squeeze by cutting (? poor english)

[jaguelin]

Dear Sir,

 

I’ve been training often with your bridge software Bridge base 2000, especially for its educational quality. I happened to play a very interesting deal, from my point of view, which could be of interest for you.

 

 

Pair tournament

Dealer : South

Vul NS

 

 

S O N E

 

1♠ 3♥ 4♠ P

P P

 

 

♠ K J 9 7

♥ K 9 7

♦ K Q 4

♣ J 3 2

 

N

♠ -------------- ♠ Q 6 3 2

♥ Q J 10 6 5 4 3 O E ♥ 8

♦ 7 3 ♦ A J 10 9 8 2

♣ 10 8 7 6 s ♣ K 9

 

 

♠ A 10 8 7 5 4

♥ A 2

♦ 6 5

♣ A Q 5 4

 

 

O leads ♥ Q, S takes with the Ace, plays the ♠ 10 taken by the K followed by the J; E covers with the Q taken by the ♠ Ace. S plays ♦6 for the K taken by the Ace by E who returns ♦ J taken by the dummy’s Q and followed by the ♣2 for the Q. S plays a small ♠ for the dummy’s 9 and the

♣3 for the K taken by the ace. S eliminates the last E trump, plays the ♣5 for the ♣J and the ♦ 4, cut with his last trump. O is then squeezed ♥-♣ for 12 tricks.

To my opinion, covering the J trump at the 3nd trick is not a good decision, provided that S needs 2 communications apart from the ♥ K to succeed in making 12 tricks. But it seems that there is a solution anyway : if E doesn’t cover, S must finesse immediately the ♣K , expecting it second and play 2 rounds of ♣, thus capturing the K. S follows with ♦ the end being the same.

 

I played this deal in S and fortunately I had the first case; in the second one, I think that a “necessity” hypothesis can find the good solution, but only if you play the slam.

 

I would be grateful to receive your advice.

 

Best regards,

 

JCIBridge Base FG

[CSGibson]

[hv=pc=n&s=sat854ha2d65caq54&w=shqjt6543d73ct876&n=skj97hk97dkq4cj32&e=sq632h8dajt982ck9&d=s&v=n&b=15&a=1s3h4sppp&p=hqh7h8hasth3sks2]399|300[/hv]

 

I've created a diagram, including removing the 14th card from south (the 7 of spades, duplicated in the north hand), and made the play up to leading the ♠J from dummy.

 

"W leads ♥ Q, S takes with the Ace, plays the ♠ 10 taken by the K followed by the J; E covers with the Q taken by the ♠ Ace. S plays ♦6 for the K taken by the Ace by E who returns ♦ J taken by the dummy’s Q and followed by the ♣2 for the Q. S plays a small ♠ for the dummy’s 9 and the

♣3 for the K taken by the ace. S eliminates the last E trump, plays the ♣5 for the ♣J and the ♦ 4, cut with his last trump. W is then squeezed ♥-♣ for 12 tricks.

To my opinion, covering the J trump at the 3nd trick is not a good decision, provided that S needs 2 communications apart from the ♥ K to succeed in making 12 tricks. But it seems that there is a solution anyway : if E doesn’t cover, S must finesse immediately the ♣K , expecting it second and play 2 rounds of ♣, thus capturing the K. S follows with ♦ the end being the same.

 

I played this deal in S and fortunately I had the first case; in the second one, I think that a “necessity” hypothesis can find the good solution, but only if you play the slam."