THEORY OF THE PLAY OF A SUIT AT BRIDGE
Joël BRADMETZ
This article deals with the playing conditions which determine the choice of
a particular strategy in the play of a suit at bridge. In particular it exposes
the consequences which must be drawn from certain intrinsic characteristics
of the yield of a given strategy and of the scoring systems in pairs tournaments
and matchplay. It does not deal with the techniques which allow strategies and
their yields to be determined nor with the algorithms used to write a program
adapted to that task. This information is given by the author in other articles
(http://www.scanbridge.net).
NB The capital T is used to represent the 10 card throughout this article.
There are different ways of playing a suit depending on the objective which
one is pursuing. In order to present them, we have chosen a classic exposition
of the situation:
Strategy 1 | Strategy 2 | Strategy 3 | |
---|---|---|---|
World 1 (.20) | |||
World 2 (.30) | |||
World 3 (.15) | |||
World 4 (.35) | |||
Yield |
In the columns, we have three strategies (synonym of plays but closer to
the terminology of game theory). In the rows we have four possible worlds, which
represent the adversaries' unknown hands. The probability of each of these worlds
is indicated, and, according to custom, it is not expressed as a percentage but
using international notation, for example: .27 ( = 0.27 or 27 %). The probabilities
of the four worlds must add up to 1.
The cells of the table give the number of tricks which will be won by using the
strategy which is at the head of the column in the world indicated at the left-hand
end of the row. The bottom line shows the overall yield of the strategy.
For example, strategy 1 yields (2 x .20) + (3 x .30) + (3 x .15) + (4 x .35) =
3.15 tricks.
If one plays a large number of hands, the best result is given by strategy 3 which
yields 3.4 tricks. This is the maximum yield, or max yield according to
the terminology used by The American Encyclopedia of Bridge. For
a given hand, if a bridge-player was to remember only one strategy, it should
be that one.
Nevertheless, there are other subtleties of which one must be aware.
First, one might choose a particular objective corresponding to the number of
tricks that one wishes to take. If, with AJ32-K954, one needs to take three
tricks in order to ensure winning the contract, one would lead the ace and return
low to K9, which will ensure three tricks in every case. If one requires four
tricks, this strategy will yield them with a probability of only .1978, much lower
than the probability associated with the strategy which consists of playing low
towards the jack in order to try and finesse the queen in second or third position
in East (.3674).
But this maximal strategy for four tricks is not maximal for three, where
it fails against a singleton queen in West (p = .9717).
Look at the table again. The best strategy to obtain three tricks is the first,
which only fails in world 1 (p = .80); the best strategy to ensure four tricks
is the third, which succeeds in worlds 1, 3 and 4 (p = .70); the best strategy
pour five tricks is the second which succeeds in world 2 (p = .30).
We call these strategies n-optimal, in other words optimal for a number
of tricks n. Bridge-players refer to a security play.
The max strategy is best adapted to pair tournaments where it is necessary
to maximise the total number of tricks made. We will see that it is well-adapted in the vast majority of cases.
N-optimal strategies, on the other hand, are best adapted to matchplay,
where it is essential to make contracts, because the penalties are very heavy
when a contract is made at one table and lost at the other, and the score for
overtricks is negligible in the majority of cases, except where contracts are
doubled or redoubled.
Consider an individual game where the score is established by simply totalling
the points in each column. It is clear that one must try to maximise the number
of points scored, but that no one strategy emerges because depending on the scores
at any given moment, very different risks may be taken and luck will play an important
role. It is above all in competition situations, where players are confronted
with the same hands, that a study of strategies proves worthwhile.
Strategy 1 | Strategy 2 | Balance | |
---|---|---|---|
World 1 (.20) | |||
World 2 (.30) | |||
World 3 (.15) | |||
World 4 (.35) | |||
Yield |
To be exact, the only thing which counts in pairs tournaments is not to make a
maximum number of tricks (which is of little interest if everyone achieves it)
but to do better than the others as often as possible.
The scoring of such games has the distinctive feature that it reduces the cardinal
scale of points to an ordinal scale, thus eliminating intervals and the relationships
between them. The scoresheets
The consequence is that, in order to beat an opponent, it is better to hope
for ten more points at p = .50 than 400 more points at p =.49. Let us consider
a comparison that is easy to understand.
