Seven Notrump

In which some people who play bridge blog about it.

Tuesday, July 03, 2007

Defense basics: the Rule of 11

Every bridge novice learns that it's her moral duty to lead the fourth down in her longest suit when defending against a no-trump contract, and some even learn why, but in the heat of battle, remembering to put the why into practice is often overlooked.

For those who forget, the Rule of Eleven allows you to deduce thusly: assuming the first card led is the fourth down in a suit, subtract its face value from 11; the difference is the number of cards higher than the led card in the other three hands. So, if partner leads the 8 of hearts, subtracting 8 from 11 means that there are exactly 3 hearts higher than the 8 in your hand, dummy, and declarer put together. Doing that math should become automatic for a defender -- sometimes it won't help you much, but in contracts like the following, which I defended a few days ago, it makes all the difference.

Contract: 3 NT 
Lead: ♠ 4
♠ A 10
♥ J 8 3
♦ K 10 8 7
♣ Q J 5 3
♠ 9 8 5 4 3 2
♥ Q 10 7 5
♦ 6 3
♣ 4
♠ K Q 7
♥ K 9 2
♦ 9 5 4 2
♣ K 10 6
♠ J 6
♥ A 6 4
♦ A Q J
♣ A 9 8 7 2

Imagine you're defending this contract, sitting East, and your partner leads the 4 of spades, which dummy wins with the ace. The natural, quick play is to throw your losing 7 on it, secure in the knowledge that your K and Q are going to win two tricks. But let's look at what happens next. Say declarer finesses in clubs twice, leading the Q, on which you play low, and then the J to your K to the ace. Now your 10 is the highest club, and it wins the next round. You play your two winning spades, and that's the end of your run. Declarer has four diamond tricks, three club tricks, a spade, and a heart for the win, no matter what order he takes them in.

However, if you do a quick calculation on the first trick, the outcome could be different. Your partner led the 4 of spades. Eleven minus 4 is seven: there are seven spades higher than the 4 in the other three hands. Dummy has two of them, you have three of them; therefore declarer must have exactly two. Your best chance to defeat the contract is to throw your Q of spades under the ace on the first trick. Then, when you win a trick later, you can cash your K of spades and then play the 7 to your partner's 9. All of partner's remaining spades are good, and as he plays them off, the contract goes down by two tricks at least.

The same applies if your partner is the one to get the lead, with his Q of hearts. If you're hanging onto your K and Q of spades, he can't run his good spades: the suit is blocked. But if you threw off one of your honors on the first trick, keeping a low one to lead back, there's no problem.



  At Friday, July 15, 2011 12:10:00 AM, Anonymous Anonymous said:

The only thing that you didn't consider is that your pd is holding 9 8 5 4 2 and declarer is holding Jxx :-)

  At Saturday, July 23, 2011 12:28:00 PM, Anonymous Anonymous said:


  At Thursday, September 15, 2011 2:35:00 PM, Anonymous Anonymous said:

u r wrong - declarer holds Jx. pd
has 6 cards - and the 7 will be overtaken.

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