|Calculating Contract Odds
Assessing the relative merits of a bridge contract entails two types of calculations:
There was an example of each query in tonight's session:
#1: As west, you played the following hand in
Your contract failed. In the postmortem, you wondered just what your
chances really were. With a normal
3-2 trump split, either a winning
diamond finesse or 3-3 spades would have seen you home. Had the trumps
proved to lie 4-1, then you would pretty much have needed both diamonds and
spades in your favor. What was the overall likelihood of success?
#2: Still sitting west, you declared this board in
3NT against a normal spade lead:
Your contract hinged upon bringing in four diamond tricks. What was the chance of that? Was your partner justified in inviting game?
Perhaps that second round of margaritas was to blame; but no one in your group offered an answer to either question. Not all was lost, however; grabbing a pencil and your companion's soggy napkin, you quickly dispatched the solutions to the amazement of all. How were you able to manage it? Were those geometric scribbles of yours some sort of magical incantation? No, you had merely utilized a visual aid in your algebraic quest.
Such is the potential value of graphical concepts such as Venn Diagrams, which long have served as representations of logic problems and their solutions. For our investigation, however, we will venture into the discipline of logic only insofar as it is applicable to analysis of bridge hands.
In beginning, I will beg the indulgence of any reader skilled in math or logic, who doesn't require all the instructional details provided. Bear with me.
This enquiry will not teach you how to calculate such things as the likelihood
that a suit will split
2-2; those sorts of numbers are derived from a complicated
formula involving factorials and permutations; so they are not the stuff of
late-hour computations at the pub. In any case, such figures are readily
available from tables in the Encyclopedia of Bridge or elsewhere.
Note: those a priori values are totally accurate only in the context of all possible distributions. In real life, the odds are skewed by factors gleaned from the bidding and discovered in the play. For example, is the
3-3heart split really a 36% chance? Answer: it depends. If your left-handopponent has eight red cards to your right-handopponent's four, then the odds are roughly 9 to 5that RHO has any particular card in a black suit (there being that ratio of room in the hand for non-redcards). This skews the chances of an even division of the hearts. The stated 9-to-5ratio is inexact as well; the actual odds involve those aforementioned formulas as applied to the permutations of each opponent's non-spades.For our purposes, however, the published figures are plenty close enough, and we shall treat them as such. What's a percentage point or two among friends?
What we can do is to determine a guideline for combining already known values, as specified by an appropriate (logic) statement describing the conditions for success of a contract. Translating logical relationships into mathematical terms is what this section is all about. Every logical statement has an algebraic equivalent. Three such relationships are essential to our study.
Everyone knows that the chance of winning a finesse is 50%; all other things being equal, that pesky honor is equally likely to lie on either side of the table. Most also know that the likelihood of winning a pair of finesses is 25%; but where did they get that figure? Well, somewhere it was learned that when two events are mutually dependent, the likelihood of both occurring is equal to the arithmetic product of the individual chances. In this case, .50 × .50 = .25. In the language of logic, this event is termed, "spade finesse -and- diamond finesse". The logical "and" function proposes that "both are true", or in this case that both finesses will win.
Rule #1: The logical (A and B) translates algebraically to (a × b). In simpler terms: P(A and B) = ab.
This relationship can be illustrated by a Venn Diagram:
But first, a mini-tutorial. The area inside the box represents the Universe, or all possible outcomes of an experiment; nothing outside the box has any relevance. The area within the oval represents outcomes that meet specified criteria. Outside the oval is all the trials which "failed" the experiment. It is useful to think of these areas as being filled with small "dots", as shown, each of which represents one element, or one possible outcome.
For example, the Universe might be comprised of all possible distributions of the
opponents' hands, and the oval area would include those layouts in which the queen
of trumps is singleton. But such a
one-variable picture is of
instructional value only. More useful is the next diagram, which depicts
Let's conduct a hypothetical experiment, using numbers familiar to us.
In this case, our Universe is a standard pack of 52 cards. We will shuffle
the deck, then draw a random card. In the diagram, the
circle represents all the clubs, and the right-hand circle includes all the
honor cards. Intuitively, the yellow area, or "intersection"
of the areas, "contains" those cards that match both criteria.
Outside the circles lie all the remaining cards.
What is the probability of drawing a club honor? One way to determine that would be simply to count the elements in the yellow section, if possible, and divide by the known Universe of 52. But what is the value of this intersection? Viewing the diagram in a new perspective:
Since the purple area represents the elements wherein both a and b are
true, we know from Rule #1 that that area is numerically defined
simply as ab! As there are 13 clubs and 20 honors in the
deck, P(club) = 13/52, and P(honor) = 20/52; therefore,
P(club -and- honor) = 13/52 × 20/52, or 5/52.
