nullve

If * T is the trump suit; * L,M,H are the lowest, middle and highest suit outside T, respectively; * 'KC' means 'number of key cards' then it's often possible to play the following instead of Exclusion RKCB: 4T+1 = L void or no void ...4T+2 = LA ......4T+3 = even KC .........4T+4 = TQ ask ............5T = no TQ ............[5T+1]+ = TQ ......4T+4 = odd KC and no TQ ......5T = odd KC and TQ ......[5T+1]+ = * ...4T+3 = no LA and even KC ......4T+4 = TQ ask .........5T = no TQ .........[5T+1]+ = TQ ...4T+4 = no LA and odd KC and no TQ ...5T = no LA and odd KC and TQ ...[5T+1]+ = * 4T+2 = M void ...4T+3 = even KC outside M ......4T+4 = TQ ask .........5T = no TQ .........[5T+1]+ = TQ ...4T+4 = odd KC outside M and no TQ ...5T = odd KC outside M and TQ ...[5T+1]+ = * 4T+3 = H void and even KC ...4T+4 = TQ ask ......5T = no TQ ......[5T+1]+ = TQ 4T+4 = H void and odd KC and no TQ 5T = H void and odd KC and TQ [5T+1]+ = * * like 5T, but confident that 2+ KC aren't missing The point, of course, is to always be able to stop in 5T when too many key key cards are missing. As an illustration: If T=♥, the structure becomes 4♠ = ♣ void or no void ...4N = ♣A ......5♣ = even KC .........5♦ = ♥Q ask ............5♥ = no ♥Q ............5♠+ = ♥Q ......5♦ = odd KC and no ♥Q ......5♥ = odd KC and ♥Q ......5♠+ = * ...5♣ = no ♣A and even KC ......5♦ = ♥Q ask .........5♥ = no ♥Q .........5♠+ = ♥Q ...5♦ = no ♣A and odd KC and no ♥Q ...5♥ = no ♣A and odd KC and ♥Q ...5♠+ = * 4N = ♦ void ...5♣ = even KC outside ♦ ......5♦ = ♥Q ask .........5♥ = no ♥Q .........5♠+ = ♥Q ...5♦ = odd KC outside ♦ and no ♥Q ...5♥ = odd KC outside ♠ and ♥Q ...5♠+ = * 5♣ = ♠ void and even KC ...5♦ = ♥Q ask ......5♥ = no ♥Q ......5♠+ = ♥Q 5♦ = ♠ void and odd KC and no ♥Q 5♥ = ♠ void and odd KC and ♥Q 5♠+ = *. * like 5♥, but confident that 2+ KC aren't missing

Partner's overcall style could also make a difference. My partners would never overcall 2♦ without 6+ D or 4+ C in this position, so I expect at least 19 total trumps* on average. If there are as many as 20, then LoTT suggests bidding. If there are only 19, then hopefully the void is worth an extra total trick. (Voids tend to make total tricks > total trumps.) * Our trump suit will be either spades, clubs or diamonds, depending on partner's hand. As eagles123 said, 4♠ implies diamond support, so partner will correct to 5♣ or 5♦ without spade support. EDIT, 22 August: 'at least 19 total trumps on average' may be too optimistic if 1♥ or 4♥ is frequently bid with only 4 H.

4♠

Seems good enough if 2♣ promises opening strength, since slam is unlikely. I like this! It can be found on page 351 in the Kokish-Kraft system notes.

Do you still play lebensohl, for example? If so, how does that work when opps' suit is clubs?

From yesterday's Free Daylong Tournament (MP) http://tinyurl.com/ybhz7geo 2N= was worth only 30.3 %, but 2N+3 would have given me all the points. Yes, finessing spades at trick 8 looks greedy in hindsight.

I voted Stayman, then 3N over 2♦. I'm prepared to look stupid.

Welcome back to the forums! :)

How can you pass 2♠ at MPs if you think the field is in 2♥?

