Match 2 - Board 37

Board 37
Our side vulnerable

♠ Q 9 2 ♥ A 7 3 ♦ K 9 7 ♣ 9 7 6 2

Two passes to me. I pass. LHO passes. Oh, well.

	NORTH ♠ K 8 7 3 ♥ K J 9 ♦ 5 2 ♣ Q J 10 5
WEST ♠ 6 ♥ Q 10 4 2 ♦ Q 10 6 4 ♣ A K 8 4		EAST ♠ A J 10 5 4 ♥ 8 6 5 ♦ A J 8 3 ♣ 3
	SOUTH ♠ Q 9 2 ♥ A 7 3 ♦ K 9 7 ♣ 9 7 6 2

West	North	East	South
	Pass	Pass	Pass
Pass

The opponents can make three diamonds, but it's hard to get there. If RHO opens, as he would have done in the old days (he does have two and a half-plus honor tricks after all), he would probably wind up in two notrump going down. Every other table passes it out as well, so we wind up with six matchpoints.

Now that we have some time on our hands, let's go back to last week's deal. I glossed over the play of this suit, and it bears some further discussion:

	NORTH ♥ K Q 9 8 3
	SOUTH ♥ 7 2

In the post, I said simply that I led the deuce of hearts from my hand. What I didn't say was it would have been an error to lead the seven. To see why, note what happened. West took the ace, and East, holding ten-four and not wishing to waste his honor, played the four. West was now unable to read the lie of the heart suit. East might have ten-four or he might have ten-seven-four. Had I led the seven, West would have known the count. He might not know whether his partner held ten-four or four-deuce. But he would know his partner held at most two hearts. There is no three-card holding East would play the four from. This was no accident; I led the deuce precisely for this reason.

In this particular case, concealing the heart count from West made no difference. But it might matter on a different deal, so it's worth discussing how one knows that the deuce will be less revealing than the seven. One possibility is to tabulate all of East's possible holdings and figure out which play makes East's card unreadable in the greatest number of cases. Of course, this method is laborious and time-consuming. An easier way (and one less apt to generate late-play penalties) is to follow a few simple rules:

Rules for Scrambling the Opponents' Count Signals

In the ensuing discussion, I will use the name 'Fred' to refer to the defender from whom you wish to conceal information. I will use the name 'Ethel' to refer to his partner, that is, the defender whose holding you are trying to obscure. Sometimes you know ahead of time which defender is which; sometimes you don't. Sometimes Fred may be, for example, whichever defender holds the ace.

Rule 1 - If you wish to represent a particular holding for Ethel, signal as if you held that holding.

Most players know this rule when it comes to attitude signals. Declarer should play high to encourage and low to discourage (or the opposite if the opponents signal upside down). Many players do not know, however, that the rule works with count signals also.

	NORTH ♣ Q J 10 9
	SOUTH ♣ 7 6 2

Fred leads the king of clubs against a heart contract; Ethel plays the four. If Ethel has four-three doubleton, you want Fred to think that it is possible she has three clubs. So you signal as if you had three clubs (which, by sheer coincidence, you do). Since the opponents play standard count, you play low. ("Low" means any card lower than the card Ethel played. If you have choices, you should choose among them at random.)

Suppose, however, you want Fred to continue clubs. Perhaps you know--for whatever reason--that Ethel has three clubs (presumably eight-five-four), and you want Fred to waste a tempo cashing his club ace. You want to represent a doubleton in Ethel's hand, so you signal as if you had a doubleton. Since the opponents play standard count, you play high. Any card higher than Ethel's will do. By concealing the deuce, you leave open the possiblity that Ethel has four-deuce.

Rule 1 is fine if you have a specific objective in mind. But what if you don't? What if you just want to play the card that has the greatest chance of scrambling the opponents' signals? In that case, you need Rules 2 and 3.

Rule 2 - With three or more cards, play your lowest or second lowest card, choosing whichever is closest to middle in rank. If the opponents signal upside down, choose similarly between your highest or second highest spot card. 

For example,

	NORTH ♣ K Q J 10
	SOUTH ♣ 8 5 2

Which card should you lead toward dummy to scramble Ethel's count signal? Assuming standard signals, your choice is between the deuce and the five. The five, being a middle-ranking card, will work more often. Playing the deuce is right only if Ethel has four-three doubleton, since it leaves open the possibility that she has eight-five-four. Playing the five is right anytime Ethel has three clubs, since it leaves open the possibility she has a doubleton.