Surely enough, there are
5 club honors.
Let's try another one. If you need three things to go your way, the procedure
is similar. You simply take the product of the individual probabilities.
2-2 trump split, a finesse, and a 4-3 division in a side suit,
you would multiply .40 × .50 × .62 = .124, or about 12%.
The logical statement here is: "2-2 split -and- finesse
-and- 4-3 split".
It makes no difference how many events are involved, as long as they
dependent — that is, all must occur.
Rule #1 can be expanded to: P(A and B and C ... and Z) = (a × b × c ... × z)
This formula conforms to the Associative Law — an axiom of
mathematics, meaning that when using the -and- function, a
statement's elements may be combined in any pattern or sequence.
For example: (P and Q) and R is equivalent
to P and (R and Q). Intuitively, you already knew
that — it makes no difference in what order you multiply a series
On the next board a preempt by your opponents caused you somehow to miss your
9-card spade fit; so you find yourself declaring in diamonds.
Your contract will succeed if either trumps split 3-2 or
the spades divide 2-2. How does one calculate that?
In logic (as in English) the operative "or" is inclusive; that is, it means "either or both". When you speak of "diamond finesse or spade split", that includes the possibility of both things working. Numerically, it means "at least one is true". Now I will tell you that the chance of "probability−A -or- probability−B" is equal to their sum, less their product.
Rule #2: The logical (A or B) translates to: (a+b−ab)
For visual clarification of this, let's try a different experiment with our
shuffled pack. What is the chance of drawing a club or an honor?
13 clubs plus 20 honors = 33 total cards. Is that the correct answer?
No, because the club honors have been counted twice! As seen in the diagram,
counting the dots in each circle and combining the totals results in a
double-count of the yellow area.
Let us label the pink circle (clubs) as c, and the green circle (honors)
as h. The graphing of
(club -or- honor) includes the
combined areas of the two circles. As the yellow section has a previously
determined value of ch, the double-counting of the club honors can be
corrected by subtracting that value.
(c+h−ch) = (13/52 + 20/52 −
5/52) = 28/52, the correct answer.
Returning to our misbegotten diamond contract, we need a trump split
or a 2-2 spade division (40%). Placing actual percentages into a diagram
can provide a useful picture of the situation:
The centermost area ds is calculated first:
.68 × .40 = .27. Section d
then becomes .68 − ds , and section
s = .40 − ds. Adding .41 + .27 + .13, we come up
81%, the likelihood of fulfilling our contract. (Just for the
record, the actual value for a 2-2 split is 40.6% — closer to 41%; but it
is so much easier to use the round number! Do what you feel is best.)
Upon further analysis, you realize that an onside club ace also offers the
potential for an extra trick, and that's a
50-50 proposition. Now our
contract requirement becomes: trump split -or- spade split -or- club
finesse. It so happens that the Associative Law also is applicable to
this construct; therefore, it can be stated as (D or S) or C,
or as S or (C or D), or any possible arrangement. Let's try the
first one: (.68 or .40)= .81. Then,
(.81 or .50) = .90, or about a 90% chance overall.
Had you opted for the second pattern, you would have calculated
.40 or (.68 or .5), for the same result (within 1%, due to possible
Rule #2 Expanded: The logical (P or Q or R) can be restated as ((P or Q) or R), or any other choice of variable sequencing.
There is a nifty shortcut for making this calculation, using the logical -not- function; however, as is it technically not required for our purposes, I have placed it at the end of this section, just for the dedicated analyst. See the Addendum.
This time you have a 5-3 heart fit; but the long trump suit is tapped
at trick two. There is other work to do; so you cannot survive
a poor trump split. Assuming that hearts divide
your game is a make if you can win the diamond finesse or split the
spades 3-3. How good is this contract?
The logical statement here is (heart split -and-
(diamond finesse -or- spade split)). The trump split being
an absolute requirement of success, that factor is not stated as part of a
contingency such as (trump split or other). It must be a
stand-alone term of an -and- statement.
In any algebraic equation such as a(b+c),
the processing hierarchy is "inside out"; that is, expressions
inside parentheses are evaluated first. B and C are added, then
the sum is multiplied by A. Subsidiary to that, any multiplication
and division are processed before addition or subtraction.
The same sort of hierarchy applies in the world of logic.
It is essential to format your statement
that is, to "say exactly what you mean".
