Nine-Never and Restricted Choice
Mathematical models require assumptions, and it is the nature of the assumptions that limit their applicability to the real world. The Nine-Never Rule is based on a mathematically sound argument, but the assumptions behind it are not always appropriate to the situation at hand. In the recent Prince Takamatsu Cup tournament held in that beloved oasis of polite bridge in the heart of heartless Tokyo known as the Yotsuya Bridge Club, a slam hand arose which provides a rare example where theory is well-matched to reality, yet the seasoned BBO commentators did not give any clear direction to the observers on how the hand should be analyzed. In fact, the analysis was somewhat confused, the simple conclusion being that here was another example where the Nine-Never Rule failed. Why it failed was not explained, and it was not clear whether the declarer had actually made a mistake when he played for the drop rather than finessing for the queen. This leads me to conclude that it is worthwhile to go through the play of the hand in detail in order to demonstrate the mathematical assumptions that lies behind the rule, and, more importantly, the procedures that a declarer should follow when making a decision with regard to the play in a suit where 9 cards are held with the queen missing. Eventually the well-known Principle of Restricted Choice is invoked.
The Basic Suit Configuration We shall assume South is the declarer faced with the problem of finding the ♦Q in the following configuration:
(South) ¨AJ9752 opposite ¨KT6 (North)
The ♦K is played from dummy to which East follows with the ♦4 and West with the ♦3. Next the ♦T is led to which East follows with the ♦8, and the time of decision has arrived. Should declarer finesse or play for the drop? There are 2 possible holdings remaining, namely,
|Diamonds were dealt 1-3||Diamonds were dealt 2-2|
|(West) ♦ 3 opposite ♦ Q84 (East)||(West) ♦ Q3 opposite ♦ 84 (East)|
On the play within the suit there is no reason to choose one configuration over the other as they share the same number of plausible plays, so the question boils down to the question of which is more likely, a 1-3 split or a 2-2 split? Declarer should play for the drop when it can be assumed that an even split in diamonds is more likely than an uneven split. This assumption is supported by the a priori odds, but it may not hold after cards have been played in another suit.
The Effect of an Uneven Split in Another Suit
In the Prince Takamatsu Cup hand hearts were trumps. The opening lead from West was a trump, and it was immediately discovered that the hearts were split 3 in the West and 0 in the East. Does this information point to an uneven split in diamonds? Yes.
|W E||W E|
|Hearts||3 – 0||3 – 0|
|Diamonds||1 – 3||2 – 2|
|Blacks||9 – 10||8 – 11|
The weights are an expression of the number of combinations in the black suits that are available under the 2 diamond splits. The ratio of combinations yields the odds of 11:9 in favor of the 1-3 split. Hence declarer should finesse under these circumstances, when nothing is known concerning the splits in the black suits. This is roughly the situation when trumps are drawn and diamonds are broached immediately thereafter.
The same odds are available from a vacant place analysis, following Kelsey’s Rule, which allows the use of the current vacant places when only the location of the queen remains in doubt. This is no surprise, as the assumption of maximum uncertainty with regard to the black suits is common to both methods.
|Current Vacant Places|
|Initial Vacant Places||10||13|
|First Round of Diamonds||9||12|
|Second Round, East Follows||11|
The odds are 11:9 that the ♦Q sits in the East. This is the basis of the criticism the BBO commentators directed towards declarer when he played for the drop. Given the hearts had been seen to split 3-0, it seems as if the odds favor an uneven split in diamonds. However, life is seldom that simple. At the table, diamonds were not broached immediately, and several rounds of both spades and clubs were played before the critical decision was made in diamonds, so more information was available than in the simple model indicted above. To gauge the effects let’s look fully at the play of the cards.
What Really Happened
|♥||K J 10 8 7 3|
|♦||K 10 6|
|♣||9 8 6|
|♠||K 8 7 3 2||♠||10 9 6 5|
|♥||A 9 4||♥||—|
|♦||3||♦||Q 8 4|
|♣||A Q K 2||♣||K 10 7 5 4 3|
|♠||A Q 4|
|♥||Q 6 5 2|
|♦||A J 9 7 5 2|
The bidding was very much to the point and showed commendable faith in partner’s non-vulnerable weak two preempts. No doubt there were hopes of a helpful lead, hopes that were not realized when Shimizu, in the face of such uncertainty, chose a passive trump lead. The ♥4 rather than the ♥A indicated he had no fear of losing his ♣A in the early rounds. Let’s look at some common distributions of sides with an established 3-0 heart split.
