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Show How to Flay Kiv BRIDGE I "W Wynne Fergus oa yVyW'L Author of PRACTICAL AUCTION BRIDGE CopyrUbt, 1929, by Hoyle, Jr. ARTICLE No. 18 Don't take your game of Bridge too seriously. The result of doing bo is dearly shown by the following: "Because ber Bridge partner played the wrong card twiois Mrs. Mary Lee Bishop shot and killed her. Mrs. Bishop was sentenced to life imprisonment in the Detroit bouse of correction by Judge John V. Brennan yesterday. Mrs. Bishop killed Mrs. Rosa Lee Henderson, 4402 Beaubien Street, December De-cember 27, 1927". Detroit Free Press. The humorous aspect of this unfortunate un-fortunate occurrence is shown by the following comment of a well-knov.n player: "Evidently the Jur4ln this case was not made up of Bridge players, else it would have returned a verdict of 'justifiable homicide.' " In the preceding article a number of "End Plays" were given for analysis by our readers. The writer's analyses 1 follow: Answer to Problem No. 14 Hearts none Clubs A, Q, 9 Diamonds none A ' Spades J, 8 Hearts none Hearts 10, 8 Cmbs J, 7, 5, 3 : Y : Clubs K, 10,4 Diamonds 8 : A B : Diamonds none Spades none : Z : Spades none k I Hearts 7 I Clubs 8, 6, 2 Diamonds none Spades 9 Spades are trumps and Z is in the lead. How can Y Z win four of the five tricks against any defense? Solution: Z should lead the seven of hearts and discard the nine of clubs from Y"s hand. B is thus forced into the lead and must lead either (1) a heart or (2) a club. If B leads a heart, Z should trump with the nine of spades and Y should discard the queen of clubs. Y must then win the balance of the tricks. If B leads a club at trick two, Y must win the balance of the tricks. Jj Answer to Problem No. 15 4 ' , Hearts none .' Clubs none Diamonds none Spades A, Q, 7, 3 Hearts none - Hearts none Clubs none : Y : Clubs 10, 5 Diamonds J, 8 :A B: Diamonds none Soadea 8, 6 : Z : Spades K, 10 i ' ' Hearts none . Clnbs 9 Diamonds none Spades J, 9, 5 There are no trumps and Z is in the lead. How can Y Z win two of the remaining tricks against any defense? Solution: Z should lead the nine of dubs, thus forcing B to win the trick. After B has also made the five of dubs, he is forced to lead spades up to the ace queen in Y's hand. End Plays in Problems 14 and 15 are fine examples of "throwing" the lead and forcing the adversaries to lead away from good cards to their disadvantage. Answer to Problem No. 16 V . Hearts J, 9, 6 Clubs J, 7 Diamonds 6 Spades none Hearts none Hearts none Qubg K, 9 8 : Y t Clubs 3 Diamonds K, 8, 7 : A B : Diamonds 10, 9, S, 4, 3 Spades none : Z : Spades none I Hearts 8, 4 Clubs A, Q, i Diamonds A, J Spades none There are no trumps and Z is in the lead. How can Y Z wm all of the tricks against any defense? Solution: Z should first lead the ace of dubs and then place Y in the lead with a low heart. V should then lead : the jack nine of hearts and A and B must make three discards and Z one discard. B's discards are immaterial as he has no winning cards. Z should discard the queen of clubs. A, however, is forced to discard to his disadvantage. His first two discards should be the nine of dubs and the seven of diamonds but what will his third discard be? If he discards the. king of clubs, Y's jack of clubs and Z's ace of diamonds will win the last two tricks. If A discards the eight of diamonds. dia-monds. Y will lead a diamond and Z will wm the last two tricks with the ace jack of diamonds. No matter what A's third discard is, Y Z must win all of the remaining tricks. This forcing of an opponent to discard winning cards to his disadvantage is the so-called so-called "squeeze" play and should be thoroughly understotxl by all players. j Answer to Problem No. 17 Hearts 8 Clubs none Diamonds none Spades A, Q. 8 Hearts 7 Hearts 5,3 dubs none Y : Clubs none Diamonds 4 :A B: Diamonds J 6 Spades K, J t Z ; Spades none Hearts A, 4, 2 ' Clubs none . Diamonds A Spades none Spades are trumps and Z is in the leadVHow can Y Z win all of the tricks against any defense? Solution: Z should lead the ace of diamonds and trump in Y's hand with the eight of spades. Y should then lead the eight of hearts and win the trick in Z's hand with the ace. Z should now lead the deuce of hearts and no matter what A plays Y must win the balance of the tricks. Y's play at trick one, of . . trumping his partner's winning card, the ace in this instance, is the so-called "Grand Coup". Its object is to shorten his trump holding bo that he will not be forced in the lead and forced to lead trumps up to A's hand. Suppose at trick one Z should lead the ace of diamonds and Y should d Li-card Li-card the eight of hearts. That would leave Y with three trumps and thus force him to trump the second trick. He would then be forced to lead trumps and A would thus make the Icing of spades. This play comes up very seldom, but try not to miss it if it does. |