East opens 1, vulnerable, in third seat. You try 2. West passes, partner raises to 3and East persists with 4. You bid 4and everyone passes. West leads the Q. East encourages, but West cannot dig up another spade and shifts to the 2. East plays the jack and you win the ace. At trick three, you play the A and East obliges with the queen. By now you should have enough information to see your way through to 10 tricks. Plan the play.
Solution:
The full deal:
7 5 3
J 8 2
6 5 4
A Q J 9
Q
10 9 7 3
K 10 9 2
8 7 6 5
A K 8 4 2
Q
Q J 8 7 3
3 2
J 10 9 6
A K 6 5 4
A
K 10 4
It looks for the world as if you have four losers: three spades and a heart. Yet, if you count your tricks, you could easily have 10: three hearts, the A, two diamond ruffs and four clubs. By the way, if you count losers (4) and winners (10) and the total doesn’t come to 13, you are supposed to count your cards!
Eddie Kantar
This deal is the exception to even that rule. You do have a count on this deal. West is known to have one spade and four hearts. He appears to have four diamonds given the bidding and the lead, so he also has four clubs.
Play three rounds of hearts ending in dummy and ruff a diamond. Now back to dummy with a club and ruff dummy’s last diamond with your last trump. Now take three more club tricks: tricks eight, nine and ten – piece of cake. You have taken 10 tricks.
They took the first trick, still have the top spades in one hand and a winning heart in the other for three more tricks! Those three tricks, however, will crash on top of each other on the last two tricks. West will be forced to ruff one of his partner’s spade winners!
7 5 3
J 8 2
6 5 4
A Q J 9
A, two diamond ruffs and four clubs. By the way, if you count losers (4) and winners (10) and the total doesn’t come to 13, you are supposed to count your cards!