If you play a belote match with a given number of games of 1000 points each,
there are two methods of scoring: the winners are either the team which has
scored the greatest total number of points, or that which has won the greatest
number of games.
For example, in a match of three games, it is better in the first case to lose
all three at 980 points than to win two at 1020 points and to lose one at 250,
whereas the opposite is true if the second scoring method is applied..
In the table above, we see that strategy 1 is clearly better than 2 in terms
of yield, but if the two strategies are compared simply in terms of the frequency
with which one outclasses the other, the second wins in 55% of cases and the
first in only 45%.
The result is that in a tournament where the scoring system means that one must
win more often than one's opponents regardless of by how many points, one should
adopt the second strategy.
This situation is found in pairs tournaments at bridge where the ranking established
for each deal is purely ordinal. Thus in this case, there will be no more consideration
of objectives nor of security play, one must simply adopt the strategy which,
statistically, beats the opponents the most often, we stress once again, regardless
of by how much.
One might think that the max strategy would be the best candidate in
this situation. In effect, it is, apart from rare exceptions which amount to
approximately 1% of hands.
If, for example, you hold AJ94-K32, the max strategy yields four
tricks at .2901 by leading the King and then low towards the Jack, whereas leading
the King and then low towards the 9 only yields four tricks at .2683 (the two
strategies are equivalent for three tricks: .7826). Despite this, paradoxically,
it is not the first strategy that is to be recommended in a pairs tournament,
but the second.
Why is this? Examination of the two strategies shows that the max strategy
outclasses the other in 3 worlds, i.e. .1833 whereas it is outclassed in 4 worlds,
i.e. .2260. How can this be possible given that max is also always n-optimal?
Simply because against the world T8-Q888 (p = .0646) max yields four
tricks and its competitor only two. This overtrick (three are sufficient) is
counted in the yield calculation (+ .0646 trick) but does not count in a ranking
system. (This hand may be analysed with the demonstration version of ScanSuit).
It seems hardly possible to draw firm, practical conclusions from these characteristics
because they are based on the strong hypothesis that all competitors make the
same contract. Only analyses on a very large scale or computer simulations could
provide a clear answer.
There are approximately 1% of the positions in the American Encyclopedia
of Bridge where the recommended strategy in pairs tournaments is not
the max strategy. The interested reader can find them using the ScanSuit
program.
In matchplay the objective is to beat the opposing table at the same contract.
Compared with pairs tournaments, there is both a simplification and an increasing
complexity. The field is reduced to two players, which is simpler and makes
the hypothesis that the same contract is played in both camps more realistic.
Scoring is a bit more complex because it is applied to the
conversion of the difference in scores into IMP, which is not simply an ordinal
scale as in the case of pairs tournaments and, moreover, this conversion of
the differences in points is not the same as would be the difference between
the conversion into IMP of the scores of each team, because the IMP scale is
not regular. The consequences that may arise from this fact will emerge later.
We know that overtricks yield little in matchplay (and mainly in doubled or
redoubled contracts) and that the important thing is to make one's contract.
If n tricks are necessary to make a contract, it is logical this time
to consider not the max strategy but the n-optimal strategy for
the required number of tricks.
As in the case of pairs tournaments, the optimal strategy for a given number
of tricks may be outclassed by another strategy.
In pairs tournaments, the reason for deviance is related to the ordinal reduction
of differences. The reason is different in matchplay: it is related to the fact
that the outclassing strategy is much more efficient at level(s) n+k
or at level(s) n-k than the optimal strategy
of level n.
Consider the table below in order to illustrate the search for the best strategy.
South, vulnérable, is playing 4 spades and must make three tricks in
a particular suit in order to make his contract. He has two possible strategies,
1 and 2 (below) against four opposing positions.
Strategy 1 | Strategy 2 | Score 1 | Score 2 | Score 1-score 2 | IMP | Weighted IMP | |
---|---|---|---|---|---|---|---|
World 1 (.22) | |||||||
World 2 (.18) | |||||||
World 3 (.30) | |||||||
World 4 (.30) |
The total of the last column (+.18) gives the mean of the IMP gained in relation
to the probability of the worlds, that is to say the expectation of gain associated
with employing strategy 1 rather than strategy 2. Because the total is positive,
it is strategy 1 that should be used. The fact that this strategy does not yield
any overtricks in the fourth world (unlike its rival) is insufficient to compensate
for the difference of 4% which separates the first two worlds.
