Nine-Never and Basic Bridge Probability

There are many bad ideas floating around the bridge world concerning probability theory. Generally there is an over-reliance on the application of the a priori odds that apply to the deal of the cards before any action is taken. The oft-quoted Nine-Never Rule has this flaw. The prior odds are appropriate when one is in a state of maximum uncertainty, or if you prefer, in a state of minimum information with regard to the placement of the cards. We do know something about the probable consequences of the dealing of the cards, but the actions at the table are going to tell us a lot more about the particular hand that is being played. Naturally, as more is revealed, the odds will change. The question we address here is how to replace the a priori odds with odds more in keeping with current circumstances. Our examples are in reference to the flawed Nine-Never Rule, but the methodology is applicable to any card play situation.

Treating the problem of changing odds in a theoretical manner often leads to misunderstandings, besides which that approach requires some assumptions that get hidden in the mathematical jargon. The purpose here is to present a general framework for the calculation of odds, and to achieve that purpose we work through a simple, but realistic, example using only basic concepts which every player can understand. Along the way we shall make reference to familiar concepts, like vacant places, which may not be fully understood. Lack of basic understanding often leads to misapplication, and that in turn may lead to distrust of methods that are put forward under the guise of science. True science, first and foremost, reflects reality by complying with observation.

A Flannery Hand

For our example we assume the opening bidder (West) advertises a hand with 4 and 5♥, 10-15 HCP. We chose this bid because of the large amount of information it reveals about the distribution of the cards. If South ends up declaring the hand, that information can be used for the planning the play. This is also true: if an opponent overcalls on a weak hand and his side doesn’t end up declaring the hand, the distributional information provided can be used by the eventual declarer to the overcaller’s disadvantage. This hand provides an extreme example of that process at work. Here are the South – North hands.

South	North	West	North	East	South
♠A 5 4 3	♠9	2 ♦	(Dbl.)	3 ♠	(5 ♦ )
♥10 9 5	♥A 7	Pass	(6♦)	All Pass
♦A J 9 3	♦K 10 8 7 2
♣K 3	♣A J 10 9 5	Lead ♠ K

 

South wins the ♠A and contemplates how best to draw trumps. The usual slogan is ‘Nine Never!’ Meaning that declarer should play off the ♦A –K hoping to drop the ♦Q doubleton. Not thinking beyond that, a novice may cash the ♦A and play towards the ♦K. A more knowledgeable player would play low towards the ♦K to guard against a 0 – 4 split when West is void in diamonds, then play to the ♦A when East follows twice with low diamonds. Both approaches are wrong because they are based on a blind rule, Nine Never, that has its origins in the a priori odds. We’ll examine first the effects of the bidding on the recommended play in diamonds.

Vacant Places First we register the distribution of the sides as revealed when the dummy appears. For NS the suits are numbered 5=5=9=6, so the NS side is numbered 8=8=4=7. The bidding tells us futhermore that the distribution in the majors is as follows:

	W – E
♠	4 – 4	Given what we know about the majors,
♥	5 – 3	the vacant places are 4 in the West and 6 in the East
♦ & ♣	4 – 6

 

The relative probability that the ♦Q being on the right (QR) rather than on the left (QL) is in the ratio of the number of vacant places available. That is, the ♦Q is more likely to be on the right in the ratio of 3:2. This assumes that neither clubs nor diamonds have been played, and that the minors suits can be lumped together as indeterminant cards. This is the odds of the deal with regard to those two suits. The missing diamonds and clubs can be reshuffled and dealt 4 to West and 6 to East without affecting the information currently available with regard to their content.

We chose to emphasize the location of the ♦Q, but the same relative probability would apply to any missing club or diamond regardless of rank, the ♣Q, for example. However, if one assumes the ♦Q is on the right, that assumption changes the number of vacant places to 4 and 5, so the odds for the ♣Q also being on the right has been reduced, although the vacant places still favor that, now in the ratio of 5:4.

So given the information possessed concerning the probable location of the ♦Q, declarer should not play for the drop, but should plan to finesse for that card in the East hand by beginning the diamond play with a low card to the ♦K in the dummy. This is not merely a safety play against a very rare void in the West, but a percentage play of greater relevance. What we want to calculate is the probability that declarer will succeed in his planned finesse.