Our equation, (H and (D or S)), translates to h(d+s−ds) = .68 × (.5 + .36 − .5 × .36) = 46%.
On the next deal, your spade game is contingent upon winning the
trump finesse or splitting the clubs
2-2. If both
things work, you score up an overtrick. But your postmortem has
evolved toward the esoteric. You are wondering —
not whether you could make your contract, but what the chances were of winning
precisely ten tricks. In order for that to happen, you
either would have to win that finesse or split the clubs, but not
both. The logical statement is "A or else B";
that operative is the "exclusive or", meaning
"either A or B, but not both". The numeric
equivalent is "exactly one".
Rule #3: The logical (A or else B) translates to (a+b−2ab)
This proposition is represented by the shaded area in the diagram.
As we have seen, adding the outcomes of circle s to the
outcomes of circle c counts the intersection
twice. This time, however, area sc is unwanted;
to delete those outcomes, it is necessary to subtract twice
its value. Let's do the numbers: s=.50
and c=.40. Calculating,
s+c−2sc = .5 + .4 − 2(.5 × .4) = .5, or
50%, the total of the blue areas. The likelihood of an overtrick
is represented by area sc, or 20%. The overall
chance of going plus is 70%. One can see that when solving
a problem, it might be easiest simply to draw a Venn Diagram and fill in the
If no common outcomes can exist, such as in the conditions of (diamond void in west -or else- diamonds 2-2), then ab=0.
Rule #3a: If A and B are discrete events, then the logical (A or else B) translates simply to (a+b).
Now we have the tools to solve Play Problem #1 at the beginning of
this article, which features contingencies for differing outstanding
trump distributions in a heart game. Since events with
3-2 trumps and those with 4-1 trumps have no common elements
(they cannot simultaneously occur), they conform to Rule 3:
3-2 spades -or else- 4-1 spades.
The entire conditions of the deal can be stated as:
(hearts 3-2 -and- (diamond finesse -or- 3-3 spades))
(hearts 4-1 -and- (diamond finesse -and- 3-3 spades))
In numerical terms, we have: (h32 + (d+s−ds)) + (h41 × d × s).
As calculated in the "Mixing And with Or" section, the
3-2 trumps portion evaluates to 46%. For 4-1 trumps,
(.28 × .5 × .36) = 5%. Total: 51%.
Well, I declare! If trumps split 5-0, there is no chance of taking ten tricks,
and squeeze chances are slim (although they really should be included); therefore
good-looking contract — one that nearly every pair
would reach — actually is only slightly odds-on. Now you
can show your partner (if he still can read your scrawls after that third margarita)
why it was not so unreasonable that you went down.
Next time, play in
3NT. That is a far better contract,
and its analysis is a good way to finish this section:
On Problem #1, how often would the 9-trick game succeed against a club lead?
Well, a simple 3-2 heart split would yield nine tricks irrespective
of all else; so that seems like a good place to begin the formation of
our logic statement. As distributional options other than
hearts would be discrete events subject to Rule 3, our initial statement
can be: hearts 3-2 -or else- (other).
That 68% chance already confirms the great superiority of a notrump contract;
indeed, one might be satisfied with that cursory observation and move on to the
next board. But no — you are anxious to further demonstrate
your math skills, even though your partner is too far gone to appreciate them.
So the analysis continues.
What if hearts split 4-1? Then you will have to give up a heart trick to
set up the fifth card. If clubs divide
4-4, the opponents will not
have enough winners to defeat you. However, four heart tricks brings your
total only to eight; so you will need the diamond finesse as well. This portion
of our statement becomes: hearts 4-1 -and-
clubs 4-4 -and- diamond finesse.
What if the clubs are 5-3? Holding up for two rounds would isolate that suit; then you could safely give up a heart trick, provided that the hand with the long clubs did not also have the long hearts. As your contract cannot possibly succeed without a fourth heart trick, you must set up that suit, and you still would need the diamond finesse.
Expressing that contingency logically: hearts 4-1 -and-
(clubs 5-3 -and- hearts and clubs
not in the same hand).
A 6-2 club division is a possibility which technically should be considered; however, this exercise is complicated enough already, and your own eyes are beginning to blur from inebriation; so you choose to ignore that unlikely contingency just this once.
