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Show " "llv How. to Play iff BRIDGE frXfesj ' ijerUs 192930 by I kkw' Wynne Ferguson gyy Author ef 'PRACTICAL AUCTION BRJDCB," Copyright, by Boyle. Jr. ARTICLE No. 20 One of the most popular forms of indoor amusements is the Auction Bridge problem. You will see interested players in the theatres, aubway trains, at the seashore and other Summer resorts, re-sorts, poring over some double dummy Bridge problem. Just to be in the fashion this article will deal entirely with such problems. The first problem Is one for you to think over during the coming week. Talk it over with your friends and remember that the correct solution must be one that is good against any defense. Don t make the mistake of figuring out a way to make the requisite requi-site number of tricks when your solution solu-tion is possible only because of the bad play of the adversaries. Such a solution is not a correct one. A correct . solution must be one that is possible against any or the best defense. Problem No. 1 Hearts Q, 6, 5 ' Clubs -none Diamonds A, K.7, 6 Spades Q, 4 Hearts K ' Hearts A, T, 4, 9 Clubs Q, 7,6,5 I Y I Clubs 10,9 Diamonds 4,3 ,1 A Bl Diamonds 8,5 Spades 10,fc J Z 1 Spades J Hearts 10.9,8,7' Clubs K, 8. , 2 Diamonds 9 Spades none Diamonds are trumps and Z is in the lead. How can Y Z via eight of the nine tricks against any possible defense? Solution in the next article. The following problem was given in the preceding article. The solution it given here. If s-ou haven't already tried to solve it, do so now before you read the solution. Answer to Problem No,18 Hearts K. J, 9 Clubs 4,3 Diamonds K, 10 c . Spades none Hearts 0, 10, 8, 7 ' , Hearts none Clubs 10V8.6 t Y l' Clubs J9, 5,2 Diamonds none I A B l' Diamonds Q, 9 Spades none ' ' 1 Z 1 Spades A jf' - ' K Hearts 6 Clubs K,Q. Diamonds o4 - Spades K . If spades are trumps and Z is in the lead, now can Y Z win six of the seven trick against any defense? Solution: Z should lead the king of spades. A can safely discard the seven of hearts, Y must discard the ten of diamonds and B is forced to win the trick with the ace of spades. B now has the choice of two leads: (1) the deuce of clubs; (2) the nine of diamonds. (1) Suppose B leads the deuce of clubs. Z should win the trick with the queen of clubs, alt following suit. Z should now lead the six of Hearts, A should play the eight, and Y should win the trick with the nine. B is forced ' to discard. He cannot discard a diamond dia-mond or Y's king of diamonds and Z's nix of dtamon4s will win two tricks, B is forced, therefore, Co discard the Ave of clubs. Y should now lead the king ot hearts and B again is forced to discard. dis-card. He still cannot discard a diamond for the same reason given in the preceding pre-ceding trick; so he is forced to discard the nine of clubs. Z should then discard the four of diamonds and A follows tuit. Y should now lead the king of diamonds. B and Z follow suit arid A is forced to discard. He cannot discard the queen of hearts or Y's jack of hearts will be good; therefore he is forced to discard the eight of clubs. Y now leads the four of clubs and Z must win the next two tricks with the king tnd seven of clubs. (2) Suppose at trick two B leads the nine of diamonds. Z follows suit and A is forced to discard. He cannot discard another heart or all of Y's hearts will be good; so he Is forced to discard the six of clubs. Y wins the trick with the king of diamonds. Y should now lead the trey of clubs and win the trick ia Z's hand with the queen, all following suit. Z should now lead the six of hearts. A should play the eight and Y should win the trick with the nine. B la thus forced to discard. He cannot discard the queen of diamond or Z's six of diamonds wili be good; so he ia forced to discard the five of dubs. Y should now play the king of heart and B is again forced to discard. He cannot discard the queen of diamond or Za ix of diamond will be good; so he ia forced to discard the nine of clubs. Z should discard the six of diamonds and A follows suit. Y should now lead the four of clubs and Z must win the next two tricks with the king and seven of ' clubs. Thus in both cases, Y Z win six of the seven tricks against any defense. It is a pretty problem in the forcing of discards. Both A and B are forced to discard clubs and thus enable Y 2 to make three club trick when only two tricks in that suit appear to be winners. The forcing of discards is one of the most common way that good players adopt to win tricks that the average player would loee. Study this problem and the methods adopted very care' fully. It will well repay you, 1 |