This calculation illustrates the best general method for estimating the best
strategy for a given contract. When there are n competing
strategies, there are (n x (n-1)) / 2 two by two comparisons
to be made. We will come back to a particular feature of this method later.
Now let us look at four concrete examples of bridge play.
AKQJ93-2 has an optimal strategy for five tricks (finesse) which only
fails if East is void and which yields six tricks at .4925. The alternative,
non-optimal strategy (lead from the top) fails if both East and West
are void but, on the other hand it yields six tricks at .8640. One wins .0075
at five tricks but loses .3715 at six tricks. In all contracts the second strategy
is, therefore, better. This is an example where the strategy is better at levels
n+k, in other words at levels superior to the number of tricks
sought.
AKQ43-T2 has a strategy X which makes four tricks at .8640 and
five tricks at .3553 and a strategy Y which gives .9199 et 0. Y
gains a little more than 5% for four tricks but loses more than 35% for five
tricks. In the majority of contracts, when 4 tricks are necessary X outclasses
Y, except, however in "basic" contracts (neither doubled nor
with unfavourable vulnerability).This is another example of superiority at levels
n+k.
Because, obviously, the sensitiveness of the scoring system means that the same
contract, depending on whether it is doubled or redoubled, vulnerable or not,
will lead to a different score which could influence the final choice of strategy.
AK9832-J invites two competing strategies. The max strategy (let
the Jack win a trick) yields respectively for 3, 4, 5, 6 tricks: 1, .9851, .7025,
.0161 and the 6-optimal strategy (lead from the top) yields: .9925,.8882, .6581,
.0323. Even when six tricks are necessary, in all contracts at all levels, the
max strategy is recommended because the chances of winning with the alternative
strategy are very low whereas max limits the possibility of losing
heavily. This is a clear example of a better strategy at levels n-k.
AKQ82-T9 also invites two competing strategies: max (let the 10
win a trick) yields 1, 1 and .50 for three, four and five tricks. The alternative
strategy yields 1, .9394 et .5410 for the same numbers of tricks. When 3 tricks
are necessary, it is obviously a consideration of the higher levels which will
determine the best strategy because both guarantee three tricks at 100%. The
second strategy will be retained because the gain which might be hoped for at
five tricks is greater than the gain which the first leads to hope for at four
tricks.
The table below details, by way of example, the contract of 4 hearts non-vulnerable
doubled which will succeed if, with AKQ43 - T2, declarer makes four tricks.
As we have seen, this hand has two strategies. The first involves leading from
the top, the second involves playing low to the 10. The second strategy is 4-optimal
(.9199 versus .8640 for strategy 1).
Strategy 1 | Strategy 2 | Score 1 | Score 2 | Score 1-score 2 | IMP | Weighted IMP | |
---|---|---|---|---|---|---|---|
0 - J99999 (.0075) | |||||||
J- 99999 (.0121) | |||||||
9 - J9999 (.0606) | |||||||
J9 - 9999 (.0807) | |||||||
99 - J999 (.1615) | |||||||
J99 - 999 (.1776) | |||||||
999 - J99 (.1776) | |||||||
J999 - 99 (.1615) | |||||||
9999 - J9 (.0807) | |||||||
J9999 - 9 (.0606) | |||||||
99999 - J (.0121) | |||||||
J99999 - 0 (.30) |
The total of the last column is positive: +.395. This advocates strategy 1 and
not 2.
The table below gives the number of contracts with deviant strategies using
a very representative sample of simple positions (the 2013 positions from Roudinesco's
dictionary of suit play).
NV - ND | NV - D | NV - RD | V - ND | V - D | V - DD | TOTAL | ||
---|---|---|---|---|---|---|---|---|
PARTIAL | Minor | |||||||
Major | ||||||||
No trump | ||||||||
Total | ||||||||
GAME | Minor | |||||||
Major | ||||||||
No trump | ||||||||
Total | ||||||||
SLAM | Minor | |||||||
Major | ||||||||
No trump | ||||||||
Total | ||||||||
OVERALL |
In the 2,013 positions which we have analysed, there are 534 hands which have
at least one deviant strategy, i.e. 26.52%.