Note that if dummy had the nine of clubs instead of the ten, the five would work in even more cases. It would work against against any three-card holding and against any even holding that includes the ten (provided Ethel is unwilling to signal with the ten), since it leaves open the possibility she has ten-eight third.

	NORTH ♣ K Q J 10
	SOUTH ♣ 9 8 4

Now the choice is between the four and the eight. Again, the middle-ranking card, the four, is the percentage play. The eight will create an ambiguity only if Ethel has seven-six-five. The four will create an ambiguity any time Ethel has an even number.

Rule 3 - With a doubleton, give correct count with one exception: If Ethel is apt to have a card she doesn't want to part with, give false count. 

Why should you usually signal an even number with a doubleton? Because, in general, it is easier to scramble Ethel's odd signal than it is to scramble her even signal. To scramble her odd signal, all you need to do is to conceal one card lower than the card she plays (or higher if they signal upside down). To scramble her even signal, you must conceal two cards higher than the card she plays (or lower if they signal upside down).

	NORTH ♣ K Q J 10
	SOUTH ♣ 9 2

Signal as the opponents would. Play the nine if the opponents play standard signals; play the deuce if they play upside down. You hope that Ethel has an odd number of clubs, in which case playing the proper card will make it possible that she has a doubleton. It is impossible to create an ambiguity when Ethel has an even number of clubs.

	NORTH ♥ K Q 9 8 3
	SOUTH ♥ 7 2

In this layout, there are two cards outstanding that Ethel might well choose not to signal with: the jack and the ten. Ethel is apt to have one or the other of them, so the exception applies, and you should give false count.

Why does this situation create an exception? Because there are now many three-card holdings from which Ethel's play is automatically ambiguous:

♥ J 6 5

♥ J 6 4

♥ 10 6 5

♥ 10 6 4

Ethel will play low from any of these holdings, but she will also play low from:

♥ J 5

♥ J 4

♥ 10 5

♥ 10 4

(Actually, it doesn't matter whether she will play low from these doubletons or not. All that matters is that, from Fred's point of view, she might play low.)

Normally, against standard carders, you want to retain the deuce when you hold a doubleton. That way, if Ethel plays low from three, it will be possible she holds a doubleton. But if Ethel holds the jack or ten, you don't need to do that. It is already possible that she holds a doubleton. Her signal from three cards is already ambiguous without your having to do anything.

Therefore, you need to shift your attention to scrambling her signal when she holds an even number. And, as we have seen, the way to do that is signal an odd number. If Ethel plays her highest spot card from:

♥ J 6

♥ J 5

♥ J 4

♥ 10 6

♥ 10 5

♥ 10 4

♥ J 6 5 4

♥ 10 6 5 4

then retaining the seven leaves open the possibility that she has three. (If they signal upside down and she plays the lowest spot card, then retaining the deuce leaves open that possibility.)

One addendum: Declarer's choice of equals rarely matters. So, whatever card the above rules instruct you to play, you are free to play any other card of equal rank. Randomizing among equals will help to keep Fred from  exploiting these rules to divine your holding.

Am I being too cautious in saying "rarely" instead of "never"? No. I can't think of a case where declarer's choice among equals matters for an even-odd ambiguity. But it can certainly matter for a two-or-four ambiguity:

	NORTH ♣ K Q J 3
	SOUTH ♣ 9 8 2

You lead the king from dummy, and East plays the seven, presumably from seven doubleton. By Rule 1, if you want to represent three cards in East's hand, you should play low; if you want to represent four cards (so that you might have a singleton), you should play high. But, if you play high, you must play the eight. The nine will not work, since East would play the eight from eight-seven fourth. This falls under the heading of "playing the card you're known to hold."

So there you have it. A simple set of guidelines for scrambling the opponents' count signals. What's particularly nice about these rules is that the opportunity to apply them typically comes up several times per session. So, if you didn't know these rules already, your expected per session score has just gone up perhaps a couple of matchpoints.

Score on Board 37: 0 (6 MP)
Total: 303 MP (68.2 %)

Current rank: 1st