In a teams event, you reach a fair contract of 6♥. West leads the ♦9, suggesting that the suit is split 7-2 (West would have led low with three). How will you get to 12 tricks on this deal?
Both South players in a team game reached 6♥ and received the opening lead of the ♦9. The first declarer was a good, if somewhat unimaginative, player. After winning the first trick with the ♦A, he cashed the ♥A and led a second one. When East threw a diamond, declarer had to concede two trump tricks to West, finishing down one. The other declarer was both more experienced and a student of the odds. After winning the first trick with the ♦A, , he crossed to hand with a spade to the ace and led the ♥J, running it when West followed with a low card.
After playing a low trump to dummy’s ace, declarer came back to hand with the ♣A and cashed the ♥K, claiming 12 tricks after conceding a trump to West’s queen.
What are the odds, given the diamonds are breaking 7-2 and there is no defensive ruff? The first declarer succeeds when trumps are 3-2 (about 58%) and also when East has a singleton 10 or queen, about a 70% chance. The second declarer makes his contract when trumps are 3-2 or when West has four trumps including the queen, about an 83% chance. The full deal: