There’s Something About Three

There are many bad connotations surrounding the number 3: ‘Two’s company, three’s a crowd’, misfortunes come in three’s’, ‘the three ravens’, ‘the three witches of Macbeth’ and more. The negative associations carry over to bridge: ‘the 3-card raise’, the 3-3 split’ and ‘a 4-3-3-3 shape’. On the other hand the number eight, a lucky number in China, has positive connotations: ‘an-8-card fit’, ‘an 8-control hand’. We’ll get to these when discussing hands below.

Italian players may have been disappointed to place third in the recent Bermuda Bowl after leading throughout the round robin largely due to 2 adverse slam swings in the last 16 boards of their quarter-final match against Netherlands. I am not that sympathetic, especially after watching members of the team performing in the Italian Mixed Teams Championship. Here is a quiz: which player found room to criticize his female partner on after the following deal.

South	North	S	N
♠ A3	♠ QT972	2♣	2♦*
♥ AK9	♥ Q43	2NT	3♥ *
♦ AK74	♦865	3♠	4NT
♣ KJ92	♣ A7	6NT	Pass

A club was led to the ♣Q and ♣K. Declarer played the ♠ A and a low spade towards the dummy. When the ♠ J appeared from West, she could claim 12 tricks. In the other room the auction was similar: declarer opened 2NT, was transferred to spades, and accepted the invitational 4NT. She too received a club lead, but soon lost her way, succumbing to an urge for trickiness.  She led low to the ♠ Q which Buratti cleverly ducked holding ♠ K654 behind the dummy. So the declarer who made the straightforward play of leading the ♠ A gained 14 IMPs. Who received the criticism?

If you guessed the player who bid and made 6NT you were correct. Her partner voiced the opinion that she should have passed 4NT. The chances of the spades producing 4 tricks is around 5 out of 8, not a worthwhile margin for a nonvulnerable slam. Of course, I think she was right to accept, because 1) she had 9 controls, 2) slam might well be bid at the other table, and 3) his spades could have been better, ♠ QJ9xx having a 59% chance of producing 4 tricks. There is no dependency on the spades splitting 3-3.

It is bad psychology to blame a partner for bidding and making a slam. Better to wait for a more appropriate time. There may have been a carry-over effect on later hand.

South	North	West	North	East	South
♠ T732	♠ 9	2♦*	4♥	Pass	?
♥ AJT8	♥ KQ97653
♦ AJ	♦KT84
♣ A87	♣ Q

The woman who had misplayed 6NT did not hesitate to raise her partner to 6♥ after he had jumped over a Multi-2♦ showing a preempt in an unnamed 6-card major. Well, that was one way to avoid confusion. The player whose partner had criticized her initiative, passed his jump to 4♥ even though she held 3 aces and very good hearts. What might her partner have for his jump to game? I think a confident partner might have found an acceptable way to move forward. Anyway, as I said, any sympathy directed towards the Italians was greatly diminished – maybe third place was where they belonged.

 

Good 3-Card Support

David Burn noted during the BBO broadcast of a 2011 Venice Cup match that a 4-3-3-3 is too often underrated by the player who holds it. True enough, as an 8-card fit is likely, but without any ruffing power the potential for a high number of Total Tricks is low. It is the hand opposite that must take up the slack, which means the ruffs are transferred to the hand with the long trumps in the classical manner of the dummy reversal.

It is common enough to raise on poor 3-card support once partner has shown a 5-card suit. This may lead to problems as a raise on xxx is really an inadequate description. On the other hand, a raise with top honors can be very useful, as in the following fanciful construction where the opener has a 3=3=3=4 shape.

Bob 1	Bob 2	B1	B2
♠ AKT	♠ Q9864	1♣	1♠
♥   KJ3	♥ AQ7	1NT	3♦
♦ K85	♦AQ64	3♠	4♥
♣ T974	♣ 2	6♠	Pass

Neither hand is especially strong, and the division of sides is a common 8-7-6-5 with 16 Total Tricks, but there is very good structure in the trump suit, and nothing wasted in clubs. The result is that there is transportation to be had in the red suits. Routine bidding will not get the pair to slam – responder must be at pains to show his shape and not be put off by a natural opening bid in his short minor. Once the opening bidder is informed of the shortage in clubs opposite, he can bid slam, because he expects tricks in the red suits and ample ruffing of the clubs. It is not easy.

Suppose a spade is led to cut down the ruffs, picking off the ♠ J. A club is lost and a second spade is led, won in dummy. Declarer can ruff a club immediately, and ruff 2 more clubs returning to dummy with a diamond and a heart. A second heart is won in dummy, the last trump is drawn and 2 more diamond tricks and the ♥ A bring the total to 12. Thus we have come to 12 tricks on 28 HCPs. The loser count is 13, so this process gains a trick over the normal expectation of 11 tricks, which is what you get if you draw trumps early, losing a club and a diamond when the diamonds split 4-2 as expected. (Note that even if declarer stops in 4♠ , the reversal process can gain many matchpoints against routine play.)

