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D, and nine rounds played; and also that eight trumps are out; and further, suppose A has one trump only, and his partner B the ace and queen, and the adversaries C and D have the king and knave of trumps between them. A leads his small trump, C plays the knave. B should play his ace of trumps upon the knave ; because D having four cards remaining, and C only three, it is 4 to 3 in B's favour that the king is in C's hand: reduce the number of four cards to three, the odds then are 3 to 2 : and reduce the number of three cards in a hand to two, and the odds then are 2 to 1 in favour of B's winning another trick, by putting on his ace of trumps. By the same rule play all the other suits.
3. Suppose you have the thirteenth trump, and also the thirteenth card of any other suit, and one losing card, play the losing card ; because, if you play the thirteenth card first, the adversaries, knowing you to have one trump remaining, may not pass your losing card, and therefore you play 2 to 1 against yourself.
4. Suppose you have the ace, king, and three small cards, in any suit not played, and that it appears your partner has the last trump remaining, lead a small card in that suit, because it is an equal chance that your partner has a better card in it than the last player ; if so, and there be only three cards in that suit in any one hand, you win five tricks; whereas, if you play the ace and king, it is 2 to 1 that your partner does not hold the queen, and consequently you win only two tricks. This method may be taken in case all the trumps are played out, provided you have good cards in other suits to bring in this ; and observe you reduce the odds of 2 to 1 against you |
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