jimmyg010

OK. What about 1C-1NT and 1C-2C, which are forcing to game with 5C+ and 5D+ respectively, can these contain 4 card majors in your system?

Maybe you're right. Suppose opener has 5D-4S and responder 5D-4H. It might go something like 1D-2D-2S-3H-3N/4D etc. Whereas if you bid the major first, 1D-1H-1S-..2C? Then I guess diamonds on the next round. This might be better and keeps the bidding lower. Swap the majors around so opener has 5D-4H and responder 5D-4S. The auction would be 1D-1S-2D-...? 2H looks like a real suit, so 3C is your only bid. Now you're very high and haven't shown any diamond support. However going through 1D-2D-2H-2S... makes it clear there's no fit much sooner, so diamonds must be the trump suit and you're already game forced. What to do :P

Thank you all for your replies. Phil: "Over your 1c, it makes no sense to play IM, since 1C is really just noise. Just make a forcing call in your longest suit, even if it's clubs. " Since 1C is only 2+ we employ different methods: 1D=H; 1H=S, 1S=no major; 1NT=5C+ GF; 2C=5D+ GF. Both the latter bids may contain 4 card majors. Opener completes the transfer with a weak NT and 2 or 3 cards, everything else is natural: 1C-1NT-2M-3M finds a major fit (and a 9 or 10 card club fit!). Our 1D opening is always 5 unless 4441, with responses 1M=nat; 1NT=6-9 no major; 2C=5C+ GF (or 3334), may have 4 card major; 2D=inv+ 4D+ (3D if 3334 not GF). Just a question of whether 1D-2D allows a 4 card major or not. Though standard is to deny a 4 card major, I don't see a problem with this when GF. Currently we have no agreements after 1D-2C and 1D-2D; everything is natural and I don't see any problems (though I haven't played this much yet). Shouldn't be too hard to miss a major fit: 1D-2C-2M-3M and 1D-2D-2M-3M finds them quite easily. We don't even have a meaning for opener's rebid of 2NT yet as he can't be balanced. --- I suppose one issue is after 1D-2D-2M, how does responder show the difference between a invitational hand and GF? I guess after 1D-2D-2M, 2NT and 3D are always invites and the two remaining suits are GF and possibly not real suits.

Thanks for your reply, Badger. We do play that inverted minors are invitational or better, so may or may not have GF values. With a 4 card major, invitational hands always bid the major first regardless of a longer minor (either the other minor or opener's), which I believe is standard. However, I was wondering about those specific hands where you have: - a 4 card major and a strictly longer minor, - that minor is opener's minor, - GF values.

We play transfer Walsh (1C=2C+ bal, may have 5D or 5C+/(441)=4 unbal) and our 1D opening promises an unbalanced hand with 5D+ or 4441. Playing inverted minors, should 1D-2D always deny a 4 card major? I would say yes, but am uncertain about hands with 4M-5D and GF values. What about playing better minor? The same question goes for both minors: can 1m-2m contain 4M-5m with GF values?

In fact, the following squeeze answers all 3 questions at once! When clubs is trumps the AC must be cashed first, and so: - West is squeezed first, - The last winner to be cashed outside the common suit must be the left suit, - South needs an entry in the common suit. Solved![hv=pc=n&s=skha2d2ca&w=shqjtdkqc&n=shk63dajc&e=sah987dc2]399|300[/hv] Observe that in no-trumps, by cashing AD followed by AH the hand is just an ordinary simultaneous squeeze.

Many thanks for all your help, Lovera. I've managed to answer question 2 and construct an example where the last left suit winner must be cashed last. [hv=pc=n&s=skh2daca&w=shkqdkqc&n=sha9dj2c&e=sahjtdc2]399|300[/hv] Playing in no-trumps, South can cash either ace first. Cashing AD first reduces the position to an ordinary simultaneous Type R double squeeze. However if clubs (the free suit) are trumps, then the AC must be cashed first, squeezing West before East and meaning the last winner outside the common suit to be cashed is the left suit. I have decided to classify all Type R double squeezes as follows: - Simultaneous: last winner outside common is free suit. - Sequential: one winner outside common and free suit that will squeeze one defender before the other - Reciprocal: winners in left and right suit and none in free suit, choice of which defender to squeeze first. - Recessed: the first squeeze card is held by North (really just an extended Sequential type where East is squeezed by North before entering the South hand to squeeze West).

