10 October 2011

Bidding problem

Your hand is:
K T 9 7 3
Q 8
K J 8 7 6

The auction starts off like this:

1 1 1 2
Pass Pass ?

Presumably the 1 the first round isn't contentious.

Yes you play support doubles and yes partner would always make one holding three trumps.


  1. Fsirly easy double and remove 3♣ to 3♦

  2. I don't play that support doubles are mandatory at this level. Do you really need to double with a balanced 12 count and three bad spades?

  3. I like the compulsory support double on the understanding that partner shouldn't bid 1S here with a poor hand.

    Currently I have the agreement that 1S shows 5 (otherwise double) but even so lousy 6 or 7 counts should pass.

  4. We must have a diamond fit unless partner opened 1C on a 12-point hand with a long club suit. Losing Trick Count would suggest that with Diamonds as trump, North has 7 losers while South has 7 losers. So, 4D should make.

    If partner has the right 15-count, 3NT may also make. Double keeps 3NT in the picture, but if partner bids 3C (see caveat in first line), we have a problem. Not as easy to correct to 3D as Paul suggests -- partner may have a 2-3-1-7 with no heart stopper.

    So, it comes down to: which is more likely? a 6/7 bagger club suit or a balanced 15 count?

    I would also double but correct 3C to 3NT, hoping for a half-stopper with South. But not an easy decision for me.

  5. Playing good/bad 2NT might help, as that eliminates the possibility of partner owning a weak hand with a long club suit. Double by responder should be "DSI" (do something intelligent), showing a desire to compete further. Even though I do play that partner's pass denies three spades, I would follow Paul's plan: to double and then remove partner's 3C call to 3D. If partner bids 2S over the double,that presumably shows Hx in spades and I would pass that call.

  6. I like compulsory support X, I prefer 1S to show 4+, so that we can bid 1S with length in both majors and X for takeout with (3)4 spades and tolerance for the other suits, bit yeah 1S auto in this auction.

    Partner can't have a 15-count. If they have a 15-count with clubs they'd have bid 3C; if they have a balanced 15-count they'd open 1NT. P has a weak hand with long clubs or a weak NT with 3 hearts. Yes we quite possibly have an 8-card diamond fit ... so what? Even if we wanted to play in our 8-card fit at the 3-level (which we probably don't) we're not going to be able to have an auction to do so without showing a better hand than we have.

    We have an easy lead to 2H, against which we're probably going plus, so why would we want to play 2S opposite Hx, or 3D opposite 3-card support? I don't understand.

  7. I think you've won this one James.

    Partner could conceivable have a 15 count say 1336 with poor clubs but if so you don't want to declare the hand.

    2H may make but probably not.

    At the table: 2H fails by 2 tricks, 3D makes but no other contract has any play.