## How often you get Trump Fit with your partner

(C) boco_san (2004/09/01)

«  Probabilities you should know when you're bidding  »

Introduction

In bridge, it's important to be, at least roughly, aware of probabilities.
We often find, in bridge books or on web sites, tables of probability, for example, for

How 7 cards are likely to split in the hands of defenders,
when you are playing as declarer.   This kind of probability is so popular to bridge players and will need no discussions.
What I'm about to talk here is another kind of probability, which you have to be aware when you're bidding.  Bridge is certainly a game of probability.
The problem of utmost importance in bridge bidding is how to find a possible fitting in a trump suit with partner.   Knowledge of fitting probabilities will help you in this process.   Since I've never seen this kind of probability tables on any books nor on web sites (except one), nor in WikiPedia, I calculated them for myself.   The results are shown below in five tables and one figure, which I'll explain in turn.
Indeed, fitting probabilities tell us quite a lot, most of which have been left unknown.   You'll encounter below novel concepts like Bergen count and Double-Fit Rule and their usefulness.   Even the concept of fitting probabilities will sound novel, since they haven't been studied systematically.
Learned readers are invited to mathematical details expounded in four appendices.
Probability of Trump Fit ( Single-suiter hands)

The first question, we ask on fitting probabilities, is, for example,

When you've 6 cards in Spades, how often (with what probability) does your partner have 2 cards in Spades ?
or likewise
When you've 5 cards in a certain suit, how often does your partner have 3 cards in that suit ?
Here and below, the word fit is agreed to mean 8 (or more) cards in the same suit in opposite two hands, as usual.   Mathematical details concerning the above sort of fitting probability are given in Appendix for the sake of interested readers.
Numerical result obtained from that mathematical formula is shown in Table (1) below.   Please notice that this table (as well as the ones to follow) is not a result of any computer simulations.   It gives mathematically exact values of the probabilities.

 Your cards Fit proba -bility Number of partner's cards in the same suit 0 1 2 3 4 5 6 7 8 9 10 11 12 13 0 1.2 0.1 1.5 7.4 18.7 27.5 24.8 13.9 4.9 1.0 0.1 0.0 0.0 0.0 0.0 1 3.4 0.2 2.6 10.6 22.9 28.6 21.6 10.1 2.9 0.5 0.0 0.0 0.0 0.0 2 8.5 0.5 4.1 14.5 26.7 28.1 17.7 6.7 1.5 0.2 0.0 0.0 0.0 3 18.1 0.8 6.4 19.2 29.6 25.9 13.3 4.0 0.7 0.1 0.0 0.0 4 33.7 1.5 9.6 24.2 31.1 22.2 9.1 2.1 0.3 0.0 0.0 5 54.4 2.5 13.9 29.2 30.6 17.4 5.4 0.9 0.1 0.0 6 76.3 4.3 19.5 33.4 27.8 12.1 2.7 0.3 0.0 7 92.9 7.1 26.2 35.7 22.8 7.1 1.0 0.1 8 100 11.4 33.8 35.2 16.1 3.2 0.2 9 100 18.2 41.1 30.8 9.0 0.9 10 100 28.4 46.2 22.2 3.1 11 100 43.9 45.6 10.5 12 100 66.7 33.3 13 100 100
With this table, we can answer the questions above.
2 spades with probability 33.4 %,
3 spades with probability 27.8 %,
4 spades with probability 12.1 %, and so on.
If you add them up, considering that you have a fit when you've 8 cards or more in your two hands, then you get the fitting probability  76.3%.
What does this mean ?
It means that you may expect 2 or more cards in partner's hand, with a high probability 76.3%.   In other words, when you've 6 cards in some suit, your side will have a fit in that suit in 3 deals out of 4.
Suppose now that you've 5 cards in a major suit and open your hand with 1 or 1.   In this case, your partner will support your bid (with 3 cards or more) with probability 54.4%, i.e., half times you open.
Did you know these probabilities ?

