How many tricks does a bid of one heart require a partnership to win?
Card editor of the London Field W.H. Whitfeld published this bridge problem in 1885. South is declarer and has the lead with hearts as trump. With a sophisticated finesse, South can win every trick. South begins by leading the ace of diamonds, which, depending on what the opponents discard, opens a possible finesse of North's jack of diamonds. Next, South passes the lead to North with a spade that North trumps. North then leads the last heart, and South discards the 10 of clubs. With the lead of the last trump and then the ace of clubs, the defenders are presented with an insurmountable dilemma. East must hold two diamonds or South takes the last two tricks in the suit by discarding a spade. However, in order to hold on to two diamonds, East must discard the jack of spades, which in turn would force West to hold the queen of spades. Since West also needs the queen of diamonds and the jack of clubs to avoid losing a trick, a discard from any of the three suits will allow South to win all of the remaining tricks by an appropriate discard.
In the game of contract bridge, a bid of one heart requires a partnership to win seven tricks.
In bridge, a bid represents a contract to win a specific number of "odd tricks," which are the tricks won in excess of the first six. Therefore, a bid of one heart signifies a commitment to win one odd trick plus the base of six, totaling seven tricks with hearts as the trump suit.1