So you don't think Pass has the highest expected value on the hand type I gave? FWIW (not much, I guess), here are 100 random weak-or-GF Ekrens 2♦ deals where Responder has 12 hcp, 2263:

I play Ekrens 2♦ as weak or GF in one partnership, but the opening is still NOT forcing. I'd pass 2♦ with 12 hcp, 2263, for example. Why wouldn't you? :)

What are the continuations after 1♠-1N; 2♣-2♦; 2♠ in "standard" Gazzilli? I wasn't sure, so I had a look at Ambra, the Bocchi-Duboin system, Fantunes and Nightmare, which all use this sequence similarly. Here's what I found: 1♠-1N; 2♣-2♦; 2♠-?: Ambra (2004 version on bridgeguys): [nothing] Bocchi-Duboin system (as described in th book Il sistema Bocchi-Duboin): [nothing] Fantunes (notes by Dan Neill): [nothing] Nightmare (translated by Dan Neill): 2N = "bal, no 3♠" 3♣/♦/♥ = "natural" 3♠ = "5♥-4♣" 3N = "5♥-4♦". Maybe continuations were considered too obvious to be worth mentioning in these Ambra, B-D and Fantunes references? Even in Nightmare we see that Responder's ranges are left out, which might give Responder a problem on the OP hand, J QJ9864 KQ76 52, if he doesn't know if 3♥ is weak or invitational. (I'd bid 3♥ regardless, but...) But maybe Responder would have bid differently with 2-S6+H and invitiational values, e.g. by * responding 2♥, "2/1 except rebid" style, instead of 1N * responding 3♥, IJS, instead of 1N * rebidding 3♥, if assigned this hand type, instead of 2♦, so that 3♥ now denies invitational values? --- I play a version of Gazzilli over 1♠-1N, too, but I wouldn't have had this problem. One reason is that I kind of switch the 2♦ and 2♥ responses to 1♠ (like e.g. Bas-Drijver do, IIRC), which allows a hearts-showing two-over-one response also on weakish hands with 6+ hearts. So in my system the auction would actually go 1♠-2♦ 2♥-P. Another reason is that like most Norwegians who claim they play "2/1", I don't play the 1N response as even semi-forcing, and I would certainly pass a 1N response not only with AQ643 3 AJ5 9862, but any hand with 10-12 hcp (I use rule of 19 openings) and 5S3-H4C, both because I don't think it will hurt me in the long run even on this hand type, and because of benefits on other hand types. For example, if were to play an otherwise standard Gazzilli (I'm not), then 2♠ over 1♠-1N; 2♣-2♦ could promise either "13-15" or a fifth club, with continuations something like P = 8-10, 2 S 2N = INV+ relay ...3♣ = NF, 5(+) C ...3♦+ = GF, NAT 3♣/♦/♥ = to play (...).

I don't think 2N would be forcing in this system (= Acol?).

I voted 3N. (No, I don't have a diamond stopper.)

It's probably true that 2♠ = "spades" puts more pressure on opps than 2♦ = "spades", but in your system 2♠ is already a Weak Two, so the question is whether the latter puts less pressure on opps than 2♦ = "diamonds". Details? Btw, why should the knowledge that Opener has 4+ M deter opps from reaching 4M? Do you believe Ekrens 2♦ gives opps two 2M cuebids for free? The best defence against Ekrens that I've been able to come up with, is actually a generic 2♦ defence where X is basically takeout of diamonds so as to be able to handle situations like (2♦)-P-(P), (2♦)*-X-(P)** * Ekrens ** long diamonds and (2♦)*-X-(P)**. * canapé preempt in spades ** wants to play 2♦X opposite diamonds

