It might seem that if the number of losers declarer has is added to the number of winners he has, they ought to add up to 13. However, sometimes they don’t.

Consider this case where South is in four spades. West leads the king of clubs, which is allowed to hold, and then the queen of clubs, taken by South with the ace.

Declarer counts his likely losers, and they appear to be one club, two diamonds and one or two trump tricks, so it seems he will go down one or two depending on how the trumps are divided. But South’s prospects are actually much better than that.

After winning the second club lead, declarer plays the A-K of trump, learning that the opposing trumps are divided 4-1, which is not surprising in view of West’s bid. This development confirms that South has five losers, but it does not necessarily mean that he can’t make the contract.

Declarer continues by cashing the A-K of hearts and ruffing a heart. He then proceeds merrily along, trumping a club in dummy, ruffing another heart, leading a diamond to the ace and ruffing dummy’s last heart.

The upshot of all this is that South has accumulated 10 tricks consisting of the A-K of spades, A-K of hearts, ace of diamonds and ace of clubs, plus three heart ruffs in his own hand and a club ruff in dummy.

So, despite starting with five losers, South finishes with 10 tricks. The explanation of this anomaly is that, after 11 tricks have been played, South’s Q-9 of diamonds simultaneously succumb to West’s K-J of diamonds and East’s Q-J of trump on the last two tricks. Four of his apparent losers are thus compressed into two actual losers, which allows South to get home safely with his contract.

Tomorrow >> Famous Hand.

— Steve Becker