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

Number of Cards	Distribu-tion	Proba-bility	Combin-ations	Individual Probability	Approxim-ation
2	1 – 1	0.52	2	0.2600	1/2
	2 – 0	0.48	2	0.2400	1/2
4	2 – 2	0.40	6	0.0678	1/3
	3 – 1	0.5	8	0.0622	1/2
	4 – 0	0.1	2	0.0478	1/6
6	3 – 3	0.36	20	0.0178	1/3
	4 – 2	0.48	30	0.0162	1/2
	5 – 1	0.15	12	0.0121	1/6
	6 – 0	0.01	2	0.0075	0
8	4 – 4	0.33	70	0.0047	1/3
	5 – 3	0.47	112	0.0042	1/2
	6 – 2	0.17	56	0.0031	1/6
	7 – 1	0.03	16	0.0018	0
	8 – 0	0	2	0.0008	0

 

Number of Cards	Distribu-tion	Proba-bility	Combin-ations	Individual Probability	Approxim-ation
3	2 – 1	0.78	6	0.1300	3/4
	3 – 0	0.22	2	0.1100	1/4
5	3 – 2	0.68	20	0.0339	2/3
	4 – 1	0.28	10	0.0283	1/3
	5 – 0	0.04	2	0.0196	0
	4 – 3	0.62	70	0.0089	2/3
7	5 – 2	0.31	42	0.0073	1/3
	6 – 1	0.07	14	0.0048	0
	7 – 0	0.01	2	0.0026	0
	5-4	0.59	252	0.00234	2/3
9	6-3	0.31	168	0.00187	1/3
	7-2	0.086	72	0.00119	0
	8-1	0.107	18	0.0006	0