Minimal bidding space consuming strategy for question-answer sequences in bridge. Fibonacci grouping

Authors  Delfí Querol,  Francesc Bofill

All bidding systems include artificial asking calls requiring partner to inform about some features of his hand. The method will often include relé rebids from questioner requiring further information after his partners partial answers. Suppose one in the partnership (PA) will always ask and partner (PB) will always answer. In this article we restrict our attention to these kind of bidding sequences. Let us call a sequence, builded in this way, a question-answer sequence (QAS).

A common example are the sequences produced using Blackwood convention or its variations.

One of the problems the system must envisage is that the final bidding level after the QAS's doesn't exceed the number of tricks available to the partnership.

So it is often important to plan a small level consuming bidding strategy.

In this article we describe the BEST theoretical lower level consuming strategy for QAS's. Best in the sense that  none of the admissible QAS will exceed some  "lowest" level.

Let's return to the Blackwood example. Traditionally the first asking bid is 4NT, but most systems will manage to take a lower starting level. But for our purposes the choice of 4NT will do. Suppose trump is already agreed to be spades. We consider the following version of Blackwood:

After 4NT from PA, PB will use the following levels to answer. For sake of simplicity suppose that an agreement of partnership is that, to use Blackwood, PA must have at least one ace himself. And let us suppose a bad but very simple formulation of the convention: Agreed answers of PB will indicate:

1         5: 0 aces without the trump K.

2         5: 1 aces without the trump K.

3         5: 2 aces without the trump K.

4         5: 3 aces without the trump K.

5         5NT: 0 aces with the trump K.

6         6: 1 aces with the trump K.

7         6: 2aces with the trump K.

8         6: 3 aces with the trump K.

And suppose that the convention finishes here. So 1 to 8 would be ending answers.

If PA asks and PB has 3 aces with the trum K. Bidding would be

PA          PB

4NT        6

And PA will then decide the final contract.

We are NOT concerned with this or any other particular choice of Blackwood conventions. But we ARE concerned to decide if given a particular choice of 8 ending answers the question-answer strategy proposed is or it is not minimal level consuming.

We shall see that of course it is NOT. In one of the cases (partner has 3 aces and trump K) the final level to complete Blackwood will consume 8 bidding levels.

The case is that ANY strategy which considers 8 final responses (as it happens with our example) will bring at least to the level 5NT in some case or cases. (The strategy described brought us higher than 5NT in three cases).

But THERE IS a unique optimal strategy to organize this requirements and deliveries of information, that keeps bidding below or at the 5NT level in all 8 cases. That is a 5 steps level consuming strategy. 5 is the smaller upper bound with this property.

This strategy si as follows. The group of 8 final answers is to be partitioned into 5 subgroups and each subgroup would correspond to one different call of PB. Like this:

5: It includes the    3   possibilities above numbered 1, 2, 3. (PB has 0, 1, or 2 aces and no trump K). PA will have to make a rélé bid of 5D to ask PB to precise which is the case between 1, 2 or 3.

5: It includes the     possibilities 4, 5, above.

5: It includes only   1   possibility 6.

5: It includes only   1   possibility 7.

5NT: It includes only   1   possibility 8.

For later use remember the sequence of  numbers 3, 2, 1, 1, 1 in bold which give the amount of elements in each subgroup. The grouping will be written as

1 2 3/ 4 5/ 6/ 7/ 8.

For instance after a 5 answer of PB (indicating that his hand is within the first subgroup), a 5 from PA is relé and PB precises if he is in case 1, 2 or 3, by bidding 5, 5 or 5NT respectively.

If PB has 2 aces without the trump K (case 3), the bidding will go

PA                PB

4NT               5 (first subgroup)

5 (relé)       5NT (option 3 in the subgroup)

And this sequence doesn't surpass 5NT. As stated.

We let the reader complete the remaining cases. And note that they all will NOT surpass 5NT.

.........

Before proceeding to the discussion of the general theoretic case, note a couple of thinks

a) For the case of 8 final answers  procedure just shown is the UNIQUE that could keep all the QAS's  and, thus, the bidding under or at 5NT level.

In another example with, say, 10 final answers (instead of 8) the strategy will need accepting the level of 6 in two cases at least, but this can be now done in several ways. The greater number of final answers that lets all  QAS's remain below or at 6  is 13. And for 13 the grouping can be achieved only in one way.

b) If we accept, as in the  Blackwood version considered at the beginning, to raise as much as to the 6 level, that is, if we accept to use up to 8 level consuming bidding steps for some of our QAS's  then the maximum number of final answers that could be taken into account would be 34 instead of 8. Other bridge variables, chosen with bridge criteria, could then be considered or added to our initial version of Blackwood incrementing the number of final answers: trump Q, voids, kings in other suits, etc....

