Hand of the Week
| West | North | East | South |
| 3♥ | Dbl | ||
| Pass | 4♥ | Pass | 4♠ |
| Pass | 5♠ | Pass | 6♠ |
| All Pass |
The North hand is far to strong to be content with a simple 4♠ in response to your takeout double, so North must cuebid 4♥. Once the spade fit came to light, North showed his extra values with a raise to 5♠, while denying a heart control. With the control in hearts, you had an easy raise to 6♠.
West led the ♥6. East won the ♥K and continued with the ♥A.
You ruffed with a low trump, as West followed with the two of hearts. All seemed easy until you cashed the ace of trumps only to find East discarding a heart. After checking with East that he was sure he didn’t have a trump, how do you plan to make the slam (assuming West began with exactly two hearts)?
Solution
You can always make 12 tricks as long as West has exactly two hearts and at least two cards in each minor. Suppose the full deal is:
Your first move is to lead the three of trumps. If West plays the ♠9, dummy will win the trick with the queen. Then you cash the ♦A and play a diamond to your queen, continuing with the ♣A and ♣K. Next you play the ♣Q. When West follows, you discard a diamond from dummy and persist with the ♣J. West ruffs and you overruff. As West is 5-2 in the majors with three clubs, his original distribution must have been 5=2=3=3, so it is safe to cash another diamond. You have taken 10 tricks and make the last two on a high crossruff.
The play is similar when West holds four clubs, in which case you will discard two diamonds from dummy and only then lead a third diamond. West will ruff and dummy will overruff cheaply. You will score the last two tricks on a crossruff.
It would not matter if West could ruff the third club, for then his original distribution would be 5=2=4=2. You would overruff and cash the two remaining diamond winners before taking the last two tricks, as before, on a high crossruff.
The overall chance of success when West began with five spades and two hearts is a surprising 87%.