The Magic of Vacant Places From what does the magic of vacant places derive? If we are going to use it, we should understand it. Then we see there is really no magic involved. Probability is no more than a ratio of combinations. It is assumed that the 10 missing minor suit cards can be placed at random between the West and East hands. How many card combinations exist with the ♦Q assumed to be on the left, and how many with the ♦Q on the right? The ratio of the card combinations give us the odds, assuming the cards are randomly dealt. Here there are 4 vacant places on the left and 6 on the right. There are 2 possibilities for the distribution of the 9 cards other than the ♦Q:

	Others		Others
♦ Q on the left	3 – 6	♦ Q on the right	4 – 5
Combinations	84		126

 

The ratio of the combinations is 6:4, which is also the ratio of the vacant places.

Thankfully one doesn’t need to know the individual numbers of combinations in order to obtain the ratio. One just lines up the splits as follows and chooses the last number of the more even split and the first number of the less even split. Thus,

5 – 4 vs 6 – 3 gives a ratio of 6:4.

This works for all adjacent splits. For 5-4 vs 7-2, one needs to treat 3 adjacent splits.

Split	5 – 4	6 – 3	7 – 2
Ratios	1	2/3	3/7

 

Ratio of combinations 5-4 to 7-2 is 2/7

Thus, the ratio of combinations for a 5-4 split relative to a 7-2 split is 7:2, not 7:4.

The Distribution of Sides Once the dummy appears declarer can count separately the number of diamonds and clubs held by the opponents: 4 diamonds and 6 clubs. It is now possible to state explicitly all 5 possible combinations.

	♠ 4 – 4	♠ 4 – 4	♠ 4 – 4	♠ 4 – 4	♠ 4 – 4
	♥ 5 – 3	♥ 5 – 3	♥ 5 – 3	♥ 5 – 3	♥ 5 – 3
	♦ 2 – 2	♦ 1 – 3	♦ 3 – 1	♦ 0 – 4	♦ 4 – 0
	♣ 2 – 4	♣ 3 – 3	♣ 1 – 5	♣ 4 – 2	♣ 0 – 4

 

Weights	90	80	24	15	1
	43 %	38 %	11 %	7 %	_ %
A priori	41 %	25 %	25 %	5 %	5 %

 

The a priori odds refer to the diamond splits. The even 2-2 split is more or less the same as the a priori value and the sum of probabilities for the 1-3 and 3-1 split is 49%, very much the same as expected initially for those splits. The difference is that the 1-3 split is much more likely than the 3-1 split due to the imbalance in vacant places.

With regard to the club splits, the 6-1 and 5-1 splits are rendered impossible by the conditions imposed by the Flannery opening bid. The 3-3 split has roughly the same probability as initially (36%) as does the sum for the 2-4 and 4-2 splits (48%), but with an imbalance in favor of the longer holding in the East hand, as expected. This illustrates that the a priori odds have some value as approximations if used in the proper sense. (If that weren’t the case, they would have gone out of use long ago.)

The Calculation of the Weights The calculations of the weights from which the probabilities are derived are merely the relative number of card combinations available on a random dealing of the cards. These weights apply before a card is played in either minor suit, so they are the probabilities of the deal only. We shall now illustrate how they come about from the numbers of card combinations available in the diamonds and clubs considered separately. It is convenient to line up the diamond splits in consecutive order and place underneath the corresponding club splits that preserve the 4-6 split in vacant places.

Diamonds	4 – 0	3 – 1	2 – 2	1 – 3	0 – 4
Clubs	0 – 6	1 – 5	2 – 4	3 – 3	4 – 2

 

Combinations	♦	1	4	6	4	1
	♣	1	6	15	20	15
Product	♦ & ♣	1	24	90	80	15
Total = 210

 

The numbers of combinations are entries in the Pascal Triangle, familiar to many from their schooldays. (Was that too long ago?) The greater the number of combinations, the greater is the probability that that particular configuration has been achieved. The individual percentages are merely the number of combinations divided by the total number of combinations possible (210). It is important to remember this arrangement because once cards are played in a suit, diamonds first, some combinations are ruled out of the realm of possibility. That reduces the number of combinations available in the diamond suit, while the number of possible club combinations remains unaffected. However, the changes in probability are not solely governed by the numbers of card combinations remaining. That is where some come to grief: the probability of the play is not the probability of the deal as not all cards are treated equally under the rules of bridge – some are significant, some aren’t, depending on the circumstances. That matters.

The Diamonds in Play Here explicitly are the combinations possible in the diamond suit (excluding the 4-0 and 0-4 splits) with the number of club combinations associated with each possible diamond combination. The number of club combinations serves as the weighting factor for each individual diamond combination taken 1 by 1.