Probability of Trump Fit ( 2-suiter Hands)

A second question occurs, when you have a 2-suiter hand,

When you've  5  Spades and  4  Hearts, how often (with what probability) will your partner support you, either with  3  (or more) Spades, or with  4  (or more) Hearts ?
Again, mathematical formula is shown in Appendix.
Numerical result obtained is as follows:
 Your 2-suiter Hand Your Holding 3-3 4-3 4-4 5-3 5-4 5-5 Probability of 8-Card Fit  or more (%) 35.24 48.91 60.31 66.13 74.16 83.52

We learn something from this table.

•     The first member 3-3 in this table stands for the 3-3-4-3 balanced hand, which you'll open with 1NT, if you've an appropriate HCP.   Partner will then support you (possibly through Jacoby transfer) with either of the major suits, having 5 cards.  This happens almost once in three deals.
•     In the second, 4-3 holding, the fitting probability remains as low as 33.7% (back to Table (1) and see the bold-face one), so long as you bid your first 4-card suit.   However, there remains an additional possibility (ca.18%) that partner will have 5 cards in your second suit.   So, in this holding, you'll find fitting about half times you open.
•     In the third 4-4 case, if you (or partner) have a chance to bid your second suit, then the fitting probability almost doubles up from 33.7 to 60.3%.   The result shows very clearly the need of bidding twice in this holding.
•     The need becomes even more remarkable in 5-4 and 5-5 hands.   So long as you bid your 5-card (first) suit, the fitting probability remains 54.4% (back again to Table (1)), which will jump up to 74.2 and 83.5%, respectively, when you can show your second suit.   The result validates usefulness of Michaels cuebid and unusual NT overcall, which tells your 2-suiters with a single call.
For the sake of reference, and in case you overlooked my statements above, comparison will be made in the following tables.
 Distribution Rule of 20 Requires Fitting Probability 53xx 12 HCP 54.4 % 54xx 11 HCP 74.2 %
Holding a 53 hand, you open with your 5-card suit.   Partner will have 3 cards in that suit with probability 54.4%.   That is all for the fitting probability, in this case.
In 54 hands, you can show your 4-card suit on your second turn of bidding, which raises the fitting probability to 74.2%.   Marty Bergen duly appreciates this difference by reducing the requirement for HCP by 1 point.

 Distribution Rule of 20 requires Fitting Probability 4333 13 HCP 33.7 % 44xx 12 HCP 60.3 %
A second biddable suit is even more important in the 44 holding, which almost doubles the fitting probability.

Did you know these probabilities ?

Probability of Trump Fit ( for all hand-patterns )
and What Rule of 20 means.

Having got acquainted with mathematics, I entered more complex situations (the result is simple, however).   As many bridge players know well, the 4-3-3-3 card holding is worst, whatever you bid.   Less bad is the most popular distribution 4-4-3-2.   These two types occur almost once in 3 deals.   On the other hand, everyone loves 5-5 holdings.  The object here is to quantify (rather than qualify) this difference in the language of fitting probability.   In other words, a third question we ask is,

