In a no-trump contract, we are nearly always active rather than passive and 95% of the time we lead our long suit.

Either:- a high card if we have three or more KQJ32, QJ1072, KJ107, AQJ32.

S 43 H J97 D KQ1075 C A64

The sequence KQ10, is near enough to KQJ to make the King the best lead here.

Or:- fourth highest if we have only one or two high cards in the suit.

S Q J 6 4 3 H 5 4 D A 6 2 C Q 6 2

After 1NT-3NT, the 4 of spades is best. You have a useful entry in the diamond Ace.

Don't forget the fairly simple convention when an honour is led, that you are promising the honour underneath it, and denying the honour above it. Note in passing also that this fourth highest lead often tells you how many cards of that suit are in the hand that has led---often a useful pointer as to the value of persisting with suit. Don't forget that if you are declarer you can do your arithmetic as well. Examples on this issue are:-

1)  the lead of a two shows that the hand has only four cards in that suit (there are three higher and none lower).

2)  you draw the same conclusion from  the lead of, say the four, if you can see the three and the two (in your hand or dummy).

3)  if the six, say, is led and you can't see any of the smaller cards, this may be from a suit with four, five, six, seven or eight cards in it. Not much use! 

POSSIBLE exceptions to the above proposed leads  are if:-

• Partner has bid
• The opponents have bid your long suit
• Your hand is so poor that you are better off trying to guess your partners suit rather than trying to establish your own.

RULE OF ELEVEN

The other piece of information that derives from the opening lead of the fourth highest is the location of the high cards in this suit. This is called the rule of eleven and works like this.

We know that the person on lead has three cards higher than the one she has chosen to lead. Take away from eleven the spot number on the card that is led. ie

if the card led is the six of spades, 11 - 6 = 5.

if the lead was the four of spades, 11 - 4 = 7.

The answer to this simple sum is the number of cards in that suit, higher than the one led, outside the hand that has led ie shared between the other three hands---dummy, declarer and the partner of the hand on lead.

The lead of the six shows there are five ( 11 - 6 = 5) cards  higher than the six shared between dummy, you and declarer. Since you can see all five of them, you can deduce that declarer has none. So, if dummy plays the S 4, the S3 from you is good enough. Partner's six will hold the trick. This revelation occurs all too rarely. (Incidentally, we note in passing that this lead may be from a five-card suit. If this is the case, then declarer started with a singleton in this suit. Is this likely in the light of the no-trump contract?)

The lead of the two shows that there are nine (11 - 2 = 9 cards higher than two outside the hand that has led. You can see seven. So, declarer has two. Not particularly earth-shattering or helpful. If dummy plays the three, you must play your queen even if is likely that it will be hit by the ace. Don't forget you do have a partner who will most likely have some high cards in this suit herself.

Incidentally, the lead of the two shows that partner has only four cards in this suit.

Personally, I don't find the rule of eleven as helpful as the knowledge of how many cards partner has led from.

S Q43 H KJ642 D 94 C Q86	After partner has overcalled in diamonds, when opponents nonetheless have reached 3NT, lead the 9D in preference to your mangy heart suit.
S Q43 H KJ65 D Q43 C Q43	If opponents have bid hearts on their way to a 3NT contract, you must try and guess partner's suit. Prefer a S to a C or a D.
S Q7543 H J87 D J87 C J87	After 1NT-3NT, you have no likely entries even if you do set up your spade suit, so look for partner's suit again. Prefer a H to a D or a C.
Try your hand at this quiz which incorporates questions on trump and no-trump contracts, and also looks at active and passive leads.

Historical Snippet