|♠ 4 – 5||♠ 4 – 5||♠ 5 – 4||♠ 5 – 4||♠ 3 – 6|
|♥ 3 – 0||♥ 3 – 0||♥ 3 – 0||♥ 3 – 0||♥ 3 – 0|
|♦ 2 – 2||♦ 1 – 3||♦ 1 – 3||♦ 2 – 2||♦ 2 – 2|
|♣ 4 – 6||♣ 5 – 5||♣ 4 – 6||♣ 3 – 7||♣ 5 – 5|
Even though the hearts are split 3-0, the even 2-2 diamond split is more likely than the uneven 1-3 split over this small selection. Why? Because the imbalance in the vacant places is more readily compensated for by uneven splits in the longer black suits. These 5 cases represent a snapshot rather than a panoramic view, however, as cards are played, it is to be expected that the play will ‘zoom in’ on the most likely possibilities shown above. Of course, the discovery of uneven splits in the black suits may alter our view.
The play in slam contracts is often a long journey of discovery. Declarer may postpone the critical play in the diamond suit until he has learned more about the splits in the black suits. Let’s follow the perilous path as it occurred at the table.
Both defenders have played 4 spades and only the ♠K is now missing. It could sit on either side. East has played 5 clubs, ♣10-7-5-3-2, and the ♣K-Q-J are still missing. From the play in the club suit we may assume the clubs were not dealt 5-5, for that would mean West began with ♣AKQJ2, which is contrary to the evidence of the lack of bidding, the opening lead and the subsequent play. Thus it is reasonable to assume clubs were split 4-6, and it is only the split in spades that remains in question at trick 11. At this point the split in diamonds depends solely on the split in spades. Here are the 2 live possibilities:
Under Condition I
|♦||A J 9|
Under Condition III
|♦||A J 9|
Of the 5 most likely distribution of sides listed above, only Conditions I and III remain as possibilities at the critical point in Trick #11. The problem has boiled down to this: does West or East hold the ♠K? The answer will determine who holds the ♦Q.
So far the numbers indicate that declarer should play for a 2-2 diamond split as on the deal Condition I is the most likely distribution of sides in the proportion of 3:2. However, as Shakespeare once wrote, ‘the play’s the thing’; there is the evidence of the play to be considered. The problem in probability is this: given the defender’s sequence of plays, is it more likely East holds the ♠K and, if so, by how much? This card play problem is related to the Principle of Restricted Choice.
At Tricks #4 and #5 EW followed suit with low spades. At Trick #7, when declarer ruffed the ♠Q the defenders’ ♠K109 were equivalent cards. If East had been dealt ♠K10965 he could have followed with any of the 3 equals, but if he had held only ♠10965 his choice would be restricted to 2 cards. In retrospect, Conditions I and III had become equally probable when East played the ♠9. At Trick #9, both defenders had to find a discard on a trump lead. The ♠8-♠10 play was 1 of 8 possible plays under Condition III and 1 in 9 under Condition I. Thus, at Trick #10 Condition III becomes the more likely in the proportion of 9:8. The BBO commentators, along with everyone else who could see all 4 hands, were correct in maintaining declarer should have finessed at Trick #11.
At the other table North opened with a weak 2♥ bid and ended up in slam. West first competed in spades at the 3-level on a poor 5-card suit, then in the end, egged on by East, doubled 6♥. With this wealth of information made freely available to him North was able to play the diamonds correctly and claim his doubled contract for a gain of 16 IMPs. Lest this be construed as a condemnation of hyper-aggressive bidding practices, we hastily add that the aggressors’ team won their semi-final match handily, only to lose by a narrow margin in the final.
We next present a mathematical description of the play process.
Bayes’ Theorem at Work
Bayes’ Theorem is a simple idea easily expressed in the form of a mathematical equation. The difficulty experienced by many in understanding the equation lies primarily in a lack of previous exposure to the mathematical notation. It is worthwhile to go through the formulation for the above problem, as Bayes’ Theorem is fundamental to our understanding of how probabilities change due to the exposure of cards during the play.