How often, or how less often, will you find trump fit with your partner, holding such and such distribution ?
So, I calculated fitting probabilities for all hand-patterns.   Leaving again mathematical details to Appendix, I'll show the result below:
 Hand Pattern Fitting Probability(%) Bergen Count Frequency(%) 8 or more 9 or more 4-3-3-3 76.4 25.6 7 10.54 4-4-3-2 79.1 29.0 8 21.55 4-4-4-1 83.1 34.0 8 2.99 5-3-3-2 82.5 34.5 8 15.52 5-4-2-2 84.5 37.5 9 10.58 5-4-3-1 85.8 39.2 9 12.93 5-4-4-0 89.3 44.4 9 1.24 5-5-2-1 89.5 46.7 10 3.17 5-5-3-0 91.0 48.9 10 0.90 6-3-2-2 90.2 49.8 9 5.64 6-3-3-1 91.0 51.2 9 3.45 6-4-2-1 92.1 53.6 10 4.70
This table covers all hands with frequency greater than 2%.   You observe above that the top two (4-3-3-3 and 4-4-3-2) hands have remarkably lower fitting probabilities than the others.
Bergen count in the table (so I call) obviously means
Bergen count = sum of the lengths of two longest suits.
Actually, Marty Bergen teaches beginners to open (in first and second seats) at one level, if
HCP + Bergen count 20
is satisfied.   It has now become a de facto standard under the name Rule of 20.   You'll observe above that the fitting probability is in good correlation with Bergen count.
The figure on the right clearly visualizes this correlation.   The abscissa gives the Bergen count, while the ordinate the fitting probability.   Crosses in red and wine are for 8-card fit (or more), while green and blue ones for 9-card fit (or more).   Overlapping crosses are colored somewhat differently.    I'll leave you this figure whatever for your interpreta-tion, but mine is this:
During I'm bidding, I don't yet see partner's cards, and I'm quite uncertain if my cards will fit with partner's.   However, if I obey the Rule of 20, even without knowledge of partner's hand, I'll more often open those hands that are more likely to fit with partner's hand.
At this point, I took notice that I had misunderstood the Rule of 20.   The point may look trivial, but never.   I had long believed, "the Rule is useful, because if I have a fit with partner, I would be able to utilize the long suit to draw, ruff, or run in NT, etc."   Now, the above IF clause has turned out to be unnecessary.   Every time I use the Rule to open, my hand is more likely to find fitting with partner's.   That's what Marty Bergen's rule implies.
[NB] The above table downgrades the 4-4-3-2 hand against expectations.   Surely, it works better than 4-3-3-3 in card-plays.   Fitting probability doesn't consider such features.

How often and how many trumps your side gets

A fourth question is quite simple.   You may be concerned,

How often 10 trumps are dealt to our side
(to myself and my partner) ?
The answer is (see Appendix for mathematics):
 Number of Cards 7 8 9 10 11 12 13 Frequency(%) 15.7 45.7 28.1 8.7 1.6 0.16 0.01
The table shows that 8 cards are most likely (45.7%) to be dealt to a pair.   Nine cards follow next, and these two cases occupy 73.9% of total.   Ten cards, which might be of great concern of many players, are dealt only once in 11 deals.

Correlation between Trump Distributions

The fifth, final, problem is the correlation between the number of trumps held by the two sides.   You may be concerned to ask, for example,

When we know that we've 10 trumps in our hands,
how often will they have 10 trumps in their hands ?
or, even ask as far as
When we know that we've 10 trumps,
what would their hands be like at least ?
Is it a simple 8-card fit, or more ?
This kind of questions emerge when you rely on LoTT and consider some adjustments.   Partial answer to this problem is found in Cohen's second book(p.148), based on 1000 deal computer simulations by Edward Schwan.
Motivated by Schwan's correlation table as well as by the double double-fit problem mentioned elsewhere, I attacked making it complete in the following two senses:
(1)  I do not rely on computer simulations, but use mathe-matics (which is always exact).
(2)  I do distinguish a simple 8-card fit from an 8-card double fit.   Difference between simple and double fits is enormous, as many know.   So, this distinction is essential.
Again, mathematics is given in Appendix, with the following result:
 Correlation between Trump Distributions [8d] = 8 + 8;  [9d] = 9 + 8 as well as 9 + 9. Our Trumps Their Trumps 7 8 8d 9 9d 10 11 12 13 Frequency 7 66.67 33.33 * * * * * * * 15.74% 8 14.81 66.67 * 18.52 * * * * * 35.41% 8d * * 32.11 47.57 * 19.03 1.30 * * 10.34% 9 * 32.43 24.32 36.04 * 7.21 * * * 20.22% 9d * * * * 55.99 33.50 9.66 0.82 0.02 7.88% 10 * * 22.68 16.80 30.45 23.43 5.85 0.75 0.04 8.67% 11 * * 8.48 * 48.16 32.10 9.67 1.49 0.09 1.58% 12 * * * * 41.00 41.17 14.94 2.69 0.19 0.16% 13 * * * * 24.30 48.60 22.08 4.64 0.38 0.01%
[A more detailed correlation table is available, where each double fit is treated separately.]

