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full system after 1NT opening

I play this system after 1NT opening with various partners. Depending on system and position, 1NT can be 15-17, 14-16 or 12-14. The system is the same, but of course you adjust as responder what hands are is GF, INV or weak.

The only thing which is somehow unusual (at least between Romanian experts) is the 2♠ response, which is transfer to clubs (normal) OR invitation to NT without four card major. This means 2♣ stayman is always with at least one four card major. This trick is borrowed from the Polish bridge school.

Answers to 1NT:

With both majors (some sequences are quite unusual, read CAREFULLY):

Slam INV after stayman, with major fit: responder bids the OTHER major on level 3:

1NT - 2♣ - 2♥ - 3♠ = slam INV in ♥, denies splinter in ♣/♦
1NT - 2♣ - 2♠ - 3♥ = slam INV in ♠, denies splinter in ♣/♦

With 5m+4M, normal stayman sequences:

1NT - 2♣ - 2♥ - 3♣ = 5 clubs, 4 spades, GF. if responder had hearts, we should gave fit imediately (see above if slam INV).

After minor suit transfers, new suits at level 3 by responder shows singletons:

1NT - 2♠ - 2NT - 3♥ = clubs + singl ♥
1NT - 2♠ - 3♣ - 3♦ = clubs + singl ♦

After the dual 2♠ response, opener acts in first instance as over BAL INV, i.e bids 2NT if he would reject invitation (minimal) and 3♣ if he accepts. then:

1NT - 2♠ - 2NT - pass = i was BAL INV, we play 2NT
1NT - 2♠ - 2NT - 3♣ = i have clubs, weak
1NT - 2♠ - 2NT - 3♠ = i have clubs, spade singl (see above), slam INV
1NT - 2♠ - 2NT - 3NT = clubs, no singleton, passable, mild slam INV
1NT - 2♠ - 2NT - 4♣ = clubs, no singleton, serious slam INV
1NT - 2♠ - 3♣ - pass = clubs, weak
1NT - 2♠ - 3♣ - 3NT = i have BAL INV, no clubs
1NT - 2♠ - 3♣ - 3♦ = clubs, ♦ singl, slam inv (see above)
1NT - 2♠ - 3♣ - 4♣ = i have clubs, slam INV, no singl

After the 4♣/4♦ super-transfer (opener just make the transfer, NEVER super-accepts with the intermediate step), 4NT is KCBW. As opposed to 2 level transfer followed by 4NT which is quantitative.

After 2♦/2♥ tranfer

after double

special running sequences (no matter what double is - this is more useful when weak NT, but we play the same over strong NT):

they overcall in 2nd position

they overcall in 4th position

1NT - (p) - 2♥ - (DBL)
2♠ = fit 3

1NT - (p) - 2♥ - (2♠)
DBL = fit 3
1NT - (p) - 2♥ - (DBL)
RDBL = ! I have 5 hearts
1NT - (p) - 2♣ - (DBL)
pass - RDBL = please answer to stayman

1NT - (p) - 2♥ - (DBL)
pass - (p) - RDBL = please do the transfer

This last position should be refined. Which is the difference between a direct 3♣ and a 3♣ after RDBL - forced transfer?