jjsb Posted April 2, 2003 Report Share Posted April 2, 2003 this deal is not really there to learn something (eventhhougth u'll see...) it's just some kind of "puzzle" funny to solve . Most of the time "puzzle" are extrememy hard to solve this one is , if not easy , "makeable " and suitable in my opinion for intermediate. contract 7nt South (the auction is really not a necessity) lead J of S AQKQJJ9876545 KAAKQAKQ109876 funny deal rigth . oh and yes don't tell me you have "20 tricks" ok :o ;D regardssyl (jjsb) Quote Link to comment Share on other sites More sharing options...
jjsb Posted April 3, 2003 Author Report Share Posted April 3, 2003 I've got some good answer for this one thansk for those to answer me :o . so what's wrong with this hand !! the C are coming i have 21 top tricks (i counted 20 before lol ) well yes and what if C are 4-0 ?? mmmhhh well ... Diamond are block so i'm in trouble ... well maybe before playing those C i can add a "tiny" little chance . and that chance is the 10 of D singleton . yes i swear it's true !! . see how u can play for adding that "tiny" little chance. U like to throw high cards? it's the hand for that , u'll throw 11 points there !! . first trick you take with the As of S play Q of S discard that Ace of heart bothering you so now you're ready to play your KQJ in H discarding ... AKQ in D and now... u're ready to play that that J of D . the 10 fall?? good things for your very nice play . It doesn't ? well as long as u don't have sure count to finesse in C , u play now your Top club hoping they are not 4-0 , they are 4-0 , you stop bridge for some days :( regardssyl Quote Link to comment Share on other sites More sharing options...
JRG Posted April 13, 2003 Report Share Posted April 13, 2003 The sequence of plays is right, but there is actually an additional chance that occurs. As well as the 10 of diamonds falling singleton or the clubs coming down, the same person may have been dealt the 10-x-(x) of diamonds and J-x-x-x of clubs. If so, they are actually squeezed in the minors and although neither suit splits favourably, you still make the slam. Nice puzzle. Quote Link to comment Share on other sites More sharing options...
jjsb Posted April 14, 2003 Author Report Share Posted April 14, 2003 The sequence of plays is right, but there is actually an additional chance that occurs. As well as the 10 of diamonds falling singleton or the clubs coming down, the same person may have been dealt the 10-x-(x) of diamonds and J-x-x-x of clubs. If so, they are actually squeezed in the minors and although neither suit splits favourably, you still make the slam. Nice puzzle. maybe i don't see it correctly but i must admit i don't see the squeeze here it looks to me the count is not reduce yet ... anyway i'm nto yet awake completely :) regards syl Quote Link to comment Share on other sites More sharing options...
inquiry Posted April 14, 2003 Report Share Posted April 14, 2003 maybe i don't see it correctly but i must admit i don't see the squeeze here it looks to me the count is not reduce yet ... anyway i'm nto yet awake completely :) regards syl There can be no minor suit squeeze simply because you can only cash 5 major suit winners. Thus, either opponent can hold onto two diamonds (Tx) and four clubs (Jxxx) in the 8 card ending, with plenty of room for two extra cards to boot. To put it in Clyde Love's vernacular, the opponent with CJxxx and DTx would be busy in both suits. The club is x to the ACE would serve as the entry, and since the threats are split, U must be right. The problem is that L (loser) is all wrong... you have have only 9 winners (2S+3H+1D+3C), so BLUE is violated. Ben Quote Link to comment Share on other sites More sharing options...
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