I have found an interesting Type R sequential squeeze where one can choose who is squeezed first! [hv=pc=n&s=saj2h2dac&w=shqjtdkqc&n=shak6dj2c&e=skqh987dc]399|300[/hv] If South cashes AD first, East is squeezed. If AS is cashed first, a diamond or heart is pitched depending on West's discard (he is positional squeezed). This means my answer to question 1 is yes. As for question 2, I can't construct a hand where the left winner MUST be cashed last. Indeed in the hand above it can be cashed last, but this isn't required.

Unfortunately I'm not sure which sections of the book you're referring to as I have the new edition and the exercises seem to be numbered differently. I'm also afraid I don't know how your most recent responses help towards my first and second questions!

Hi Lovera, Assuming you mean this post: "Indipendently from the Exercise 31 let's study this position: Q 9 in North against J 10 in West in spade, K 8 and A 6 2 in North South against Q x x and J x x in West East in heart, diamond J in South against Q in East, Ace in South in club and 2 in North. When you play club Ace West pitch heart otherwise you have the extra trick, then little heart to King but when you play spade Queen East must discard heart or diamond and you 'll have the extra trick. " I have rotated the hands so that South has the single threat (the right threat) to keep it consistent with my original set of questions. [hv=pc=n&s=sq9hk8dc2&w=shj43dqc8&n=sha62djca&e=sjthq75dc]399|300[/hv] I believe you've answered question 3, so thank you! South does indeed need the KH entry and North has the squeeze card. --- That leaves questions 1 and 2. I have no idea about 2, but my suspicion for 1 is that there are no Type R sequential double squeezes where West is squeezed before East (indeed, in the hand above East is squeezed before West again). I think the reason is because the squeeze against West is positional: you need to play the squeeze card from South, and upon West's discard you must throw either a threat or a small card needed as an entry from North - this destroys the squeeze against East. Perhaps this is correct? It's not exactly a proof though!

Thanks for your reply, but unfortunately it doesn't quite answer the question as those aren't Type R squeezes. I need the single threat hand to have the right threat, but in your examples that hand (North) has the common threat.

Hi, apologies if this is the wrong forum - not sure what qualifies as expert or not. I'm reading C. Love's "Bridge Squeezes Complete" and have some questions regarding the Type R Double Squeeze: 1) Sometimes a double squeeze is sequential - the simple squeeze against one defender is extended while the other defender still has idle cards. In all examples in Love's book and hands I can construct myself, East is always squeezed before West (where South is the single threat hand - holding the right threat). Are there any examples where West gets squeezed first? If there aren't any, why not? 2) Love says, "Positions can be constructed where the left suit must be cashed last, but since you are unlikely ever to encounter one, the wording of the Rule ("cash left suit winners early") as it stands it satisfactory for practical purposes". Any examples of where this rule fails? 3) "When the hand opposite the common threat [south] has no winner in the common suit to serve as an entry, the squeeze card must be played from that hand". This makes sense, but can someone construct a Type R squeeze where, if South has an entry in the Common suit, North has the squeeze card and South's entry in the common suit is actually needed after the squeeze trick? Many thanks for any help.

Thanks, and that's fine - it's been enjoyable reading both the constructive and not so constructive responses :)

Apologies if this does not belong in the expert-class forum. I was reading about 5NT pick a slam recently and wondered how this might affect the (admittedly much rarer) cases where you want to force to 6NT and invite to 7NT. E.g. P opens a 15-17 1NT and you hold a flat 20 count with no 4 card major. Playing 5NT pick a slam, how would you find out if partner is a maximum for 7NT? Thanks.

Multi-Landy and penalty double against weak NT. Multi-Landy and Woolsey double (4M-5m) against strong NT.