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XYZ auction

#1 User is offline   jillybean 

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Posted 2011-August-04, 12:15

1 (X) XX (P)
1M (P) xyz?

I think xyz should be on here, how do you play it?

And over 1 (X) XX (P) 1 (P) 1N is to play?
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#2 User is offline   aguahombre 

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Posted 2011-August-04, 13:37

Since so much depends on what hands would redouble, this is hard to answer. There are some of us who don't let that double, which doesn't take up any room, interfere with our normal response and rebid structure, with these exceptions: 2C no longer inverted, but rather a minimum response with 4 clubs ---and 2N = a good club raise (5+, invite+).

This would leave only a 10+ hand with only four clubs (or perhaps 3-3-4-3) for the redouble; anything else would have responded as normal. In that context, 1M rebid by opener is a weak opening bid with longer clubs than M. Opener did have the opportunity to pass around the redouble with a decent opening bid.

Conclusion: xyz not needed in that world.
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#3 User is offline   Flameous 

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Posted 2011-August-04, 15:01

This has sort of been bothering me.

Is it better idea for opener to bid with weak shapely hands with clubs where you most certainly want to play 1XX, but the opponent will mostly remove it. (Of course they might not have clear agreement what pass shows...) Here we are just giving partner the information about the major we hold and that we are weak.
Other option is of course bidding with short (2-3) clubs as sort of rescue.
Is it best to switch according to what pass shows?

Apart from that, I don't think there is any need for invitational checkback, but some GF probe you might want to use, be it the other minor or whatever. I think 1NT should show something like 10-11. Former decides whether this is invitational or not.
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#4 User is offline   aguahombre 

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Posted 2011-August-04, 15:06

Yeh. Another way of putting it would be: if you want to start a possible xyz auction, don't throw an "R" into it.
"Bidding Spades to show spades can work well." (Kenberg)
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#5 User is offline   Phil 

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Posted 2011-August-04, 18:26

I would assume xy would be on over:

1x - x - xx - p
1y - p - ?

The xx essentially replaces the 'y'.
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