The mechanics of the Stayman convention are as follows:
Opener Responder
1 NT 2 ♣ = “Please name a four-card major”.
2 ♦ = No four-card major
2 ♥ = Four hearts
2 ♠ = Four spades
Question: What does opener reply if he has both four-card majors?
Answer: 2 ♥.
Responder assumes that a 2 ♥ reply does not contain four spades in addition to the four hearts. On those occasions where opener does have four spades, it is opener’s duty to bid spades. It is not responder’s duty: if he does not support the hearts, he is known to have four spades or he would not have bothered to bid Stayman.
South Deals N-S Vul |
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West | North | East | South |
1 NT | |||
Pass | 2 ♣1 | Pass | 2 ♥ |
Pass | 3 NT2 | Pass | 4 ♠3 |
Pass | Pass | Pass |
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4 ♠ by South |
Lead: ♣ 9 |
A 3 NT contract would almost certainly have lacked a ninth trick. But 4 ♠, reached via Stayman, was an excellent contract. West led the nine of clubs, and declarer played low from dummy, East winning the king and returning the two. With which honour should declarer win this trick?
Declarer’s basic plan was to cash the two top trumps, then trump his two losing hearts in dummy. In order to carry this out, he needed an extra entry to hand. It was vital that he won dummy’s ace of clubs (key play), preserving his queen as an entry.
Declarer won the ace, cashed the ace-king of spades, then, leaving master queen of spades out, cashed the ace-king of hearts, trumped a third heart, crossed to the queen of clubs (West discarding - but trumping with the queen, had he held it, would not have hurt), and ruffed his last heart. East overruffed with his queen, but declarer later scored dummy’s ace of diamonds and his two long trumps. Game made.