(a) Asterisk (*) means zero probability, as in Schwan's table.   Such an event never occurs.
(b) 8d means an 8-card double fit [8+8], as distinguished from a simple 8-card fit, [8].
(c) Similarly, simple [9] is distinguished from 9d, which counts two kinds of double fit, [9+8] and [9+9].
(d) This table should be read row-wise.   When summed in a row, total amounts to 100%. Column-wise reading is meaningless.
(e) A few words for readers who doubt about lack of symmet-ricity:   If you multiply every row by its frequency on the right, and stop distinguishing double fits, then the table will restore symmet-ricity and validates Schwan's 1000 deal simulation.
You'll enjoy this table from several aspects.   I'll raise the following two points for the sake of LoTTers.
• When you've 10 trumps in your two hands, as you asked above, opponents will have at least an 8-card double fit [8d].   They never will have a simple 8-card fit [8].
• The [8d] row tells us something so interesting, and I would put it as a Double-Fit Rule, which says,
[Double-Fit Rule]   When you've an 8-card double fit [8+8] in your two hands, opponents will necessarily have an 8-card double fit [8+8], or a [9+7] double fit, or [10+6] ….
Their hands can never be a simple 8-card fit [8].
This Rule solves the double double-fit problem.   Actually, both authors (Vernes and Cohen) of LoTT didn't have to stress "double double", because deals are automatically double-double, if one side is double.   Observe that 8d and 8 has no intersections in this table.
The double-fit rule is explained plainly:
You have an 8-card double fit, holding 88ab in two hands.   Opponents will then hold its complementary 55xy,
where  x + y = 26 − 5 − 5 = 16, because they have 26 cards in their hands.   If x = 8, then y = 8, and if x = 9, then y = 7, etc.   QED.

Appendix

Most likely, many bridge players will have no interest in mathematics.   Nevertheless, for the sake of completeness, I'll show below how I derived the mathematical expressions so that interested readers may confirm them and may even do computations by themselves.
You'll see everywhere below the symbol NCp , which stands for binomial coefficients in mathematics, or the number of combinations in statistics to choose  p  events (deals, in bridge) out of  N  events,   NCp =  N ! / p ! (Np)! .

Calculation of Probability [ Table (1) ]

Calculation of probabilities starts with counting the number of events (deals) of interest.   The rest is easy.   Probability is obtained by dividing it by the total number of events.   It is here naturally assumed that every event is evenly likely to happen.
Now, we call, for brevity, the 4 players  N, E, S, and  W.
In Table (1), we asked a problem like

When player N has 6 spades,
how often does his partner S have 2 spades ?
which is generalized here to
Suppose that player  N  has  a  cards in spades, then
find the probability P (a, x ) with which
his partner  S  has  x  cards in spades.
The answer will turn out to be
 13Cx   26C13−a−x P (a, x) = −−−−−−−−−−−−−−−−−−. 39C13−a

The proof goes as follows:

Just concentrate on the number of spade cards in the 4 players.

1. Out of the 13 cards in his hand,  N  has  a  cards in spades,  13−a  cards in the other suits.   The number of such combinations is  13Ca .
2. Out of the 13 cards in his hand,  S  has  x  cards in spades,  13−x  cards in the other suits.   The number of such combinations is  13Cx .
3. Out of the 26 cards in their hands,  E  and  W  have  13ax  cards in spades (remember that there are only 13 spades in the world), and  13+a+x  cards in the other suits (because they have 26 in total).   The number of such combinations is  26C13−ax .
Math is simple in this case.  Multiply the above three combinations to obtain
N (a, x)   =   13Ca × 13Cx × 26C13−ax
for the number of events which satisfy the required condition.   The probability  P (a, x)  we're after is obtained by dividing
N (a, x) by its sum over x;

 x  N (a, x) =  13Ca  ×  x  13Cx × 26C13−a−x =  13Ca × 39C13−a
where, I made use of the useful summation formula for binomial coefficients,
x   nCx   mCpx   =   n+mCp .
Here, the sum is to be taken over all possible value of  x, such that the factorials inside do not have negative arguments.
In the above calculation, the factor  13Ca  cancels itself, since it does not depend on  x.  Note that   is a given constant, while   is a variable.
Table (1) gives the result based on this formula for  P (a, x ).

