Nasty lead problem from DC LM pairs

[Humper]

[hv=d=s&v=n&s=sjthqjxxd9xxcaxxx]133|100|Scoring: MP

Opps bid unobstructed:

1d - 1S

2NT - 3d*

3NT** - 6NT

 

* major suit inquiry

** 2 spades, 2-3 hearts[/hv]

 

0-1500 LM pairs, day 2 or 3 (can't remember). You're on lead against this sequence where your RHO has showed 18-19 balanced, 2 spades, and 2-3 hearts. Your LHO has leaped to 6NT hearing this description. You and partner have passed throughout.

 

What do you lead?

[xcurt]

You are so ripe for a squeeze on this hand. CA.

[aguahombre]

Is the Club ace the best way to get partner off a squeeze, or is the spade jack and another spade when in with the club? seems like club ace rectifies at trick one.

[andy_h]

Small heart. 2nd chioce ♠J. Club ace is too crazy for me, who says we can't beat the contract?

[effervesce]

Safe leads normally vs 6NT. I lead the J of spades. 2nd choice a diamond.

[xcurt]

Is the Club ace the best way to get  partner off a squeeze, or is the spade jack and another spade when in with the club?   seems like club ace rectifies at trick one.

Small heart. 2nd chioce ♠J. Club ace is too crazy for me, who says we can't beat the contract?

 

How many HCP do you think partner has? If dummy has the ♠9 (3:4 against if 5 spades to HH in dummy, and there's also the chance of the ♠9 in declarer's hand and the ♠8 in dummy) and they have 5 diamond tricks and the ♥AK that's 12 tricks. If dummy has a 6-card spade suit and they have even 4 diamond tricks you're also going to get squeezed. Unless partner has ♥Txx and can figure out to keep all of them.

[gnasher]

Is the Club ace the best way to get  partner off a squeeze, or is the spade jack and another spade when in with the club?  seems like club ace rectifies at trick one.

Are you trying to break up a squeeze when partner has ♠9xxx and ♦J10xx, and declarer has 11 tricks in the form of three spades, two hearts, three diamonds, and three clubs after knocking out the ace? If so, that isn't going to work unless there's a blockage in diamonds.

[gnasher]

I'd cash ♣A. This sounds like they've bid it on high cards, in which case it's a fairly normal contract and partner has nothing at all. They could well have five spades, two hearts and five diamonds on top, leaving me vulnerable to a squeeze. (Especially if I've led a low heart. Where was Justin when that suggestion was made?)

 

If they do have five spade tricks, they'll know that that after two rounds, which they can cash without any risk. It would be different if I had ♠Jxx, when declarer would probably set up his club tricks instead of risking a third spade.

 

And assuming that those "x"s really are small, ♣A will cost only if partner has ♣J and a stop in one of the pointed suits, and declarer wasn't going to find his way to 12 tricks in some other way.

[Humper]

This sounds like they've bid it on high cards, in which case it's a fairly normal contract and partner has nothing at all.

I agree with how it sounds.

 

At the table, it turns out that they have SEVEN spade tricks and SIX diamond tricks on top.

 

Partner has the ♠Q, ♥A and ♣K. Go figure.

[bid_em_up]

I really think the Ace of clubs stands out by a mile. This is matchpoints. Assuming the opponents are competent and actually have their bids, partner will be broke and we have absolutely no reasonable expectation of being able to defeat this contract (imo). We can expect them to easily take 5-6 spades, 2 hearts, 4-5 diamonds and at least one club. Thats 12. Our objective here should be preventing any possible overtricks, and not worrying about beating the contract.

 

The fact that the opponents have bid ridiculously and the club A lead sets them is just icing on the cake.

 

jmoo