Windboe

First: its not true, that line 2 also works if East has long !H and the♦King. In the four card ending[hv=pc=n&s=sh5da9ck&w=shtdj7cq&n=shk9dqtc&e=shqjdk2c]399|300[/hv] Nord plays ♦Queen or 10, East covering or not proper. On the ♣King East can discard according to Nord's discard. I cannot say which Squeeze is working, because there is no indication about. The line of running the ♠ without cashing the♣King in the hope of getting helpful signals means, the opps have one extra idle card. This might not help but confuse you. That W didnt return wether a ♥ nor a ♦, might lead to speculations but not to a conclusion of worth. I have another line which seems me slight better and the bonus of non confusion: Win the ♠in hand, discard a ♦ on the ♣King, cross to the ♥King and run the ♠. You will win automatic again West AND East if anybody they has the long ♥+ ♦ King AND Jack. Its not a very good line, but I see nothing better.

There are many squeeze possis: a) A simple !C/!H squeeze against W or E, if they have 3!H and !C Queen. b) A double squeeze if W has 7!S and E 3!H c) a triple squeeze against E, if he has stops in all suits. Surely E can convert it to a). d) A simple !S/!C squeeze against W has 7!S and !CQueen. e) A simple !S/!H squeeze Only if W has 3 !H but only 6!S and E the !CQueen third, there is no squeeze and you will go down. Your problem is not wether there is a squeeze. Your Question is: Which squeeze works, if there is one? Might they make mistakes? So try your best by card reading as far as possible. Run the !D, Trick 2-7 and watch the discards. You have much to count, special the outstanding cards in 3 suits. Look who hesitates, there is the one who has problems. After 7 tricks you have: in hand 4!H 1!D 1!C; dummy 2!S 2!H 3!C(with one card to discard). Helpful things will happen: 1. W has discarded 5 !S from his 7-card holding. You discard a !H - not a !S - in dummy, cross via !H and ruff the !S high. 2. E refuses to throw away his third !S, nor he discards a !H. That means he has 2!S 3!H 1!C in his hand. Wether he is a world class player or he has no !CQ. What do you think? 3. My human opponents usually discards !S at East, because they have 3!H and!S discards are so easy. Now you have them. Play a simple squeeze as a double squeeze; cash the !H - ending in hand - and table the last !D. Not that it works automatic. If W has 4 !h I might miscount. But the chance of making is great.

I doubled and played a spade. Was good enough but trump was better from principle. Heart attack seemed me bad. :rolleyes:

To recognise 6 NT as superiour, means, you are not only able to bid your cards, but additional the process of card playing - not bidding. I never have seen any judgement or rule or bidding system, which will be able to achieve this. :)

I agree Fluffy. I do not understand this argue. You have a secure spade entry, and squeeze chances, if the opps don't defend correct. At least you will see some distribution before you answer the diamond question. Ducking the first diamond was the right play. We do not play bridge because we are able to count ten tricks by open card play.

Technically its clear: A trump shortening. S has to ruff 2!D, eliminate the side, reach a 2 card ending and play a side suit from dummy and the trumps of E are catched. That is technically easy. In praxis E might come to the idea to ruff my side suits in which I try to reach the dummy. So I have the next question: what is the distribution of E? Can I read it? In books E has 4-3-3-3, so no problem, but the real life is not so nice. Although I cannot look trough the cards, there are 2 things, which can help me: 1. Its the tendency of the nature to flat everything, to find the middle of all. That means, if W has only a single in !S, there is - after my observation - the nature, God or the universe - will give W exact one more card in the other suits. That is not possible here, so the nature will find the next likely distribution, 1-4-5-3 or 1-3-5-4 with W; respectively 4-2-4-3 or 4-3-4-2 with E. 2. The first free discard of W might help. He discarded a !C and that means his !C are not shorter than his !H. He is more likely to have 5!C than 5!H. So it seems right to start with 4. !D ruff 5.!C to the 9 6.!Druff 7.!H to the Jack 8.!D with !H discard 9.!D with !C discard 10.!H with !C discard But woe to me, my last cards are 2 trumps and a high !C and dummy has two high !C but a !h loser and I might go down, if E originally had a single !C. The right play seems to be 5. to cash the !hQ 6.!C to 9 7.!D ruff 8. !C to dummy to execute the rest like above. This play will fail, if E has a singleton !C or a singleton !H. I could have made with a singleton !C and three hearts by E (or otherwise), reaching dummy with a third round in !h but a second !C, but I have no way seen to conclude this.

:)

Can your P can have K,Jxxxx,Qxxxx,xx for an overcall? In this case I would search a new P. In the other case you have the smaller part of the blame. But the lion's share of the fault takes your P. He has to talk you out the third !D. He is the boss. He gives you the direction first, in second line the information. Worse, seeing the !C menace - not knowing that you stop it - he has to signal ...not a Q, but the urgency to switch to !H. The bid and the signal inform me, that your P likes the vagueness. Not a Partner for me.