[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]

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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."