A list of known percentages is in order. These you obtain from an index card stashed in your wallet just for this purpose:
If clubs are 5-3, then one opponent has ten non-clubs to the
other's eight. So the hand with short clubs is approximately
10-to-8 (or 5 out of 9) favorite to hold the long
hearts. Remember that this is only a rough estimate,
but it is close enough for barroom work, and it yields the requisite figure:
Now, it remains to put the data into some sort of coherent format. Let's start by reducing our statements to more concise language:
3-2 hearts is exclusive from 4-1 hearts; hence and 'or else' condition is appropriate between statement #1 and the others. Statements 2 and 3 have common terms (h41 and df); so they can be evaluated inclusively, as per Rules 1 and 2. The remaining terms, c44 and c53s, are mutually exclusive, subject to Rule 3.
The algebraic expression abc+abd can be expressed as
ab(c+d), by combining the common elements. Statements
2 and 3
can be factored in that manner to: h41 and df and (c44 or else c53s).
A reduction such as this is an important step in ensuring that the logic is valid,
and it makes calculations easier. The complete statement becomes:
hearts 4-1 -and- diamond finesse -and- (clubs 4-4
-or else-(clubs 5-3 -and-suits separate))
Evaluating from the inside out as always:
All that work for 6 percent! Adding it to 68% for 3-2 hearts,
as per Rule 3, begets a final evaluation of
74%, which is a
lot better than one's chances in a heart contract. You proudly
display that revelation on your cluttered napkin, but by this time
your partner has passed out.
The methods presented involved mathematically straightforward situations; however, at the table other contingencies arise that are more difficult to assess for probability; in fact, they might be impossible to evaluate numerically. In such cases, you might be called upon to make an estimate or two.
For example, on the last hand the defenders were under no
obligation to play that third round of clubs. Hypothetically,
at trick three an opponent, recognizing that the club suit was dead,
might switch to diamonds; however, on this deal either hearts will
run and there are no problems, or you will need the diamond finesse
to work in any case. So the aspect of a timely switch happens
not to be an issue in this case. In many other situations,
the defenders' choices of
plays — and projected likelihood of
same — might need to be factored into the equation.
Consider the logical statement "p or q",
meaning that "at least one is true".
For every statement, there is a negative, or "opposite"
meaning. The statement "not (p or q)" is
the logical equivalent of
"neither is true",
which is the same as (not-p and not-q). If your
contract is dependent upon the "spade split or club
finesse," then clearly you will make it unless both
contingencies fail to materialize — that is,
no spade split and no winning club finesse.
Okay. If a 3-2 trump split is 68%, then what is the chance of the suit not splitting favorably? Intuitively, it is 100% − 68% = 32%. In numerical terms, "no" equates to (1−'yes'), and (yes=1−no).>
Rule #2n: The logical not (A or B) translates to 1−(a+b−ab) = (1−a) × (1−b)
The validity of that algebraic identity is easily verified,
but how can we use it? Let's dredge up that forgettable
diamond contract where you needed
3-2 trumps (68%)
or a 2-2 spade split (40%). Instead of calculating
(d+s−ds), it might be easier to calculate the complement:
(1−d) × (1−s). Try it:
(1 − .68) × (1 − .4) = .32 × .6 = .19,
or a 19% failure rate.
Ergo, 1 − .19 = 81% success rate!
With less number-crunching.
This method comes in even handier when there are
more than two variables, such as when we realized that a
club finesse also was in play:
diamond split Rather than using
the nested procedure employed earlier, try calculating
(1−d) × (1−s) × (1−c)
= .32 × .6 × .5 = .096. Subtracting that
from unity gives us a 90.4% likelihood of -or-
spade split -or- club finesse. success — the correct value!
This is the suit in question from Problem #2 at the beginning of this article. The contract hinges upon winning four tricks in diamonds:
♦A K opposite ♦10 9 4 3 2
It is necessary first to determine what outstanding card combination(s) would provide winning positions. In this case, there are four:
1. any 3-3 split
2. Q J x x x x
3. Q x J x x x
4. J x Q x x x
#1. A 3-3 split is 36%, from tables.
#2. There are fifteen different combinations of two cards from a total of six. Anyone familiar with the formula for such things could calculate that the "combination of 6 things taken two at a time" is equal to (6×5) ÷ (1×2) = 15, showing that the two honors share doubleton suit 1/15 of the time.
#3-4. With this small a sample set, one could simply itemize all the possibilities, which is easy enough:
QJ Q8 Q7 Q6 Q5 J8 J7 J6 J5 87 86 85 76 75 65
Yep, the total is fifteen. It remains only to count the items that
include an honor, learning that there are nine. All
total 48%. These are discrete entities from the hands with 3-3
splits. Therefore, this contract will succeed whenever:
3-3 split -or else- 4-2 split with honor(s) dropping = .36 + (.48 × 9/15) = 65%.
