This hand arose in a club event and shows a common play problem when the ace and queen are in different hands. The auction reached the correct contract of 4S. North led the 8H and, since there was no ruff, South exited quietly with the third heart. Now declarer has all the side suit tricks and only needs to play the trumps for one loser. He played the ace of spades and then a small one towards dummy and put up the queen which lost to the king and he must now also lose a trick to the jack and go one off. The question is whether he was unlucky or misplayed trumps. What do you think? If you have memorized a table of plays then you know what to do but most people haven’t done that so how do you work it out at the table? It is not difficult. Start by assuming the trumps are 3-2. If both the king and jack are with South, one will appear on the second round and declarer cannot go wrong. If both are with North, declarer always loses two tricks.
Pairs, Love all, Dealer East
If the honours are split, the ten will win if South has the jack and the queen will win if South has the king. It is a complete guess since each play will win half the time. If trumps are 4-1 with a singleton jack or king it will fall under the ace and there is no problem. If North has KJxx then, declarer always loses two tricks. If South has KJxx then declarer must play the 10 and then lead a second trump towards the queen. No other line works in this case. So we have one case whether the 10 must be played and don’t care whether the Q or 10 is played in all other cases so it is correct to always play the 10! The general rule when the ace and queen are in different hands is to play towards the queen with 9 cards. With seven or eight cards, play the ace and lead to the Q10 and finesse the 10 as above. This rule also applies when the AQ10 are all in the same hand so, if you play AQ106 opposite 7543 by finessing the queen on the first round, you are making a mistake!