Calculation of Probability [ Table (2) ]

In Table (2), we considered a problem like

When player  N  has 5 spades  AND  4 hearts,
how often will  S  have either 3 (or more) spades,
OR 4 (or more) hearts ?
Again, we generalize it to
Suppose that  N  has  a  cards in spades  AND  b  cards in hearts, then, find the probability P (x, y)  with which
S  has  x  spades AND  y  hearts.
Math goes a bit complicated, because of complexity of the problem.   Its answer will turn out to be
13−aCx   13−bCy   13+a+bC13−xy
P (x, y )  =   −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− .
39C13
To obtain Table (2), we sum this  P (x, y )  for all possible values of  x  and  y  such that
a + x 8    or    b + y 8
i.e., for all cases where spades OR hearts in the two hands are as long as  8  or longer.
The proof goes as follows:
1. First, concentrate on the spade cards.
Of the 13 cards,  N  takes a  cards.   From the remaining  13−a  cards,  S  will take  x  cards,   (and,  E  and  W  take the yet remaining  13−ax  cards).   So, the number of combinations in the spade suit is  13Ca × 13−aCx .
2. Counting in the heart suit goes the same way, and will give the number of combinations  13Cb × 13−bC.
3. The 26 cards in diamonds and clubs will be counted in common.   Of those  26  cards,  N  will take  13−ab  cards (subtracting his spades and hearts from 13).
Now, the  13+a+b  cards (in diamonds and clubs) are left for  S,  E,  and  W,  from which  S  takes  13−xy (again subtracting).   Distribution of the rest to  E  and  W  doesn't matter, and the number of combinations for diamonds and clubs is  26C13−ab × 13+a+bC13−xy .
Multiplication of the above three combinations yields
N (x, y )  =  13−aCx × 13−bCy × 13+a+bC13−xy ,
where the constant factor  13Ca ×  13Cb × 26C13−ab   has been omitted, which does not depend on the variables  x  and  y.   (Note again that  a  and  b  are constants.)   Since we can easily sum this  N (x, y )   for all values of  x  and  y , using the summation formula,
 x y  N (x, y) =   x  y  13−aCx × 13−bCy × 13+a+bC13−x−y =   x   13−aCx × 26+aC13−x =   39C13  ,
we are led to the probability formula for  P (x, y )  above.

Calculation of Probability [ Table (3) ]

How often, or how less often, will you find trump fitting with your partner, holding such and such distribution ?
This is a more complex question.   It is too vague to work with math.   To substantiate it, put it into a more definite question, for example,
Say player  N  has  4 spades,  3  hearts,  3 diamonds and  3 clubs, then what is the probability  P  that his partner  S  does have either  4 (or more) spades, OR  5 (or more) hearts, OR  5 (or more) diamonds, OR  5 (or more) clubs ?
This problem is again generalized to
Suppose that  N  has  a  spades,  b  hearts, and  c  diamonds (and hence  13−abc  clubs), find the probability  P (x, y, z)  with which  S  will have  x  spades,  y  hearts and  z  diamonds (and the rest in clubs) ?
The answer turns out to be
13−aCx   13−bCy   13−cCz   a+b+cC13−xyz
P (x, y, z )  =   −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−.
39C13
To obtain Table (3), it suffices to sum this probability P (x, y, z) for those sets of (xyz) which achieves an 8-card (or more) fit in at least one of the four suits.   Fitting probabilities for 9-card (or more) fit can also be obtained in the same way.
The problem looks hard and almost desperate (so did it to me, at first), but, not so, if you go step by step.   Here we go.
1. Start again with spade cards.   Of the  13 spades,  N  takes  a  cards,   S  takes  x  cards, (remaining  13−ax  spade cards are held by  E  and  W ).   So, the number of combinations for such holdings is  13Ca × 13−aCx .
2. Similarly, combinations in hearts are  13Cb × 13−bCy .
3. And for diamonds,  13Cc × 13−cCz .
4. Finally, consider clubs.
Of the 13 clubs,  N  takes  13−abc  cards (again, by subtraction), and the remaining  a+b+c  club cards are dealt among  S,  E, and  W,  from which  S  will take  13−xyz   cards (here, also subtraction).   So, in clubs, we have  13C13−abc × a+b+cC13−xyz  combinations in clubs.
Multiplying the above three combinations, and omitting again constant factors (which do not depend on variables  x, y, and z), we obtain
N (x, y, z ) = 13−aCx × 13−bCy × 13−cCz × a+b+cC13−xyz .
The above probability  P (x, y, z)  is obtained by dividing it by its total
x y z   N (x, y, z) = 39C13 .