Let us denote by N the number of final answers. In our initial example N=8. And let us denote by S the level consuming steps for optimal unique lower consuming strategy. In the example S=5.

Things are as shown in the following table. We make subgroups by writing the number of elements in each subgroups. Our example will be that shown in file 5. For which N=8, S=5 and partition is  3 2 1 1 1.

N ...... S ....... Partitions

1 ...... 1 ........ 1

2 ...... 2 ........ 1 1

3 ...... 3 ........ 1 1 1

5 ...... 4 ........ 2 1 1 1

8 ...... 5 ........ 3 2 1 1 1

13 .... 6 ........ 5 3 2 1 1 1

21 .... 7 ........ 8 5 3 2 1 1 1

34 .... 8 ........ 13 8 5 3 2 1 1 1

etc...

Now we describe how this table is constructed and how it could be extended.

Of course, in each file, the sum of the numbers under the title Partitions is the number under N. Also one number under N is the sum of the two previous numbers under N. The number under S is obtained adding one to the previous number under S. Also note that the string of numbers in one file is the same string of the previous file under Partitions preceded by the number under N corresponding to 2 files above.

To make it clear we copy three consecutive files in the table and use italic and bold to recognize the mentioned components.

13 .... 6 ........ 5 3 2 1 1 1

21 .... 7 ........ 8 5 3 2 1 1 1

34 .... 8 ........ 13 8 5 3 2 1 1 1

As explained next file would be

55 (21+34) .......   9 (8+1) ...................21 13 8 5 3 2 1 1 1 (21 + 13 8 5 3 2 1 1 1)

The infinite sequence 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,....

in which each number is the sum of the two preceding numbers is called the Fibonacci sequence.

This sequence appears in many problems of nature and science. If interested you can make a search in Internet.

It also happens that the sum of all the numbers in the sequence smaller than one of them is the following number in the sequence. For instance the sum of all Fibonacci numbers smaller than 13 is 21.

So in our table the numbers under N reproduce the Fibonacci sequence from the last "1" on. And the strings under Partitions are incremented parts of the Fibonacci sequences written in reverse order.

.........

Let's come back to bridge, to Blackwood. And suppose we decided to use an at most S=8 step bidding consuming level. According to the table our most complete choice of number of final answers to include would be 34.

N ...... S ....... Partitions

34 .... 8 ........ 13 8 5 3 2 1 1 1

We chose and enumerate the 34 possibilities in increasing order according to bridge criteria. Now we write the 34 numbers one after the other spacing the subgroups. 13 in first subgroup, 8 in second subgroup, etc... as required under Partitions. This should be:

1 2 3 4 5 6 7 8 9 10 11 12 13 /   14 15 16 17 18 19 20 21/    22 23 24 25 26/27 28 29 /    30 31/   32 /   33/   34

And suppose that the final answer of the sequence starting with 4NT is that in place 15. 15 belongs to second group, so first answer of PB should be second step. That is 5.

Next call from PA is a relé 5 and asks PB to further describe where answer is in second group

14 15 16 17 18 19 20 21

This new group has 8 elements which is a Fibonacci number and thus it can be found under N in the table. There we will find

N ...... S ....... Partitions

8 ...... 5 ........ 3 2 1 1 1

And previous subgroup must be divided in 5 more subgroups according to the string under Partition. Like this

14 15 16/    17 18 /    19 /    20/    21

since 15 is in first group PB must answer next step: 5.

Then 5NT from PA would ask for which of 3 remaining numbers. And PB should respond 2 steps above: 6.

In resume, to describe final ending answer, corresponding to number 15, bidding should be:

PA                 PB

4NT            5 (2nd subgroup)

5 (relé)     5 (1st sub-sub-group)

5NT (relé)    6 (case number 15)

______________

Final note: The authors developed some computer programs which allow quite easily find an estimation of any probabilistic question on bridge. Just as an example: we can estimate very accurately the probability that with 24 high card points a partnership won't have a game available.

Besides we can analyze a very large amount of deals exemplifying this lack of existence of game. This could be done visually but we claim that, in general, also algorithmically.

This would allow us empirically find the causes of this lack of game existence and, according to probability of occurrence, decide if a bidding system should take into account or not these specific causes. "Taking into account" includes the possibility to explain through bidding the detection of some of this causes.

Combining the content of the present article and these later considerations on probability, we are planning to construct a new artificial bidding system or drastically adapt some existing one to our conclusions and to probabilities. But work can be vast. Perhaps we could organize a group to do it together. (We could partition this group in Fibonacci subgroups so that the flow of information from one participant to others could be the lowest level consuming. Laughing). We shall greatly appreciate and consider collaboration offers.