You could play 2♦ as a "canapé preempt" with spades as anchor suit. My preferred strength is "0-9(10)" (doesn't meet the rule of 19) in 1st seat and shapes include * 4S4D(41) * 4S5m(31) * 40(54) * 4S6+m * 4450 * 4S5+H * 5S6+H, but currently not 4S5m22, which I prefer to open with Pass = "0-10 BAL". 2♦-?: (with non-MAX = doesn't meet the rule of 16, but Responder assumes the rule of 13 is met) MAX = meets the rule of 16, but not the rule of 19) P = allowed 2♥ = P/C OR INV, 3-S5+H ...P = 4+ H OR hand doesn't meet the rule of 13 ...2♠ = 3- H, hand meets the rule of 13 ......P = allowed ......2N = < INV, "bid your minor!" ......3♣ = INV, 3-S5(+)H2+C ("P/C") ......3♦ = INV, 3-S5(+)H1-C ......3♥ = INV, 3-S6+H 2♠ = to play 2N = GF relay ...E.g. (what nullve-nullve play): ...3♣ = 3-H5+D OR (if possible) 4144 ......3♦ = relay .........3♥+ = A(♦). See A(m) below. ......(...) ...3♦ = 4+ H ......3♥ = relay, aksing for 5+ H unless slam interest .........3♠ = MAX, 4 H .........3N = non-MAX, 4 H .........4♣ = MAX, 4S5+H .........4♦ = MAX, 5S5+H .........4♥ = non-MAX, 5+ H ...3♥+ = A(♣). See A(m) below. 3♣ = INV, 3-S4-H2+C ("P/C") ...E.g. (what nullve-nullve play): ...P = non-MAX, (4)5+ C ...3♦ = non-MAX, (4)5+ D ...3♥ = non-MAX, 4+ H ...3♠ = MAX, either 5+ m or 4 H ......3N = to play opposite 3-H5+m .........P = 3-H5+m .........4♣ = 4441 :( .........4♦ = 4450 :unsure: ......(...) ...3N = MAX, 4S5H ...4♣ = MAX, 4S6+H ...4♦ = MAX, 5S6+H ...(...) 3♦ = INV, 3-S4-H1-C 3♥ = INV, 4+ S 3♠ = preempt (...) A(m): 3♥ = non-MAX, either 3-H5+m or, if m=♦, 4144 ...3♠ = relay, asking for 3 H unles slam interest ......3N = 2- H ......4♣+ = 3 H ...(...) 3♠ = MAX, 2-H5+m 3N = MAX, 3 H 4♣(m=♦) = MAX, 4144 :(.

There's more space to handle 20-21 BAL after 2♣(20-21 BAL or GF)-(2♦/♥/♠) than after 2N(20-21 BAL), so it's not obvious to me that opening 2N would work better in this case. For example, over 2♣-(2♥) one could play P = NF (Opener may pass with 20-21 BAL and 3+ H) X = takeout (also on some GF hands with 4+ S) 2♠+ = Rubensohl-ish, e.g.: 2♠ = NF, 5+ S 3m-1 = F1, 5+ m 3♦ = GF, 4 S, unsuitable for X 3♥ = GF, 5+ S, unsuitable for X (...) and thus be able stop in 2♠ and 3m oppsosite 20-21 BAL, which is impossible after a 20-21 2N. The structure should work reasonably well also opposite a GF Opener.

Sorry, I meant to say that the 2♣ openng can be treated as highly limited if unbalanced. So over 2♣-2♦, something like 2♣-2♦; ?: 2♥ = Kokish: "22-25, 5+ H, unBAL" OR "20-22 BAL" ...2♠ = waiting ......2N = "20-22 BAL" ......3♣+ = "5+ H, unBAL, NAT" ...(...) 2♠ = "22-25, 5+ S, unBAL" 2N = "26+ BAL" 3♣ = "22-25, 5+ C, unBAL" 3♦+ = "22-25, 5+ D, unBAL". Of course, the "22-25" range must be taken with a grain of salt, just like the "11-15" openng range in Precision, for example.

If Opener has 4540 shape, then he's about 40 times as likely to have 22 hcp as 26 hcp. Similar results hold when Opener has other unbalanced unbalanced shapes, so I think it makes sense to pretend the 2♣ Opener is highly limited. But if Opener is highly limited, then there's little reason for Responder to limit his hand with either a negative 2♥ response or a second negative.