Split	3 – 1	2 – 2	2 – 2	1 – 3
	Q 6 5 – 4	Q 6 – 5 4	5 4 – Q 6	4 – Q 6 5
	Q 6 4 – 5	Q 5 – 6 4	6 4 – Q 5	5 – Q 6 4
	Q 5 4 – 6	Q 4 – 6 5	6 5 – Q 4	6 – Q 5 4
	6 5 4 – Q			Q – 6 5 4
Clubs	6	15	15	20

 

We shall assume declarer begins properly with a low diamond to the ♦K in dummy and that West follows with the ♦6 and East with the ♦4. Here are the possibilities that remain. The club weights are unchanged as no club has been played.

Split	3 – 1	2 – 2	2 – 2	1 – 3
	Q 6 5 – 4	Q 6 – 5 4	6 5 – Q 4	6 – Q 5 4
Clubs	6	15	15	20

 

The relative probability of the ♦Q on the left (QL) and on the right (QR) can be calculated from the weights from the associated club split.

QL = 6 + 15 = 21

QR = 15 + 20 = 35

 

The odds are now 5:3 that the ♦Q is on the right. What of the vacant places? They started at 4 and 6 and now they are 3 and 5, if we count the formerly unknown diamonds that have now been exposed. That’s great for simplifying the calculation of odds, but this is a special case as we shall see when we come to investigate more closely the plausible plays in the diamond suit. The special situation is this: the play of the ♦6 and ♦4 are equally probable for all remaining combinations.

Next we shall assume that declarer next leads the ♦3 from dummy and East follows with the ♦5. How does this change the probabilities? There are only 2 possibilities remaining:

Split	2 – 2	1 – 3
	Q 6 – 5 4	6 – Q 5 4
Club Combinations	15	20

 

The odds of the ♦Q being on the right are 4:3, in exact agreement once more with the number of vacant places remaining. Obviously declarer should take the finesse against East and not play for the drop.

Finally we assume that declarer plays the suit the wrong way around by cashing the ♦A in hand and leading towards the dummy. Assume West follows to the second round with the ♦5. The remaining possibilities are as follows:

Split	3 – 1	2 – 2
	Q 6 5 – 4	6 5 – Q 4
Club Combinations	6	15
	QL = 6;	QR = 15;	so QR:QL is 5:2

 

When West plays the second low diamond, the vacant places stand at 2 and 5, which reflects the current relative probabilities, and South should play for the drop. The Nine Never Rule applies. So in one sequence declarer should play for the drop and in the other he should finesse. Here in detail are the vacant place situations described above.

Vacant Places	West	East		West	East
Before a ♦ play	4	6		4	6
West plays ♦ 6, East ♦ 4	3	5		3	5
East plays the ♦ 5	3	4	West plays ♦ 5	2	5

 

Preferential Plays and Vacant Place Calculations The accuracy of the vacant place ratio in determining probabilities depends on the way in which low cards are played, in particular, the assumption is that the low cards are chosen equally at random. This reflects the conditions of the deal. From a doubleton ♦65, say, the choice of the ♦6 has a probability of 50% as does the choice of the ♦5. From ♦654, each low card has a 1/3 probability of being chosen on the first round and any permutation over 2 rounds has a 1 in 6 chance of being chosen. This assumption is equivalent to a condition of maximum uncertainty (entropy) for which the amount of information transmitted by the sequence of plays is a minimum. Any rule that gives preference of one sequence over another transmits more information. Here is an illustrative example. Suppose that defenders always play their highest spot card, in particular,

1) from a holding of ♦65 or ♦64 they always play the ♦6;

2) from a holding of ♦54 they always play the ♦5.

Applying these rules we are left with these combinations after the play of ♦6 and ♦4:

Split	3 – 1	2 – 2
	Q 6 5 – 4	6 5 – Q 4

 

After 1 round the odds of the ♦Q on the right is 5:2, whereas the vacant place ratio is 5:3. Thus the direct correspondence between vacant place ratios and probabilities is broken. The additional information available has effected a reduction in the number of possible combinations remaining from 16 down to 2, which under the condition of maximum entropy would require, as we have seen, a third card to be revealed in order to achieve the same degree of reduction.

Permutations and Plausible Plays The order in which a defender chooses to play his cards when following in a suit represents a selection of one sequence of plays chosen from all possible permutations on the cards he holds. If all permutations are equally likely, the probability of his having chosen the observed sequence is merely the reciprocal of the total number of equal choices. This is the normal assumption, but it is subject to revision under some circumstances. The mathematics can accommodate other assumptions when appropriate. We discuss the effects later.

