CR Bridge

                                                                    Most players know the “8-ever, 9 never” rule for finding the missing queen, playing a nine-card suit.  The image below shows why this rule works.  (Click to expand. Type Esc to return, or click the “x” in the upper right corner.)

The odds of success range from about 50% to about 58% for the four strategies.  The two best options require playing one top honor and then leading up toward the remaining top honor.  The finesse and the drop strategies are less than 2% apart in their original odds of success.  However, assuming that you have now seen all three small cards, the current odds have become 12 to 11 for the drop.  The difference in present (i.e. updated) probability is 4 % or (1/23).   You arrive at this by taking 20.35% vs 18.65% for the two remaining original cases.   The current probabilities are

.2035/(.2035+.1865) =.522  (or 12/23) and
.1865/(.2035+.1865) =.478  (or 11/23) .

The finesse at this moment is no longer quite a 50% proposition even though it rated as 56% when you began.  Perhaps you wonder where the number 23 came from in the odds.

The “principle” of “vacant places,” or  sometimes “vacant spaces,” says that the odds of finding a single particular card are proportional to the number of “vacant spaces” in each concealed hand.  Each of those hands start with 13 cards.  From 13, you subtract the number of cards in each suit for which the distribution is completely known.  In the key suit, when you are down to the missing honor, having located all the other cards, you can subtract those other cards from the concealed hands. 

At the decision point, you have seen all the missing spot cards in the key suit.  This tells you that the number of vacant places is 13-2=11 in the hand in front of your honor and 13-1=12 behind your honor, so it is better to play for the drop.  Suppose you play several cards in other suits before you tackle the key suit.  If you do not learn any complete distributions of other suits, you still use 13-2 and 13-1 at the decision point.

Suppose the opponent behind your honors has made an overcall, and his partner has raised.  You infer that the suit suit is divided 5-3, with the five cards behind your honors.  At the decision point the number of vacant places in front of your honor is 13-2-3 = 8.  Behind your honor the number of vacant places is 13-1-5 = 7.  This tells you that the odds now favor the finesse by 8 to 7, or about 6.7%.  Often a preemptive bid will shift the odds in favor of a finesse.

When considering vacant spaces, you cannot include suits for which you have seen some, but not all cards played.  You must know (or choose to assume) the complete distribution of the suits you include.  In the key suit, having seen all the spot cards, you can imagine that they constitute a “fifth suit.” for which you know the distribution.

Knowing the complete distribution of  one or more side suits in which the defender behind your key-suit honors holds a total of two more cards than his partner tips the balance in favor of the finesse at the final decision point.  This assumes you have no other reason to place the missing queen with the opponent holding the 5-carder.  

Often it is necessary to combine the probabilities of multiple events.  For example, holding AKQT, missing six cards, you would like to know the chance of making four tricks. You hope to drop the jack in three rounds, but you finesse when obvious.  The latter succeeds if you drop the jack on the first two rounds (doubleton or singleton) or if the opponent behind the QT ( the RHO) shows out on the first or second round.  You have

Note that all the cases considered are “mutually exclusive,” meaning they cannot occur at the same time.  Their probabilities simply are added to form the combined probability of one or more of them happening.  If ”LHO jack singleton”  had simply been “jack singleton”, the case of jack singleton on the right would have been counted twice, erroneously, via “RHO any singleton.“ This illustrates one technique for combining cases:  Define the cases so that they are mutually exclusive and then add the probabilities.

This doesn’t work if the events being combined can occur together.  The simplest such situation is that the events are “independent.”  That is, the occurrence of one doesn’t affect the probability of the other. Though seldom perfectly true, this assumption is good enough in many cases.  For independent events, the probability that they both occur at the same time is just the product of their probabilities.  If you want the probability that at least one of the events occurs, you can’t just add the probabilities.

When you consider two events and  you want the probability of having either or both to occur, their “union,” it goes like this:

P(either A or B) = P(A) + P(B)  – P(A and B together), or,

only for independent events,  P(either A or B) = P(A) + P(B)  – { P(A )  X  P(B) }

This latter can be restated for convenience as:

P(either A or B) = P(A) + P(B)  X  {1-P(A)} ,  or else
P(either A or B) = P(B) + P(A)  X  {1-P(B)}.

Example 1 :  You make your contract if a 50% finesse succeeds or if a different suit splits 3-3 at 36%.  Your chance of success is

P (success)= .36 + .5  X .64 = .68 .  (alternatively, .5 +.36  X .5 =.68)

Using approximation rules for suit splits, you would obtain

P (success)= 1/3 + ½ of 2/3 = 2/3 or .67 . (alternatively, ½ + 1/3  X  ½  = 2/3)

You cascade three independent events in the same way.  Once you get P( A or B) you treat that combination as one event and combine it with a new event “C” using

P( any of A or B or C)  =  P(either A or B) + P( C ) * {1 – P(either A or B)}. 