1♦-1♠ 2♠-P

4♦

So in tree form, 1NT = 13*-19, either "44 majors & 32 minors or 43 majors and 33 minors which means you do NOT have a 5+ card major or a 4+ card minor" ...Pass = "0-8" ...2C = "6+ Stayman asking for a 4 card major" ......2D = "13-15" .........2HS = "6-9" .........2NT = "10" .........3CD = "11" .........3HS = "12+" ......2H = "16-18" ........."No Fit": .........2S = "6" .........2NT = "7" .........3D"[4spades]" = "8" .........3NT = "9+" ........."with a Heart Fit": .........Pass = "6" .........3C"[=4hearts]" = "7" .........3H = "8" .........4H = "9+" ......2S = "16-18" ........."No Fit": .........2NT = "6-7" .........3C"[4hearts]" = "8" .........3NT = "9+" ........."Spade Fit": .........Pass = "6" .........3D"[=4spades]" = "7" .........3S = "8" .........4S = "9+" ......3HS = "19" ........."major or 3NT game is bid" ...2D = "Transfer" "to 2H" ......2H .........Pass = "0-8" .........3C"[4hearts]" = "9-10 [or use 2NT=9-10]" .........3H = "11" .........4H = "12+" ...2H = Transfer "to 2S" ......2S .........Pass = "0-8" .........3D"[4spades]" = "9-10 [or use 2NT=9-10]" .........3S = "11" .........4S = "12+" ...2S = "9-10" ...2NT = "11" ...3NT = "12+" * "13+HCP = 12+ guarded which is any A, at least Kx, Qxx, + Jxxx=1 12+1=13 STOPPER in at LEAST 3 suits" ?