We begin with the distributions of sides of which there are many, but with hindsight we may limit our attention to Conditions I and III. Condition II is a remote possibility which we shall put aside for the sake of simplification. The probability of Condition I relates directly to the probability that spades are split 4-5 [denoted as P(4-5)] and the probability of Condition III relates to the probability that the spades are split 5-4 [denoted as P(5-4)]. The beauty of probabilities is that we need concern ourselves only with ratios which represent relative probabilities. Thus, we can state that:
P(4-5) / P(5-4) equals 105/70 or 3/2.
In their ratio, one may take P(4-5) and P(5-4) as representing the probabilities of a particular combination of spades rather than all possible combinations. Once cards are played a new type of probability arises which requires a new notation, as follows. Let S represent a sequence of plays in which the cards are revealed. The probabilities that a particular sequence would appear under Condition I or Condition III are denoted as P(S │4-5) and P(S │5-4), respectively. What one seeks is the relative probability that the spades are split 4-5 after the sequence S has been observed, and this probability is denoted as P(4-5 │S) . Bayes’ Theorem is a statement of this linear proportionality:
P(4-5 │S) is proportional to P(4-5) times P(S │4-5) ;
P(5-4 │S) is proportional to P(5-4) times P(S │5-4).
This means that the probability of Condition I after the sequence S is observed is proportional to the initial probability of Condition I before a spade is played times the probability that the sequence S would be chosen if indeed Condition I were to apply. Similarly for the probability of Condition III after the sequence S has been observed.
As noted above, probabilities vary as the sequence grows. Before a diamond is led at Trick #10, the last action of the sequence involves a restricted choice of discards in the black suit at which point:
P(S │4-5) divided by P(S │5-4) equals 16/27. Thus, we can conclude that
P(4-5 │S) divided by P(5-4 │S) equals (3/2) times (16/27) which equals 8/9.
In plain words Condition III has become more likely than Condition I once the discards have been observed. At this juncture each condition has been reduced to just one possible spade combination. It is usually the case after a long sequence of plays that restrictions in the play weigh more heavily than the restriction of the deal.
Comments on the Bayes Approach
Truth will sooner come out of error than from confusion.
- Sir Francis Bacon (1561 -1626)
Seldom can it be said that a model fits reality to perfection; models are perfect, but reality is messy. The concept of ‘best chance’ is an abstraction in itself that requires the incorporation of all the information that is available at the time of decision. Some simplification is required, how much partly depends on a declarer’s powers of observation. Although one may not be able to reach the correct conclusion through common sense alone, in retrospect a mathematically derived conclusion should be seen to reflect common sense. In simple language Bayes’ Theorem tells us the following:
(1) If a player hasn’t played a particular card, it is more likely he hasn’t got it than that he has it but chose to play an equivalent card instead.
(2) The more likely a player is to have played a particular card if he had it, the more likely it is that he doesn’t have it if he hasn’t played it.
The Assumption of Perfect Play At the table it is a player’s task to induce errors, and suppress the fear that an opponent can see through the cards, yet the mathematical model assumes the play is perfect. It is all too easy to assume a defender will not err, but then, how can one expect to win against a faultless opponent? In the above example it is in the defenders’ best interest not to give away information with regard to the distribution of the cards, which means that they should play equivalent cards at random rather than attempt to give partner the count, which would have made declarer’s task much easier.
Memory Problems Not many declarers could remember at Trick #11 the sequence of cards played by the defenders, yet a complete analysis requires one make use of this information. The task is made easier if a declarer is alerted to the main features of interest. That is why when the dummy comes down, declarer should focus the mind on the most likely distributions of sides, which are the models for what is most likely to transpire. In the above example, it should be immediately clear that a major objective will be to determine the diamond split, so declarer should make plans to achieve that objective. He goes about his business looking for clues, planning his sequence of plays accordingly.
With regard to Ino’s journey of discovery, might he not have adopted a different route to better induce a revealing play in spades by leading the ♠J from dummy at Trick #2 to see whether East would cover? Of course, East does not cover, and declarer goes about his business. Suppose that subsequently East shows up with the ♠T9. Now looking back on Trick #2 declarer may draw the correct conclusion based on the observation that if East held ♠KT9 he most certainly would have covered the ♠J at Trick #2. On this partial basis he might choose correctly to finesse for the ♦Q. This is a matter of anticipation and recall for which an understanding of the mathematical treatment prepares the mind.
Quick Japanese Yotsuya means‘Four Valleys’. Takamatsu means ‘Loftly Pine’. Bush is ‘Yabu’. Obama is a seaport in Western Japan.