How does a partnership win a rubber in contract bridge?
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 contract bridge, a partnership wins a rubber by being the first to win two games.
A game is won when a partnership accumulates 100 or more trick points below the horizontal line on the score sheet, which may be achieved over the course of one or more deals. Once a game is completed, a new game begins.
Upon winning two games, the partnership is awarded a rubber bonus:
* 700 points if the opponents have not won a game.
* 500 points if the opponents have won one game.
After the rubber is concluded, all trick and honour points are totaled for each side, and the partnership with the higher total wins the difference from their opponents' score.1