An idea I've been working on for some time: Suppose 2♣ = <range 1> BAL OR <range 2> BAL OR unBAL GF not with primarily diamonds 2♦ = Weak 2M OR <range 3> BAL OR unBAL GF with primarily diamonds and that 44(41), 4M5m22 and 5M4m22 are treated as balanced. Then it's possible to play 2♣-2♦; ?: 2♥ = "5+ H, unBAL" OR <range 1> BAL ...2♠ = relay ......2N = BAL ......3♣+ = 5+ H, unBAL, R(♥). See R(x) below. ...Anti-relays: ...2N = 2-H6+C (=> 3♦+ = unBAL) ...3♣ = 2-H6+D (=> 3♥+ = unBAL) ...3♦ = 6+S2-H (=> 3♥/3N+ = unBAL) 2♠ = "5+ S, unBAL" ...2N = relay ......3♣+ = R(♠). See R(x) below. ...Anti-relays: ...3♣ = 2-S6+C ...3♦ = 2-S6+D ...3♥ = 2-S6+H 2N = <range 2> BAL 3♣+ = "(4)5+ C, unBAL", R(♣). See R(x) below. 2♦-2M; ?: P = weak, long M 2N = <range 3> BAL 3♣+: As if 3♣+ = "GF, (4)5+ D, unBAL", R(♦). See R(x) below. except that 3♦(M=♠) = Weak 2♥ OR as in R(♦). See R(x) below. R(x): (with ab(x) = the "adjacent" suit "below" x (S,C,D,H if x is C,D,H,S, respectively) na(x) = the suit not "adjacent" to x (H,S,C,D if x is C,D,H,S, respectively) Or(x) = the other suit of the same rank as x, and lo,mi,hi being the lowest-, middle-, highest-ranking suit outside x, respectively) 3♣ = 4+ lo, not 5-5 OR 1-suited ...3♦ = relay ......3♥ = 1-suited ......3♠+ = S(x,lo). See S(x,y) below. ...(...) 3♦ = 4+ mi, not 5-5 ...3♥ = relay* ......3♠+ = S(x,mi). See S(x,y) below. ...(...) 3♥ = E(x). See below. 3♠+ = S(x,hi). See S(x,y) below. S(x,y): 3♠ = 5x4y13(04) (e.g. Mulberry over this) 3N = 5x4y31(40) (e.g. Mulberry over this) 4♣+ = 6+x4+y**. Maybe something like 4♣ = 6x4y (e.g. Mulberry without 4♣ over this) 4♦ = 6x5y 4♥ = 7x4y And if we don't want to stop there, why not 4♠ = 6x6y 4N = 7x5y 5♣ = 8x4y 5♦ = 7x6y 5♥ = 8x5y 5♠ = 9x4y ? :blink: E(x): Basic idea: 3♥ = 5x5ab(x) OR 5x5na(x)2+Or(x) OR (if x=m) 4m1na(m)44 OR (if x=M) 5M4OM22 ...3♠ = relay ......3N = 4m1na(m)44 (if x=m) or 5M4OM22 (if x=M) ......4♣ = 5x5ab(x) ......4♦+ = 5x5na(x), 2+ Or(x). In more detail: E(♣): 3♥ = 5S5C OR 5H2+D5C OR 1444 ...3♠ = relay ......3N = 1444 (e.g. Mulberry over this) ......4♣ = 5S5C (e.g. Mulberry without 4♣ over this) ......4♦ = 5H2+D5C, even # of KC(♣) (=> e.g. 4♠ = RKC(♥), 4N = ♣Q ask, agreeing C) ......4♥ = 5H2+D5C, odd # of KC(♣) (=> e.g. 4♠ = RKC(♥), 4N = ♣Q ask, agreeing C) ...(...) E(♦) 3♥ = 5D5C OR 5S5D2+C or 4144 ...3♠ = relay ......3N = 4144 (e.g. Mulberry over this) ......4♣ = 5D5C (e.g. Mulberry without 4♣ over this) ......4♦ = 5S5D2+C (=> e.g. 4♥/4N = RKC(♦/♠), respectively ...(...) E(♥) 3♥ = 5H5D OR 2+S5H5C OR 4522 ...3♠ = relay ......3N = 4522 (e.g. Mulberry over this) ......4♣ = 5H5D (e.g. Mulberry without 4♣ over this) ......4♦ = 2+S5H5C, even # of KC(♣) (=> e.g. 4♠ = RKC(♥), 4N = ♣Q ask, agreeing C) ......4♥ = 2+S5H5C, odd # of KC(♣) (=> e.g. 4♠ = RKC(♥), 4N = ♣Q ask, agreeing C) ...(...) E(♠) 3♥ = 5S5H OR 5S2+H5D OR 5422 ...3♠ = relay ......3N = 5422 (e.g. Mulberry over this) ......4♣ = 5S5H (e.g. Mulberry without 4♣ over this) ......4♦ = 5S5D2+H (=> e.g. 4♥/N = RKC(♦/♠), respectively) ...(...) * Passable with a Weak 2♥ in the sequence 2♦-2♠; 3♦-3♥. ** If Opener has shown at least 10 cards in x and y at 4♦ or higher, but nothing yet about key cards, then the lowest/2nd lowest bid by Responder outside x and y could be used as "RKC" in the lowest/highest ranking suit of x and y, respectively. [i'll fix mistakes when I see them.] Edit: Random examples (work in progress): Added, starting 3 May 2020:

Idea: After winning the first trick with the ♦A, play a spade to the A and another spade. If East wins the second spade, then blah blah. If he ducks, play a third spade and discard the ♥Q from hand. If East wins the third spade, then blah blah. If he ducks again, cash the ♦K and play three rounds of clubs and discard a spade or small heart from dummy. If clubs are 3-3, 12 tricks are guaranteed. If clubs are 4-2, in which case West is almost certainly the one with four, play a fourth club and make sure to keep ♥AJ and the remaining diamond in dummy, forcing West to give you two heart tricks unless he can exit with a third diamond.

P-1♠; 2♣-P = "No suitable call -- 6+ ♠; 11+ HCP; strong rebiddable ♣; biddable ♠; 12-13 total points" :)