Calculation of Probability [ Table (4) and Table (5) ]

Now, let us attack the final problems.
In Tables (4) and (5), we're interested in how many trumps each pair has in their two hands.   So, it doesn't matter how 26 cards are dealt within a pair.   Say,  5 and 4, or 6 and 3, or even 7 and 2, it doesn't matter.   In any case, you've 9 trumps.
The problem is, therefore,

How to deal a deck of 52 cards to TWO pairs,
26 cards to each,
which has obviously  52C26  combinations.   Probability is calculated assuming that each of these  52C26  deals is evenly likely to happen.

Now, suppose that N-S pair is dealt  s  spades,  h  hearts,  d  diamonds and  c  clubs.   Their sum is, of course, equal to 26.

s + h + d + c = 26,
and we have
13Cs × 13Ch × 13Cd × 13Cc
combinations for such a dealing, because every 13 cards (in each suit) are dealt in this way to the pairs.   In short, probability is obtained by dividing the above number of combinations by its total,  52C26 .
Let us consider in detail the case where N-S pair is dealt the least, seven, trumps.   This occurs only in the following two types of distributions:
(a)  7 - 7 - 6 - 6.
In this case, E-W will have a complementary  6 - 6 - 7 - 7  distribution.   So, both pairs have  7  trumps.   This dealing will own
13C7 × 13C7 × 13C6 × 13C6
possibilities (compare with the above general expression).   But, you have to take account of other distributions of the same type in other suits, say,
(s, h, d, c ) = (7, 6, 7, 6)
in addition to the original one
(s, h, d, c ) = (7, 7, 6, 6),
so, multiply  4C2 = 6   to obtain
4C2 × ( 13C7 × 13C7 × 13C6 × 13C6 )  =  5202 59937 50016.
In the other case,
(b)  7 - 7 - 7 - 5,
E-W pair will have a complementary  6 - 6 - 6 - 8  distribution.   So, they will have an  8-card fit.   This dealing owns
13C7 × 13C7 × 13C7 × 13C5
combinations.   Consider other dealings like (7, 7, 5, 7) and multiply  4C1 = 4,   to obtain
4C1 × ( 13C7 × 13C7 × 13C7 × 13C5 )   =  2601 29968 75008.
The relevant probabilities are obtained by dividing them with
52C26   =  49591 85329 48104,
as
Prob(a)  =  10.4908 %,
Prob(b)  =    5.2454 %.
Their sum gives the probability  15.7362%  of  7-card fitting, as seen on the top lines in Tables (4) and (5).   Furthermore, their ratio is exactly 2/1, so, when your side has a  7-card fit, opponents are twice more likely to have a seven rather than an eight card fit.

On the basis of these mathematical formulas, I prepared computer programs in Visual Basic.   Numerical results have been given in Tables (1) through (5).

References on probabilities in bridge.
Durango Bill: presents mathematical basis for probabilities in bridge.
So long as the two works overlap, frequency in my Table (3) grees with his.
Theo Groen: gives several tables (calculated or simulated, as he says), which agree with my Table (1), frequency in Table (3), and Table (5), where he does not distinguish single and double fits.