 

The Probability of a 3-3 Fit

A table of a priori probabilities is based on the possible combination of cards being dealt to 2 players. To calculate the probability of a 4-2 split relative to a 3-3 split when 6 cards are missing in a suit, one merely takes the ratio of the numbers of combinations available.
There are 2 components to take into account, the combinations within the suit and the
combinations outside the suit.
4-2 split:
 the number of combinations within the suit is 6! divided by 4!2!  (equals 10);
 the number of combinations outside the suit is 20! divided by 9!11!
3-3 split:
within the suit, 6! Divided by 3!3! (equals 15);
outside the suit 20! Divided by 10! 10!

The ratio of the numbers of combinations in favor of a 3-3 split is (3/2) times (11/10), so the probability of a 3-3 split in clubs, say, is much greater than the probability of a 4-2 split in clubs. However, the a priori tables include the possibility of a 2-4 split as well, in which case the probability of either a 4-2 split or a 2-4 split is greater than that of a 3-3 split by itself.

When one declares a hand it may be that one or the other 4-2 splits is eliminated from consideration, in which case a 3-3 split becomes the favorite. Here is an example where one wishes to estimate the chance of obtaining 3 tricks from a suit in this layout:

♣5432    opposite  ♣KQT

At the beginning the chances are slim – Roudinesco’s dictionary rates it at less than 20% – but often one is faced with making what one can with what one is given. A low club goes to the king, winning, and a second low club towards the tenace is won by the queen. What are the chances the ♣2 will set up at this point? With the ♣A and the ♣J the only 2 clubs outstanding there are 2 live possibilities to consider: AJ on the left, or A on the left and J on the right. (At my club no one behind the ♣KQ holds up the ♣A twice.) The probability of getting a third trick is the probability that the suit was dealt 3-3.

One cannot go to the a priori tables to obtain this probability. We are comparing one 4-2 split to one 3-3 split. The relative probability is the current ratio of the number of card combinations in the outside suits. Originally this ratio involved combinations of 20 cards in spades, hearts, and diamonds, but now the number of unknown cards has been greatly reduced with the comings and goings.  The number of vacant places must be taken into account. If there is no difference in the vacant places, the 3-3 split is more likely, and you’ll make that ♣2 more often than not, provided you can get back in hand to cash it. If there are 2 fewer vacant places on the left, it more likely than not that playing the ♣T will result in 2 clubs being cashed on the left. That could be good if it results in a suicide squeeze and/or an endplay.

As you may have gathered I am partial to the 3-3 split and the 3-card raise, but then I was brought up on The Three Musketeers.

 

The Division of Sides

An interested reader asked me how to calculate the division of sides. It is a basic characteristic of any bridge deal, so it is worthwhile to grasp the procedure. From Mathematical Theory of Bridge by Borel and Chéron we obtain these probabilities:

8-7-6-5   23.60%    7-7-6-6  10.49%     7-7-7-5    5.245%

The probability of one division relative to the other is the ratio of the number of card combinations that can be dealt under each condition. Let’s compare the numbers of combinations for 8=7=6=5 against those for 7=7=6=6, where the suits have been specified. We need to take into account all 52 cards.

Your Side 26! divided by  8!7!6!5!
Their Side the same,
    Or 26! divided by  7!7!6!6! the same.

The ratio is 16/9 in favor of the 7=7=6=6 division.
In addition there are the various combination of suits which may apply. There are 24 possible divisions of 8-7-6-5, and 6 of 7-7-6-6. Thus, overall with all combinations included, the 8-7-6-5 division is the more likely in the ratio of 9/4.

Next we compare the 7=7=7=5 division and the 7=7=6=6 division.

Your Side    26! divided by  7!7!7!5!
Their Side 26! divided  by 6!6!6!8!
    Or 26! divided by  7!7!6!6! the same.

The ratio of combinations is 4/3 in favor of the 7=7=6=6 division. There are just 4 combinations available to the 7-7-7-5 divisions, so the ratio overall is increased by a factor of 6/4. The 7-7-6-6 division is favored in the ratio of 2 to 1.

The above results can be verified by comparing to the ratio of the percentages given above. The absolute percentages are calculated by obtaining the probabilities of all possible divisions relative to one (say, 7-7-6-6), obtaining their sum, then dividing each element by the total, thus normalizing to a sum of 1. That is, the probability of one of these divisions being dealt must be 1, as all possibilities have been included.

Once the dummy hits the deck the division of sides is known exactly, so the a priori odds concerning the divisions are irrelevant beyond that point. There remain to be discovered the distributions of the suits within the division. The most likely splits are the most even splits that can produce the division under the constraint of what is already known, because these are the splits that have associated with them the greatest number of possible combinations.