Both Sides Now, Part 2

 Looking at the same hand as last time, but from the other defender's perspective...

	♠ J 4 ♥ K J 4 ♦ 8 7 6 5 2 ♣ K 7 2
♠ A 10 9 ♥ 10 8 6 5 ♦ Q 3 ♣ Q 9 5 4	N W E S

You	North	Partner	South
		Pass	1 ♠
Pass	1 NT	Pass	2 ♦
Pass	3 ♦	All pass

North's 1NT was forcing one round. You led the ♣ 4. Declarer played low from dummy. Partner played the ♣ J, declarer played the ♣ 6. At trick two, partner led the ♦ 4.

Who has the ♣ A? It seems unlikely that declarer would hold up the ace here, so partner has the ♣ A.

Who has the ♣ 10? Remember the Third Hand Basics: "Third hand high, but no higher than necessary, and keep control over dummy's honor." If partner had the jack and ten, partner would have played the ten, no higher than necessary. So, declarer has the ♣ 10.

If declarer had only two clubs to begin with, you can lead the ♣ Q when you get in, smothering declarer's ten. If declarer and partner both had three clubs, then declarer's ten is protected and the suit is "frozen": whichever side leads the suit gives up a trick. (Work this out on paper if you don't see it.)

The next six tricks proceed:
♦ 4 from partner, ace from declarer, three, deuce.
♦ K from declarer, queen, five, ten from partner.
♥ A from declarer, five, four, three from partner.
♥ 2 from declarer, six, jack from dummy, queen from partner.
♦ J from partner, nine, ♠ 10 from you, six from dummy.
♠ 5 from partner, two, you win with the ace of course.

What now? You know partner has the ♣ A. If declarer has only the ten remaining, you must lead the queen and get two club tricks (declarer is out of trump). If declarer has two clubs remaining, you give up a trick by leading clubs.

How many clubs did declarer start with, two or three?

Remember the auction? Declarer bid 2 ♦ after the forcing notrump. Declarer had three diamonds. If declarer had three clubs and three diamonds, declarer would have bid 2 ♣ rather than 2 ♦. So declarer started with only two clubs. 

So you lead the ♣ Q, smothering declarer's ten, to get two more club tricks.

AND, A BONUS MINUTE...

At trick two, partner led the ♦ 4 from what turned out to be the J-10-4. Why didn't partner make the normal lead of the jack?

1. Partner knows that you know that she has the ace of clubs.
2. Partner knows that you know that she does not have the ten of clubs, because she played the jack rather than the ten.
3. Partner knows that declarer has either the ten or nine of clubs, because you didn't lead the ten from 10-9.
4. Partner knows that if you have Q-10 of clubs, you can lead clubs safely
5. Partner knows that if you have Q-9 of clubs you can't lead clubs unless she had four clubs originally and that you can't determine for sure how many clubs she had.

Aha! Partner realized that you might have a problem in the club suit. She made a non-standard lead, in this case a low trump, to suggest suit-preference low.  It couldn't have been to encourage a heart lead to her unsupported queen without knowing where the heart ace is. So the  ♦ 4  must be encouraging the other low suit, clubs!

Partner did not know where your entry was; if it was in diamonds, you wouldn't have the benefit of those other five tricks to figure out the count in clubs. So she made a trump suit preference play, hoping you could figure out the clue.

-- Ray 

Better Bridge in 5 Minutes. Guaranteed! (or the next one is free)