Our new problem is 20 years old. I was in Stockholm to play bridge. I can't recall which tournament it was, and if I was succesful or not, but I clearly remember what happened the night before.
I was staying by my friends, bridge experts Bim and Max Ödlund, as I used to, and we had been upp most of the night, playing chess, listening to music, drinking wine and chatting. When we finally went to bed, Max handed me a piece of paper on which he had scribbled down a bridge diagram:
|Q 9 7 6 5 4 3|
|4 3 2|
|K 10 7 4||Q 9 6 3|
|A 10||K J 8|
|A K Q 4||J 10 3|
|J 10 9||A K Q|
|A J 2|
|9 8 7 6 2|
|8 7 6 5|
It didn't look especially interesting, since East-West have twelve easy tricks if they guess spades correctly. I was just about to say that Sture Ekberg would have gone down in slam, since he always plays "the queen to be over the jack", but I wasn't given the chance. Max said: "You are South and declare a spade contract. How many tricks can you take against the worst possible defense? And how should the play go?"
I couldn't solve the problem then, neither could I at the breakfast next day, so Max had to give me the solution.
Now I ask you, just as Max did: How many tricks can you take on North-South if the defense is the worst possible?
Some solvers have suggested ten tricks, other eleven or twelve, but the correct answer – which most of the solvers have found – is that "the worst possible defense" results in South's taking all thirteen tricks. There are some minor variations, but the start is always the same: that West leads the heart ten, to the queen and jack, and that South at trick two ruffs a heart low, East following with the king. After that, South leads the diamond two to the five (East and West ducking, of course).
Then, the simplest line is that South cashes all high hearts, while the rest of the players discard first clubs, then diamonds. Now, the clubs are high, so South cashes one club trick to reach:
|K 10 7 4||Q 9 6 3|
Now the play is ended with a cross ruff, East and West underruffing "reasonably high" all the time. On a club from dummy, East ruffs with the nine, South overruffs with the jack and West underruffs with the ten. A diamond ruffed with the four, five and three. A club to the spade queen, spade ace and spade king, and then it's all ended with a diamond, ruffed by the seven, eight and six. Thirteen tricks made!
A couple of years ago, a similar problem was published in the Swedish bridge magazine Bridgetidningen, where South should try to make 6 spades on rather slim values...
|10 8 6 4 2|
|5 4 3 2|
|K 10 7 4||A J 8 5 2|
|A K Q J 9 7 5 3||–|
|–||A K Q J|
|A||K Q J 10|
|10 9 8 7 6 5 4|
|9 8 7 6|
Once more you have to set up a crossruff, and the only difference compared to Problem 7 is that East-West now win the last trick.
West leads a low heart, which dummy wins as cheap as possible. Then, South cashes three more heart tricks, West underplaying as expensive as possible, while East discards diamonds and South discards clubs. Then, the deuce of hearts, ruffed with the spade two, overruffed with the spade three. South's diamonds are now high, so he cashes four diamond tricks, both defenders discarding.
Now, the ending is:
|K 10 7 4||A J 8 5|
|6 5 4|
A diamond ruffed by the seven, nine and eight; a club ruffed by the five, six and four; a diamond ruffed by the ten, queen and jack; and then the defenders win the last trick with ace-king of trumps.
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