set of leaves (green) or bowls (red) would out-rank a 4-set of chasen (white). Which would naturally have implications for the counter-strategy.
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Ulrich Roth
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Triboluminous wrote:
I can't find anything in the rules that suggests that the chasen (whisk) tokens are worth less than the 'suited' ones (.......). The ranking chart only references the suited chadogu, not the chasen, and there's nothing in the rules to indicate where the chasen actually rank.

Yikes - you are right, I can't find anything either. surprise
I just assumed that white was the lowest-ranking colour, and I guess I'll keep on assuming it for the time being...

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Chris
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ludopath wrote:
Yikes - you are right, I can't find anything either. surprise
I just assumed that white was the lowest-ranking colour, and I guess I'll keep on assuming it for the time being...



You and me both. It's by far the cleanest and least intrusive solution. I think the (somewhat slippery, but legally defensible) solution is to treat the following statement ...

Quote:
In the case of a tie ... whoever has the most green matcha tokens wins. If this is a tie, then the player with the most red tea bowl tokens wins, and so on down the suit rankings.


... as an explicit statement of absolutes. That is, literally, in a tie green is best followed by red. Whatever the ranking of the chasen cards is, those two are clearly stated to be 'best', so that's what we're going with.
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David Harding
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Oh dear :) another print run will have to add that in the rulebook, but as you decided by the sentence saying green is best, yes, the whisks are at the bottom in heirarchy.
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Chris
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huffa2 wrote:
Oh dear another print run will have to add that in the rulebook, but as you decided by the sentence saying green is best, yes, the whisks are at the bottom in heirarchy.


Thanks for the official clarification -- and thanks for making such a great little game. Please pass on regards to TJ Lubrano, too. The artwork is great and a perfect tonal match.

If I could ask for another clarification with regard to that Tie-breaking paragraph: Does N leaves beat N tea-scoops in that situation, or does the circularity persist, such that Leaves beats everything except tea-scoops? I'd assume that the circularity persisted (not actually had a tie situation yet personally, but the 'always mismatch' counter-strategy sort-of relies on it), but that same paragraph could be interpreted to mean an absolute ranking of Leaves downward in the case of a tie.
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David Harding
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Thanks :) no problem!

In the tie-breaker the circularity does not persist. In a tie, the person with the most leaves wins. If that's a tie, the person with the most bowls wins and so on down to whisks.

But... From memory I don't think I've seen a tie-breaker go past the leaves.
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Chris
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huffa2 wrote:
Thanks no problem!

In the tie-breaker the circularity does not persist. In a tie, the person with the most leaves wins. If that's a tie, the person with the most bowls wins and so on down to whisks.

But... From memory I don't think I've seen a tie-breaker go past the leaves.


thumbsup Thank you.
 
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Julien K
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ludopath wrote:
huffa2 wrote:
Play it and see.

We have now done so and found that - fortunately! - there is indeed at least one sure-fire refutation of the dull "Always Mismatch"-strategy:

Counter that by systematically mismatching yourself, EXCEPT one (any) colour.
(Preferably a high-ranking one, in case the opponent switches strategy, but if he really only ever mismatches, it doesn't matter which colour it is.)

Whenever you can win a card of that one colour, do so. DO NOT go for cards in other colours, because that will put you behind in the race.

Typically the game will then end in a tie of 4 tokens each in one colour (white for the opponent, the chosen colour for you), with the tiebreaker in your favour.


I'm quite interested in this discussion, since I basically thought the same thing after an explanation of Matcha's rules.

I can also see how you can refute the always-mismatch strategy, but one problem is that you don't know at first if your opponent is going for it. Do you think that there is any way of winning against an always-mismatch opponent if the first round ended with you getting two *different* items (let's say, a red and a blue), and the always-mismatch opponent getting two whites ?

I think that the game is interesting, but I can't see myself liking it if I always have to play by either using the always-mismatch strategy or by skewing my play because my opponent may play this strategy.

Regardless of the answer to the previous question, what about limiting the number of white tokens you can get in a round to one ? In this case, if in any round two items of the same color can be won, the always-mismatch strategy is at a disadvantage.
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Clyde W
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Curious to see an answer to the above question...
 
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Anthony C
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I feel like the game doesn't really WANT you to get a full set of tokens. I think you should ALWAYS be trying to get 4 tokens of a type and getting a full set is maybe a backup.

If the card flop at the beginning of the round has 2 or more of the same type of suit, play to get that suit and mismatch everything else, or pass so your opponent gets the tokens. If your opponent is JUST playing to get white whisks, you can spot that quickly enough by seeing what they play for the first round, then adjust your strategy. If you manage to get ahead in whisks by the end of the first round, well then, you could either stick to that strategy for the rest of the game or switch it up to screw with your opponent. You two will keep stalemating until one of you passes and gives the other player a token (unless you both pass the last card of the round), or until one of you decides to try and grab a higher ranking token, in which case, the other player gets a whisk.

I don't know, I'm eager to play more to get a better grasp of things. I feel like the whisk strategy is only counterable by getting 4 of any other type. If you both play to get the same type of non-whisk, then it's kinda left to luck to see who has the higher card to score the suit token. But then, you can likely stop someone from winning all of the specific ranked tokens they're going for (again, with a lucky cards draws) and then just mismatch or go for 4 of a different token to win. I don't know, it seems like there is more strategy the more I write this but like I said, I'll need to play more.
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Håkon Sønderland
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I still don't see any counter to the whisk strategy. If you don't know your opponent is going for it in the first round, she can easily get 2 or three and I see no way of stopping the win then. Especially if the player going for the whisks are not the first player and can match the passing.
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Daniel Gill
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As a counter to the whisk strategy, in the first round of the game if you see that your opponent has obtained a whisk, couldn't you just try to win by having one of every symbol?
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