Hand of the Week
| West | North | East | South |
| 3♣ | Dbl | ||
| Pass | 4♣ | Pass | 4♥ |
| Pass | 5♥ | Pass | 6♥ |
| All Pass |
Partner’s bid of 5♥ asked for a club control, so you bid the slam directly rather than show the ♣A.
West leads the ♦10 to dummy’s ace. East follows suit. As the lead marks the West with a void
in clubs, how do you plan to take 12 tricks when East has two trumps and West three?
Solution
One possible layout is:
After drawing trumps in three rounds, you can place East with seven clubs, two hearts and at least one diamond. After you cash the ♦ K, you know that East began with a 3=2=1=7 shape.
The way to overcome this is to cash your top spades and ruff a spade. East then has only clubs left. So when you cash the ♣A and then duck a club, East has to win the trick and lead away from the ♣K. You will ruff the ♣K and use the ♦ Q as the entry to the good ♣Q. You will make two spades, a spade ruff, four trumps, three diamonds and two clubs.
What would happen if East had followed to the second round of diamonds? The answer is that you would cash your top spades next to try and resolve East’s original distribution. Whenever East began with a 1=2=3=7 shape, you would extract his last diamond by cashing the queen and proceeding as above. If East started with two spades, and so an original 2=2=2=7 shape, you play on clubs in the same way as before.
You might think we have to consider what would happen if East began with a 0-2-4-7 distribution. However that is impossible, for then West would have started with the queen-jack-ten-nine in spades and would surely have led one.