[kgr]

IMP's

---

2H=8+ and 3c♠

Should we be in game, If yes: who should have done more?

Spoiler

RHO had all points and did lead ♣A, ♣K and a small ♣ after which 2♠ made +3 on a ♦-♥ squeeze

[gordontd]

You could lose a spade, a diamond, two clubs, one or two hearts. So, it might seem that on a really bad day you could go off in 2♠, although in fact the defenders are usually going to give away at least one of those tricks on the lead.

[Codo]

So you make a game where you have all valuable tens and nines and west zero points so that east has to gift you some tricks?

Yes you should continue to bid these games.... at least against me...

[Fluffy]

6322 opposite 4333 with 21 combined does not normally produce game.

[kenberg]

I doubt that you will get any votes for getting to 4♠.

About the squeeze: Were spades also split 2-1? Unless they were, I don't see how this squeeze would work. Suppose 3-1. As E helpfully plays AK and x of clubs you pitch a red and cash dummy's high spade. If E started with a singleton spot you are going to lose a trump. If instead a quack appears from E you run the ten and pick up the trump in three rounds. That's fine, but now you still have a club winner on the board and no way to get to it except via the ace of hearts. If you do that, you destroy the squeeze position. If you just leave the club there and run spades, the position is not, I think, tight enough. After three clubs and six spades, E holds KJ and KQ in hearts/diamonds. N plays a heart, E covers, dummy wins, the club is cashed, E pitches after N. Really I am not even sure how this squeeze works [edit, aas kgr points out, I am wrong here] when spades are 2-2 unless you start with the spade ten to the ace, which would not work well if E shows out. If spades are 2-2, and you start with ten to the ace, back to the King will work. Cash the club, keeping Qx of hearts and the stiff ace of diamonds. Run trump executing a criss-cross. But who would play spades that way? Maybe I have this wrong [I do, see below], but it sounds as if the squeeze was a bit of a pseudo. I can imagine the defense getting a little careless in a 2♠ imp contract when the only question is ten or eleven tricks.

Anyway back to your question. No one is reaching 4♠ on these hands. And it won't always be made if they do.

[kgr]

Thanks all for the answers. Question was probably too influenced by the result after the play.
At the start of the play I didn't really mind the bidding

kenberg, on 2013-January-06, 10:42, said:

I doubt that you will get any votes for getting to 4♠.

score card was:
3♣X -4: 1x
4♠=: 4x
2♠+3:1x
3♠+1:3x
2♠+2:2x
3♠=:1x
3♥-2:2x
(This is a off-line tournament I play against Jack; other results are from real live)
I don't know how 4 spade contracts were bid. Maybe more/other competition or a 3♠ bid by South

kenberg, on 2013-January-06, 10:42, said:

About the squeeze: Were spades also split 2-1? Unless they were, I don't see how this squeeze would work. Suppose 3-1. As E helpfully plays AK and x of clubs you pitch a red and cash dummy's high spade. If E started with a singleton spot you are going to lose a trump. If instead a quack appears from E you run the ten and pick up the trump in three rounds. That's fine, but now you still have a club winner on the board and no way to get to it except via the ace of hearts. If you do that, you destroy the squeeze position. If you just leave the club there and run spades, the position is not, I think, tight enough. After three clubs and six spades, E holds KJ and KQ in hearts/diamonds. N plays a heart, E covers, dummy wins, the club is cashed, E pitches after N. Really I am not even sure how this squeeze works when spades are 2-2 unless you start with the spade ten to the ace, which would not work well if E shows out. If spades are 2-2, and you start with ten to the ace, back to the King will work. Cash the club, keeping Qx of hearts and the stiff ace of diamonds. Run trump executing a criss-cross. But who would play spades that way? Maybe I have this wrong, but it sounds as if the squeeze was a bit of a pseudo. I can imagine the defense getting a little careless in a 2♠ imp contract when the only question is ten or eleven tricks.

Anyway back to your question. No one is reaching 4♠ on these hands. And it won't always be made if they do.

East had:
♠Q7
♥KJ982
♦KQ
♣AK84
Play: ♣AK and 8
Now ♠K and ♠ to the A. and back to dummy with ♠T.
♣Q and ♦A followed by all trumps squeezes East.

[kenberg]

Oops, yes, that is so. The case with spades 2-2 was something of an afterthought, I should have rechecked my thinking. I guess I had relocated the two red aces. Or something.


I still cannot imagine getting to 4♠. It happened, I can't explain it.

[MrAce]

kgr, on 2013-January-06, 07:12, said:

Should we be in game, If yes: who should have done more?

No

[Zelandakh]

I can think of a few routes that might occur in club bridge:

1♠ - (2♥) - 2♠ - (P);
P - (3♣) - P - (P);
3♠ - (P) - 4♠

1♠ - (2♥) - 2♠ - (P);
P - (3♣) - 3♠ - (P);
4♠

1♠ - (2♥) - 3♥ - (P);
4♠

1♠ - (X) - 3♠ - (P);
4♠

1♠ - (X) - 2♠ - (P);
P - (3♥) - 3♠ - (P);
4♠

1♠ - (X) - 2♠ - (P);
P - (3♥) - P - (P);
3♠ - (P) - 4♠

1♠ - (X) - 2NT - (P);
4♠

I am not suggesting any of these as good bidding, only trying to show how it might happen.

[rmnka447]

Opener has 2 QT's and 11 HCP. Looking at the hand, I'd say it was just on the cusp between an opener and a weak 2 bid. So, I can find no objection to opener's 2 ♠ rebid -- it's the proper valuation of the hand.

Likewise, responder's hand is also on the cusp between a simple raise and a 3 card limit raise. There are 10 HCP, but it is a 4-3-3-3 hand with 9 1/2 losers. I think the flat distribution and number of losers tip the scales toward treating the hand more like a simple raise than a limit raise. So, I see no problem with responder's bidding.

Even if responder took an optimistic view of the hand and treated it like a 3 card limit raise, I think opener's right action is still to pass the invitation.