Assume a defender holds ♦654 and is required to follow to 2 rounds of diamonds. There are 6 possible ways to follow: 6-5, 6-4, 5-4, 5-6, 4-6, and 4-5. We term these the plausible plays. If the defender chooses to play the equivalent cards at random, the probability that a given sequence will emerge from the original holding is 1/6. On the other hand, if the defender always gives true count, the only sequence possible under that restriction is 6 followed by 5 and there exists a condition of certainty. The probability of observing 6-5 is 1. Thus, in order to calculate probabilities one must specify initially the assumptions concerning a defender’s method of choosing equivalent cards.

Suppose that a defender picks up each card as it is dealt. He ends up with ♦654. The order in which he gets to observe the arrival of the cards isn’t important. The same is true if he is dealt ♦Q64. All 6 permutations on the deal are equally likely. When he gets to play the cards the order becomes important, because the play is subject to the rules of bridge. The defender is assumed to play his cards in such a way as to optimize his chances. It would be foolish in most situations to play the unsupported honor on the first round. The plausible plays over 2 rounds are 6-4 and 4-6, and the probability of observing either is _. This is not a necessary assumption. There may be situations where the play of the ♦Q on the first round is called for. A defender may choose to cover an honor with an honor, or may attempt to create an entry to partner’s hand. Declarer needs to be aware of these situations, as does anyone wishing to calculate the odds. The mathematical formulation can accommodate any assumption in this regard.

Why is it commonly assumed a defender will play equivalent cards at random? Because that is the optimum strategy when the objective is to keep declarer in the dark to the greatest extent possible. If the objective is to inform partner to the greatest extent possible, such as by giving true count or suit preference, equivalent cards will not be played at random, but will conform to prior agreement. However, there is no guarantee the defenders will keep to such an agreement if they feel it is in their best interest to be uninformative.

Restricted Choice Consider next the combination ♦QJ6. On the first round normally a defender will follow with the ♦6. On the second round he may play either the ♦Q or ♦J with equal effect. The number of plausible plays is 2, 6-J and 6-Q, and the probability of each is _. This is a common example used to illustrate the Principle of Restricted Choice. When a declarer sees an honor emerge, naturally he takes notice. Of course, the very same principle applies to the play of low cards, as we have shown above. The selection from 2 honors is merely the last element in a sequence of equal choices.

If a defender wishes to inform partner that he holds both honors, he will not play the ♦Q on the second round, he will play the ♦J. If that takes the trick, his partner can see that declarer hasn’t got the ♦Q, otherwise he would have covered. If the defender takes the trick with the ♦Q, the possibility still exists that declarer holds the ♦J. Playing the ♦Q and ♦J at random creates a condition of maximum uncertainty in the minds of both declarer and the other defender, at least to the greatest extent that is possible given the cards they can see in their own hands.

It is understandable that the Principle of Restricted Choice is treated in the bridge literature solely as applying to the choice of honor cards. That is the most dramatic application as the play involves cards that have the potential of taking tricks. Also, low cards usually appear early in the play of a suit whereas honor cards emerge later at a critical stage. A defender will not usually part with an honor card unless there is a distinct advantage to doing so, whereas it is easy to part with a low card whose only function appears to be to fill in the suit. Usually there are more low cards missing than honors. For these reasons, then, the illustrations of Restricted Choice involve honors. We shall illustrate the application of Restricted Choice using a second Flannery deal.

A Second Flannery Hand

South	North		West	North	East	South
♠ A 5 4 3	♠ 9		2 ♦	(Dbl.)	3 ♠	(5 ♦)
♥ 10 9 5	♥ A 7		Pass	(6 ♦)	All Pass
♦ A 10 9 3	♦ K 8 7 5 2
♣ K 3	♣ A J 10 9 5		Lead ♠ K

 

On this hand the ♦Q and ♦J are missing. Suppose that the ♦K is played to which West follows with the ♦6 and East with the ♦4, just as in the previous hand. On the second round, East follows with the ♦J. What is the probability the ♦Q will fall doubleton from the West hand? After the first round these are the possibilities remaining:

Split	3 – 1	2 – 2	2 – 2	1 – 3
	Q J 6 – 4	Q 6 – J 4	J 6 – Q 4	6 – Q J 4
Clubs	6	15	15	20
Plays	1	1	1	1

 

Each combination is subject to just one plausible play, so they all have the same probability of being chosen. Under the condition of equality of permutations, one can apply the vacant places to the calculation of probabilities. Adding the weights as before, we find

QL = 6 + 15 = 21 = JL; QR= 15 + 20 = 35 = JR,

and the odds of either honor being on the right is in the ratio of the remaining vacant places, namely, 5:3. The situation changes dramatically if East follows to the second round with the ♦J.