This recursive procedure continues when you have additional events to combine.

Example 2:  You make your contract if a 50% finesse succeeds or if either of two other suits splits 3-3 at 36%.  The first two events were combined as P(either A or B) in example 1.  You now incorporate the additional chance of success as

P (success)=.68 {from Example 1} + .36 * .32 = .8 (approximately) .

If using approximation rules for suit splits, you would have

P (success)=2/3 {from Example 1} + 1/3 * {1 -2/3} = 7/9 = .78

To recap: you combine the first two cases using P(A or B) = P(A) + P(B) X (1- P(A)) .  Then you take that result and combine with the next by the same rule.  And so on.

Example 3: Maybe you need either of two suits to split 3-2.  Your approximate probability would be

P (success)=2/3 + 2/3 X 1/3 = 8/9 = .89.

A much easier way to do this,  is to compute the probability of Failure and then subtract from 1 to find the probability of Success.  That is because a Failure is often an intersection (“A and B” instead of “A or B”) of independent events, and its probability is, therefore, just the product of the probabilities.

From Example 1:  Failure = ½  X  2/3 = 1/3.  P (success)= 2/3.

From Example 2:  Failure = ½  X  2/3  X  2/3 =2/9.  P (success)= 7/9 = .78.

From Example 3:  Failure = 1/3  X  1/3 = 1/9.  P (success)=8/9 = .89.

One reason for starting this article with the harder method is to clarify that you cannot simply add probabilities for events that might occur together.  The Kelsey book, Bridge Odds for Practical Players approaches the matter similarly.

You are unlikely to calculate complicated cases at the table, but you can apply the methods to bridge problems and puzzles, especially in studying card combinations.  At the table you often can estimate the chances of successive failures and subtract from one.  

Footnotes

*”Combinations” refers to the number of ways you can form the given holding. You can usually figure it out without any formulas.  “Jx” from 6 cards, for example, is 10 combinations because there are 5 spot cards you can swap in and two sides where you can place the doubleton.   Or look at it from the other side, where there are 4 x’s.  One of the 5 x’s is on the other side, and there are 5 ways to do it. Consider another case, one-sided QJx — xx from 5 cards.  This offers 3 combinations, swapping x’s.

An excel spreadsheet that will compute precise probabilities of suit splits (suit splits tab) and probabilities for combinations (combinations tab) involving honors can be downloaded here. 

**For “individual probabilities” see Basic Bridge Odds or Wikipedia .

The “Suit Splits” tables at the end of this post show the odds (casual wording) of a split of a certain number of missing cards.  The information is from https://en.wikipedia.org/wiki/Contract_bridge_probabilities but rearranged to emphasize patterns.  These “a priori” probabilities assume you haven’t played a single card and don’t know anything about the hidden hands.  If you do know something about their distribution, it can shift the odds.  One should commit to memory the 2/3 —1/3 rule for an odd number of missing cards  and 1/3—1/2—1/6 rule for an even number.  These approximations often suffice at the table but are less accurate for smaller numbers.  For two missing cards, the 2-0 and 1-1 splits are almost equally probable , or 1/2 -1/2.  For 3 missing cards, the 2-1 vs 3-0 split is approximately 3/4 vs 1/4.  See Summary and Tables sections.

The “individual probability” column is used for considering card combinations*.  The total probability of a given card placement is equal to the “individual probability” multiplied by the “number of combinations.”  For example,  suppose one wanted to know the probability of dropping the QJ doubleton, missing 5 cards.  This allows only 2 combinations of the 20 within a 3-2 split, one combination on each side of you, so the probability is 2 X .0339 =.068, or one tenth of the probability of the 3-2 split.  Another way to arrive at this same conclusion is to assume a 3-2 split and imagine drawing the two cards from five.  Chances of getting both honors in two draws from 5 cards are 2/5 X 1/4, or 1/10, leading to a probability of .068.

With 4 cards missing, including QJ, suppose you play one round and drop the Jack behind your AKT9x.  The singleton jack is one combination from the 3-1 distribution for a probability of .0622, whereas the doubleton QJ is one combination from the 2-2 distribution for  a probability of 0.0678.  You are tempted to think that you should play for the drop.  However, you must account for the possibility that the human player has made a discretionary choice between two equivalent cards.  The usual way out of this conundrum is to assume that, if the player had both honors, he would play the jack only half the time.  Hence  the combined probability of QJ and that the jack, in particular, was played is cut in half from the probability of QJ. This leads to the well-known ‘rule of restricted choice,” which states that, generally, the defender is less likely to have made a choice between equal honors than to have played the only choice he had.

