Bridge Tips

When playing in a no-trump contract, we generally try to establish our long suit, but we must be aware that the opposition have probably led their longest and strongest suit.

When we do not have enough STOPS to win the race to establish our suit, we have to try to break the opponents’ communications so that although they may establish their suit first - they cannot cash it because the wrong player is on lead.

We do this by holding up in their long suit until one of the players has none left.
A little RULE invented to help you know when to hold up is the called
The Rule of Seven:-
And it works like this:

Add up all of the cards in the opponents suit which you hold in your two hands.
Subtract this number from seven.
The answer is how many rounds you ned to hold up.

NOTE: The Rule of 7 assumes that you are worried about a five-card suit.

		♠	82			Here is an easy example:
		♥	QJ7			Knock out the ♦ A and we have 10
		♦	K876			tricks, however the opponents have
		♣	AK42			the advantage of leading and they
♠	KQJ74			♠	1095	will knock out our ♠ A first. Therefore
♥	652			♥	10983	we need to break their communications.
♦	2			♦	A105	If spades break 4-4, we will be OK, for
♣	10986			♣	QJ5	we will lose just 3 spades and ♦ A.
		♠	A63			What matters is when they break 5-3.
		♥	AK4			Using the Rule of seven gives 7-5 = 2
		♦	QJ943			So we hold up for two rounds and
		♣	73			eventually win the third round.
3NT by SOUTH						Fortunately it is East who holds ♦ A
Lead = K♠						and he has no more spades. So we can

claim 10 tricks. Had West held ♦ A we would have had no chance.

		♠	52			Here is a second fairly straightforward
		♥	954			example. We have 9 tricks once
		♦	7542			the ♦ A is knocked out, however
		♣	AKQJ			once again the defence have picked a
♠	KQJ103			♠	76	weak spot. How long to hold up this
♥	763			♥	KQJ	time? A 4-3 break doesn't matter:
♦	96			♦	A83	We will lose 3 spades and the ♦ A.
♣	542			♣	109876	5-2 is the important break: The
		♠	A984			Rule of Seven gives: 7-6 = 1. So we
		♥	A1082			hold up for one round only.
		♦	KQJ10			We win the 2nd round and knock out
		♣	3			♦ A. Once again East is spadeless
3NT by SOUTH						and we make 9 tricks very easily.
Lead = K♠						Note. If we hold up for a 2nd round

the defence can switch to hearts and take us one down. They make 2 spades, 2 hearts and ♦ A.

		♠	K92			A third example to finish: Again a
		♥	Q76			straightforward hand. We once again
		♦	8765			have a battle to establish our long
		♣	AQ9			suit. The defence have the advantage
♠	QJ1063			♠	87	as always. We have 8 tricks on top
♥	985			♥	J10432	and plenty of diamonds if we can
♦	K3			♦	A2	establish the suit. But spades are a
♣	865			♣	J1043	problem. West will win the race so
		♠	A54			we must cut East off. But how many
		♥	AK			times to hold up? The rule of 7
		♦	QJ1094			does not help. It only applies when
		♣	K72			we have one certain stop; not two.
3NT by SOUTH						When we have two stops, we must
Lead = Q♠						work out how many times we

must/might lose the lead. and then subtract one from that number. If we might/will lose the lead once - there is no point in holding up to cut communication (we might hold up for other reasons). But if we might/will lose the lead twice, then we must hold up once.
Note: on this hand if we do not hold up the first spade, then East will win the first diamond with his Ace a return a spade. Now no matter if we hold up or not. West has the boss diamond and will be then be able to cash his esatablished spade suit.
Holding up once will cut their means of communication.