Theory of the Play of a Suit at Bridge
by 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, and the ScanSuit program, will be found at www.scanbridge.net.
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)||2||3||4|
|World 2 (.30)||3||5||2|
|World 3 (.15)||3||3||4|
|World 4 (.35)||4||2||4|
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.
1. Classic yields.
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 Official Encyclopedia of Bridge. For a given deal, 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 fails only 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 for 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.
2. Scoring of individual games.
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.
3. Intrinsic irregularities in strategies in the scoring of pairs tournaments.
Consider the table below:
|Strategy 1||Strategy 2||Balance|
|World 1 (.20)||2||3||2 > 1|
|World 2 (.30)||5||2||1 > 2|
|World 3 (.15)||5||2||1 > 2|
|World 4 (.35)||3||4||2 > 1|
|Yield||3.7||2.9||.55 > .45|
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
650 ; 620 ; 620 ; 170 ; 140 ;-100 ;-200
160 ; 150 ; 150 ; 140 ; 130 ; 120 ; 110
are thus identical.
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 nine 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 10x-Qxxx (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 ScanBridge).
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 Official Encyclopedia of Bridge where the recommended strategy in pairs tournaments is not the max strategy. The interested reader can find them using the ScanBridge program.
4. Intrinsic irregularities in strategies in matchplay scoring.
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 imps, 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 imps of the scores of each team, because the imps 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 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, vulnerable, is playing four 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||Imps||Weighted Imps|
|World 1 (.22)||3||2||620||-100||720||12||2.64|
|World 2 (.18)||2||3||-100||620||-720||-12||-2.16|
|World 3 (.30)||3||3||620||620||0||0||0|
|World 4 (.30)||3||4||620||650||-30||-1||-.3|
The total of the last column (+.18) gives the mean of the imps 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 * (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.
AKQ43-102 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.
AKQ82-109 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 - 102, 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||Imps||Weighted Imps|
|void - Jxxxxx (.0075)||3||4||-100||590||-690||-12||-.0894|
|J - xxxxx (.0121)||4||3||590||-100||690||12||.1452|
|x - Jxxxx (.0606)||3||4||-100||590||-690||-12||-.7272|
|Jx - xxxx (.0807)||4||4||590||590||0||0||0|
|xx - Jxxx (.1615)||4||4||590||590||0||0||0|
|Jxx - xxx (.1776)||5||4||690||590||100||3||.5328|
|xxx - Jxx (.1776)||5||4||690||590||100||3||.5328|
|Jxxx - xx (.1615)||4||4||590||590||0||0||0|
|xxxx - Jx (.0807)||4||4||590||590||0||0||0|
|Jxxxx - x (.0606)||3||3||-100||-100||0||0||0|
|xxxxx - J (.0121)||4||4||590||590||0||0||0|
|Jxxxxx - void (.30)||3||3||-100||-100||0||0||0|
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|
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 minus 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 imps and the standard deviation is 0.0349 imps.
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 imps and the standard deviation is 0.183 imps.
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 imps on average) it is overtaken and replaced by the max strategy.
5. Special characteristics of the pairs tournament scoring system.
Consider the following table.
|Strategy 1||Strategy 2||Strategy 3||1 - 2||1 - 3||2 - 3|
|World 1 (.20)||300||500||400||-200||-100||+100|
|World 2 (.25)||400||200||500||+200||-100||-300|
|World 3 (.15)||200||300||500||-100||-300||-200|
|World 4 (.35)||500||400||300||+100||+200||+100|
|World 5 (.05)||300||500||100||-200||+200||+400|
There is no ambiguity in the choice of the best strategy in these five 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)||3||1||2||-.20||+.20||+.20|
|World 2 (.25)||2||3||1||+.25||-.25||+.25|
|World 3 (.15)||3||2||1||-.15||-.15||+.15|
|World 4 (.35)||1||2||3||+.35||+.35||-.35|
|World 5 (.05)||2||1||3||-.05||+.05||-.05|
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 five in the 2013 in the sample studied:
AK95 - J82 ; A1098 - K432 ; A974 - J832 ; KQ1084 - 32 ; K987 - Q432
6. Special characteristics of the matchplay scoring system.
Consider the points table below.
|Strategy 1||Strategy 2||Strategy 3|
|World 1 (.20)||300||400||500|
|World 2 (.20)||400||200||300|
|World 3 (.20)||200||300||400|
|World 4 (.20)||500||400||300|
|World 5 (.20)||210||300||90|
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 imps, their comparison gives:
1 vs 2 = -3 +5 -3 +3 -3 = -1
2 vs 3 = -3 -3 -3 +3 +5 = -1
3 vs 1 = +5 -3 +5 -5 -3 = -1
There is no dominant strategy. The reason this time is the fact that the IMP scale is not regular and does not even maintain a linear system of measurement. For example 20 points = 10 points + 10 points but, after conversion into imps, we have 1 = 0 + 0.
Hands which have no strategy in matchplay are also very rare. We may recall that there are 5/2,013 = .0024 for pairs tournaments. There are 15 / 2,013 for matchplay, i.e. .0072, but if we do the calculation on the basis of contracts and not of hands, these 15 hands generate 437 defective strategies, i.e.437 / 554,472 = .0007.
These 15 hands are as follows:
AK95 - J82 ; A1098 - K432 ; A974 - J832 ; KQ1084 - 32 ; K987 - Q432
AKJ9 - 432 ; AJ94 - K32 ; AK876 - J92 ; A10843 - K72 ; A1086543 - Q2
AQ108 - 432 ; AQ1096 - 5432 ; AQ432 - 10987 ; A732 - JT8 ; K954 - J32
We have noted that the first five do not have a dominant strategy in pairs tournaments either, which allows us to state an empirical law: if a hand is defective in pairs tournaments, it will also have at least one defective contract in matchplay.
A1098 - K432 should be noted as a highly exceptional hand from this point of view: it alone accounts for 197 defective contracts (i.e., 45% of all the defective contracts), including all its vulnerable redoubled contracts!
It possesses six main strategies which act as follows:
|Strategy 1||Strategy 2||Strategy 3||Strategy 4||Strategy 5||Strategy 6|
|Loses 3 tricks against||Singleton Q in W (.0565)||Singleton Q in E (.0565)||QJ76(5) in E (.0848)||QJ76(5) in W (.0848)||Singleton Q in W (.0565)||Singleton Q in E (.0565)|
|Wins 4 tricks against||Singleton Q in E (.0565)||Singleton Q in W (.0565)||QJ in O & Q in E (.904)||QJ in E & Q in OW(.904)||QJ in E (.0339)||QJ in W (.0339)|
Strategies 1, 2, 5 and 6 are 3-optimal; strategies 3 and 4 are4-optimal. They are symmetrical two by two.
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