An different approach, favored in this post, is to classify the missing cards as HHxx, where “H” denotes one of two equivalent honors and “x” denotes equivalent spot cards. In the mind’s eye, note that an “H” has been dropped but do not identify it specifically as jack or queen, as a basis for decision.  Treat the honors as indistinguishable, like same-colored marbles from a jar.   This highlights the two combinations of relevant singletons as elements in the probabilities, so the actual probability of the 1-sided fall of a singleton “honor” is recognized as being twice as large, or 0.1244 vs 0.0678 for a doubleton.

This “marbles” approach eliminates the human element in a different way than restricted choice.  Instead of replacing the defender with a 50-50 model, his choices, if any, are neutralized by declarer’s treating the equivalent cards as indistinguishable members of a class –“colored marbles”– throughout the numerical analysis.  This is easier to compute, to explain and to justify mathematically.   Restricted choice, however,  is easy to remember and is widely known and explained; and, for those reasons, it may be preferred at the table.  Many examples are given in Encyclopedia of  Bridge, 7^th Edition.

The numbers 0.1244 and 0.0678 above are the “prior” probabilities of the positions you have found.  At the decision point, they represent the only remaining cases.  The “posterior,” or “conditional,” probabilities always keep the same ratios they had before other cases were ruled out by your “experiments” and “observations.”  This means you have arrived at a point where the present “odds” favor the success of the finesse by .1244 to .0678, or 18 to ten.  The present “probability” of success, at the decision point, has become 0.64.  These tiny numbers are “amplified” at the decision point.

Consider a holding of AK1098 opposite xxx.  Five cards are missing.  You need to capture one of the two missing honors, and would be pleased to capture both.  Start by playing the A.  This identifies a 5-0 split and also drops a singleton honor on either side.  If an honor drops behind, it may be from H alone, HH or HHx, but certainly not from HHxx.  The HHx case would be a deliberate falsecard.   At trick 2, an x is led toward the strong hand.  Next comes the decision whether to finesse.  The realistic probabilities are:

H:     2 * .0283 = .0566 — Suggests finesse for 2nd honor.
HH:  1  * .0339  =.0339 — Suggests play to drop 2nd honor .
HHx: 3 * .0339 = Zero to .1017 — The second honor cannot be captured.

Whether an honor falls under the ace, finesse on the 2nd trick.  This will resolve the entire position.  Losing both honors will occur only when HHxx or HHxxx  are behind the hand, about 10%.  The possibility of a false-card does not affect the decision to finesse.

Holding a suit like AKJTxx opposite xx, it can be tempting to play off one top honor, hoping to drop the singleton Q.  However, this limits the number of times a finesse can be taken.  Two finesses are needed to pick up Qxxx.  This holding is four times (.1132) more likely than the singleton queen (.0283) because  the former represents four card combinations.  Therefore, the first-round finesse is indicated.

An example by Hugh Kelsey, of enhancing success by considering relative probabilities, is described here.

Summary:

The probability of a given holding by defenders is computed multiplying the number of card combinations in the given holding by the probability for one combination.  That probability is obtained by dividing the probability of a given split by the total number of possible combinations within that split.

Utilize these approximation rules at the table:

1)Missing an odd number of cards, they will divide evenly 2/3 of the time and one step from evenly 1/3 of the time.  (Ignore greater steps.)  Exception: 3 missing cards divide 2-1 about 3/4 of the time.

2)Missing an even number of cards, they will divide evenly 1/3 of the time, one step from evenly ½ of the time and two steps from evenly 1/6 of the time.  Exception: 2 missing cards divide 2-0 about 1/2 of the time.

If one of two equal honors drops on the first round, you should usually finesse against the other hand on the second round.  This is consistent with the principle of Restricted Choice.

Another relevant article is https://en.wikipedia.org/wiki/Vacant_Places .  Bridge Odds for Practical Players by Kelsey and Glauert is an in-depth reference for winter reading.  Finally, the principle of “restricted choice” is described here .  Related posts on this website are here and here.

Footnotes:

* To illustrate “combinations,” consider a five card suit containing H1,H2,x1,x2,x3.  For a 2-3 split, ask how many combinations there are in Hxx — Hx.  The doubleton could be either of H1 or H2, together with any one of x1,x2, or x3.  That makes six combinations for Hx or for Hxx on one side.  Swapping the sides gives a total of 12, for a probability of .0339*12 = 0.407. XX offers 3 combinations and  HH is one combination.  Swapping sides makes it 6 and 2.  Therefore the total number of combinations for a 3-2 split is 20.

Suit Splits