A fine contract!
But what if there are 56 items? Or 120?
counting could become quite tedious. Fortunately, there is
Ted's Shortcut, which obviates the size of the numbers
and requires only the multiplication of a couple of fractions.
A graphical representation of our experiment is useful; let's try it on
pattern #2. Using our same Universe of six cards:
Q, J, 8, 7, 6, 5, we will deal them to the opponents.
Envision, in the defenders' hands, slots to be filled by the cards —
two on one side, four on the other, as shown:
Because it makes no theoretical difference in which order the
cards are dealt, we will arbitrarily chose to deal the queen
first to a random slot. (Yes, this is valid.) What is the
likelihood that that card will go the to short (left) hand?
Clearly, it is
2 chances out of 6. Place the queen
on the doubleton side:
Next, deal the jack The chance of its going to the left is
1 available slot out of 5 remaining.
The other cards go, by default, to the other hand. As the first two
deals were mutually dependent events (Rule 1), the probability of
occurrence is simply:
2/6 × 1/5 = 1/15. How about that?
Now let's try pattern #3: Qx. Again, deal the queen to the short
2 chances in 6. Now deal the jack to the long side:
this will occur 4 times out of 5 slots. The overall odds are
2/6 × 4/5 = 4/15. Dealing Jx is equally likely,
for another 4/15. The overall total for the three patterns is
(1/15 + 4/15 + 4/15) = 9/15 = 3/5.
Note: Since it made no difference in the play as to which opposing hand actually held the doubleton, no further adjustment is necessary. Anytime a suit length or card placement is hand-specific (such as requiring the doubleton to be on the left), however, the probability is reduced by half.
As suggested in the Addendum section of Part 1, it often is easier to calculate the chance of non-occurrence of an event, then subtract it from the Universe of all events — that is, 1−(non-event).
Our calculation might have been quicker had we applied that technique. The 4-2 suit would not work for us only when one hand has QJxx. Let us start afresh with our six-card pack. What is the chance of dealing the queen to to the four-card holding? Clearly 4/6, or 2/3. Now deal the jack to that hand as well, which would occur 3 times out of 5 remaining slots. The placement of the spot cards is immaterial; so our probability becomes: 2/3 × 3/5 = 2/5, the likelihood of failure of the suit to come home with fewer than two losers. Therefore, 1 − 2/5 = 3/5, the chance of success. 3/5 × .48 = 29% for favorable 4-2 combinations. Adding that to 36% for 3-3, our total is 65%, as before.
We can advance our enquiry by altering our key suit somewhat, putting the top honors in the long hand (with a side entry to that hand):
4 3 opposite A K 10 9 2
Still looking for four tricks, this suit could be played as before, by cashing dummy's ace and king, which we have shown is a 65% chance. Call this Plan A. Alternatively, we could take two finesses, Plan B. Perhaps that is better. We could find out by calculating the finessing options from scratch, or we could simply compare pertinent cases between the two methods. Let's do that.
Either play works when LHO has QJ, Qx, Jx, Qxx, Jxx, QJx, Jxxx, Qxxx; nothing works when QJxx is offside, or when either hand holds five or more of the suit. We can effectively eliminate those from our comparison, leaving:.
Plan A gains when RHO has QJ
or QJx, for then both finesses would lose.
Plan B gains when LHO has QJxx, for then no honor would drop in two rounds.
Plan A: QJ doubleton has been calculated as 1/15 of
48%. Using our tricky shortcut, we determine that QJx is
3/6 × 2/5 = 1/5 of 36%.
(1/15 × .48) + (1/5 × .36) = .104.
This figure must be reduced by half to 5.2%, because these suit
combinations are applicable only to one side of the table.
Plan B: QJxx has been figured as
2/5 × .48 = .192. This we also must divide by two,
It appears that Plan B outscores Plan A, giving it a
69% overall rating. In terms of total tricks,
the disparity is reduced somewhat by the fact that Plan A would
produce five tricks even with QJ doubleton offside.
That overtrick would accrue (1/15 × .48 / 2) = 1.6%
of the time; but of course that consideration is not worth the risk
of a contract. In any case, the overall comparison is close;
but taking two finesses is the mathematical winner.
Since the two lines of play are approximately equal in merit, this is one of those times when clues gained from the bidding or play should influence declarer's plan of attack. The lead of a low club suggests longer clubs to the left, which reduces the likelihood of the magic QJxx in diamonds in that hand. The more clubs on the left, the more attractive Plan A becomes; conversely, longer clubs on the right would make Plan B the standout choice.