Split	2 – 2	1 – 3
	Q 6 – J 4	6 – Q J 4
Clubs	15	20
Plausible Plays	1	2

 

With a 2-2 split the appearance of sequence 6-4-J is a certainty, so the weight retains its full potential. However, the probability of the sequence 6-4-J given a 1-3 split is reduced by half under the assumption that it is just as likely for East to have chosen the ♦Q on the second round resulting in the sequence 6-4-Q. There are 2 equally probable sequences. The odds of dropping the ♦Q in the West has improved to 3:2. Before a diamond was played the odds favored the queen being with East 3:2. Thus, the freedom of choice of one honor over the other greatly affects the odds.

Let’s suppose that East is more likely to play the ♦J than the ♦Q, trying to keep his partner better informed. At trick 3 the situation is still unclear. If the probability of playing the jack rather than the queen were 75%, that would make the 2-2 split and the 1-3 split equally probable at this point in the play, so the odds of dropping the ♦Q in the West hand would be 50%. If East would always play the lower of 2 touching honors as a matter of general principle, alleviating the need for thought, the weight of the 1-3 split retains its full value as there is just one plausible sequence available. In that case one returns to the odds based on the current vacant places, 3:2 against the drop being successful. This is true before a diamond is played and later whenever the number of potential card sequences is the same for all remaining combinations.

What about the ♣Q? Once declarer determines the diamond split, the club split becomes a certainty and the play proceeds accordingly. Let’s suppose that against the odds according to Restricted Choice East was dealt ♦QJ4, so now there is a diamond loser after declarer plays for the drop. The slam is still assured when a heart loser can be avoided by establishing 2 tricks in clubs on which 2 losing hearts can be discarded from declarer’s hand and the losing heart in dummy can be ruffed. Declarer has been keeping count, so feels the clubs must be split 3-3. He makes his claim, showing his cards, after which the following exchange takes place.

South: I ruff the clubs good and pitch 2 hearts on the jack-ten of clubs. You can take your trump queen whenever you like, but I still make my 12 tricks.

East (showing his cards): down 1. I have four clubs to the queen.

North: Oh, Partner, you could have made it by taking a ruffing finesse in clubs!

If West holds the majors, East must hold the minors.

West: Sorry, with 6 lousy hearts it felt like a Flannery hand.

South: Grrrr….

Yes, it happens. The opening bid provided information, but it was inaccurate. Information is not exact, but one should plan according to the information available. In some situations before drawing trumps, declarer may take some seemingly pointless ruffs in a side suit in order to confirm what he thinks he knows about the distribution. In a 3NT contract declarer may duck a trick or two to good effect. These are processes of gathering information, and it may happen that this gathered information contradicts what was assumed previously. A re-evaluation of the assumptions is required.

If one is unable to safely gather more information, then it is correct to play according to what is most likely given the current information. In this deal West took an unusual action, but the probability that he did so was very low. This is generally true. It doesn’t pay to assume an unusual action and over-react in what is most likely a normal situation. More is lost by assuming aggressive opponents are trying to swindle you than is lost by assuming they are playing normally. However, it is wise to check.

Bayes’ Theorem Without stating the governing equations explicitly what we have demonstrated in these examples are simply applications of Bayes’ Theorem to card play. The mathematics involved is not difficult as it is merely a consequence of linearity, however, the notation is difficult to grasp on a casual reading. Putting the equations into words has caused confusion, beginning early in the 18th Century with Rev. Thomas Bayes himself, who was hoping to add further confirmation of the existence of God by extending the results produced by Blaise Pascal, the 17th Century co-founder of probability theory. Once philosophers got involved, matters became unnecessarily complicated. It wasn’t until the 1950’s that clarity was restored and probability was linked mathematically to information. Hopefully, all that confusion is behind us now and common sense will prevail, at least as far as the analysis of bridge play is concerned.

Warning! The application to the cases where the defenders hold an odd number of trumps is more complicated than for the examples treated above, because the number of plausible plays reaches equality at a later stage of play than for an even number of missing trumps. Expect more on the Eight-Ever Rule later.