There are 554,472 possible contracts with these 2,013 hands. Among them, 34,165
have a deviant strategy (including 437 without a strategy, as we will see below),
i.e. 6.16%.
If we estimate the percentage of contracts with a deviant strategy among the
534 hands which contain at least one, we arrive at 34,165/165,870, i.e. 20.60%.
When one examines the type of substitution of strategies,
one discovers, in the great majority of cases, that it is the max strategy
that replaces the n-optimal strategy.
Of the 33 728 contracts with a deviant strategy that we have tabulated
above (i.e. 34,165-437), 31,001, or 90.74% have their n-optimal
strategy replaced by the max strategy.
In 718 cases, (2.1%), the n-optimal strategy, which was also the
max strategy, is replaced by another.
It thus appears that a fairly simple general rule may be drawn from the table:
when an n-optimal strategy presents a yield significantly inferior
to the max strategy, it should be replaced by the latter.
We have calculated the values of these differences in yield.
For the totality of the 131,705 contracts normally resolved by the n-optimal
strategy, which, in 79.64% of cases is the same as the max strategy,
the mean difference between these two strategies is 0.011 IMP and the standard
deviation is 0.0349 IMP.
On the other hand, for the 33,728 deviant contracts, the mean difference between
the max strategy and the n-optimal strategy is 0.134
IMP and the standard deviation is 0.183 IMP.
There is thus a difference of 1 to12 between the two. This difference is very
significant statistically.
One can also observe that the n-optimal and max
strategies, which are the same in 79.,64% of cases when the contracts are normal,
are only the same in 2.12% of cases (718 out of 33 728) when the contracts call
for deviant strategies.
These facts show the strength of the max
strategy and its overall sensitivity to the scoring systems which tend to reinforce
the best mean yields. In effect as soon as an n-optimal strategy
is not max, if its expectation of winning is too different from
that of the max strategy (more than 0.1 IMP on average)
it is overtaken and replaced by the max strategy.
Strategy 1 | Strategy 2 | Strategy 3 | 1 - 2 | 1 - 3 | 2 - 3 | |
---|---|---|---|---|---|---|
World 1 (.20) | ||||||
World 2 (.25) | ||||||
World 3 (.15) | ||||||
World 4 (.35) | ||||||
World 5 (.05) | ||||||
Weighted total |
There is no ambiguity in the choice of the best strategy in these 5 worlds:
it is the third one.
It is equally clear that the differences between the totals of the columns are
equal to the sum of the differences (columns 4, 5 et 6).
Let us now convert these score into rankings according to pairs tournament logic.
Strategy 1 | Strategy 2 | Strategy 3 | 1 - 2 | 2 - 3 | 3 - 1 | |
---|---|---|---|---|---|---|
World 1 (.20) | ||||||
World 2 (.25) | ||||||
World 3 (.15) | ||||||
World 4 (.35) | ||||||
World 5 (.05) | ||||||
Weighted total |
It may be seen that strategy 1 is better than 2, 2 is better
than 3 and 3 is better than 1. No one strategy dominates
the other two, there is not, in this case, a strategy to recommend for a pairs
tournament with a position which leads to this results table. These positions
are rare: there are only 5 in the 2013 in the sample studied:
AK95 - J82 ; AT98 - K432 ; A974 - J832 ; KQT84 - 32 ; K987 - Q432
Strategy 1 | Strategy 2 | Strategy 3 | |
---|---|---|---|
World 1 (.20) | |||
World 2 (.20) | |||
World 3 (.20) | |||
World 4 (.20) | |||
World 5 (.20) | |||
Yield |
The ranking is unambiguous and, here also, the sum of the differences is equal
to the differences of the sums. If one converts the points differences between
the strategies into IMP, their comparison gives:
Strategy 1 | Strategy 2 | Strategy 3 | Strategy 4 | Strategy 5 | Strategy 6 | |
---|---|---|---|---|---|---|
Loses 3 tricks against | ||||||
Wins 4 tricks against |