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
DoubleFit 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 ( Singlesuiter 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.
« Table (1) »
Correlation between your cards and partner's [ probability in % ]
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.
Suppose you've 6 spades.
Then, your partner will have
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 ( 2suiter Hands)
A second question occurs, when you have a 2suiter 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
:
« Table (2) Fitting Probability of 2suiter Hands »
 Your 2suiter Hand

Your Holding  33  43  44  53  54  55

Probability of 8Card Fit or more (%)
 35.24  48.91  60.31  66.13  74.16  83.52

We learn something from this table.

The first member 33 in this table stands for the 3343 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, 43 holding, the fitting probability remains as low as 33.7% (back to
Table (1) and see the boldface one), so long as you
bid your first 4card 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 44 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 54 and 55 hands.
So long as you bid your 5card (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 2suiters 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 5
3 hand, you open with your 5card suit.
Partner will have 3 cards in that suit with probability 54.4%. That is all for
the fitting probability, in this case.
In 5
4 hands, you can
show your 4card 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 4
4 holding, which
almost doubles the fitting probability.
Did you know these probabilities ?
Probability of Trump Fit ( for all handpatterns )
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 4333 card holding is worst, whatever you bid.
Less bad is the most popular distribution 4432.
These two types occur almost once in 3 deals.
On the other hand, everyone loves 55 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 handpatterns.
Leaving again mathematical details to
Appendix, I'll show the result below
:
« Probability Table (3) »
Hand Pattern  Fitting Probability(%)  Bergen Count  Frequency(%)

8 or more  9 or more

4333  76.4  25.6  7  10.54

4432  79.1  29.0  8  21.55

4441  83.1  34.0  8  2.99

5332  82.5  34.5  8  15.52

5422  84.5  37.5  9  10.58

5431  85.8  39.2  9  12.93

5440  89.3  44.4  9  1.24

5521  89.5  46.7  10  3.17

5530  91.0  48.9  10  0.90

6322  90.2  49.8  9  5.64

6331  91.0  51.2  9  3.45

6421  92.1  53.6  10  4.70

This table covers all hands with frequency greater than 2%. You observe above that the top two (4333 and 4432) 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 8card fit (or more), while green and blue ones for 9card fit (or more).
Overlapping crosses are colored somewhat differently.
I'll leave you this figure whatever for your interpretation,
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 4432 hand against expectations.
Surely, it works better than 4333 in
cardplays.
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)
:
« Probability Table (4) »
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 8card 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 doublefit problem mentioned
elsewhere, I attacked making it complete in the following two senses:
(1) I do not rely on computer simulations,
but use mathematics (which is always exact).
(2) I do distinguish a simple 8card fit from an 8card 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
:
« Probability Table (5) »
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.]
Some comments are in order:
(a) Asterisk (*) means zero probability, as in Schwan's table.
Such an event never occurs.
(b) 8d means an 8card double fit [8+8], as
distinguished from a simple 8card 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 rowwise. When summed in a row, total amounts to 100%.
Columnwise reading is meaningless.
(e) A few words for readers who doubt about lack of
symmetricity: If you multiply every row by its
frequency on the right, and stop distinguishing double fits,
then the table will restore symmetricity 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 8card double fit [8d]. They never will have
a simple 8card fit [8].
 The [8d] row tells us something so interesting, and I would put it as a DoubleFit Rule,
which says,
[DoubleFit Rule]
When you've an 8card double fit [8+8] in your two hands, opponents will
necessarily have an 8card double fit [8+8], or a [9+7] double fit, or [10+6] ….
Their hands can never be a simple 8card fit [8].
This Rule solves the
double doublefit problem.
Actually, both authors (Vernes and Cohen) of LoTT didn't have to stress "double double", because deals are automatically doubledouble,
if one side is double.
Observe that 8d and 8 has no intersections in this table.
The doublefit rule is explained plainly
:
You have an 8card double fit, holding 8
8
ab in two hands.
Opponents will then hold its complementary 5
5
xy,
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
_{N}C
_{p} , 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,
_{N}C
_{p} =
N ! /
p ! (
N−
p)! .
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
 
_{13}C_{x}
_{26}C_{13−a−x}

P (a, x)
 =

−−−−−−−−−−−−−−−−−−.

 
_{39}C_{13−a}

The proof goes as follows:
Just concentrate on the number of spade cards in the 4 players.
 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
_{13}C_{a }.
 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
_{13}C_{x }.
 Out of the 26 cards in their hands, E and W have 13−a−x 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 _{26}C_{13−a−x} .
Math is simple in this case. Multiply the above three combinations to obtain
N (a, x) = _{13}C_{a} × _{13}C_{x} × _{26}C_{13−a−x}
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)

= _{13}C_{a} × _{x} _{13}C_{x} × _{26}C_{13−a−x}


= _{13}C_{a} × _{39}C_{13−a}

where, I made use of the useful summation formula for binomial coefficients,
_{x} _{n}C
_{x} _{m}C
_{p−x}
=
_{n+m}C
_{p }.
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
_{13}C
_{a} cancels itself, since it does not
depend on
x. Note that
a is a given constant, while
x is a variable.
Table (1) gives the result based on this formula for
P (
a, x ).
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−a}C_{x}
_{13−b}C_{y}
_{13+a+b}C_{13−x−y}
P (x, y ) = −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− .
_{39}C_{13}
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
:
 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−a−x cards).
So, the number of combinations in the spade suit is _{13}C_{a} × _{13−a}C_{x }.
 Counting in the heart suit goes the same way, and will give the number of
combinations _{13}C_{b} × _{13−b}C_{y }.
 The 26 cards in diamonds and clubs will be counted in common.
Of those 26 cards, N will take 13−a−b 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−x−y (again subtracting).
Distribution of the rest to E and W doesn't matter, and
the number of combinations for diamonds and clubs is
_{26}C_{13−a−b} × _{13+a+b}C_{13−x−y }.
Multiplication of the above three combinations yields
N (x, y ) =
_{13−a}C_{x}
× _{13−b}C_{y}
× _{13+a+b}C_{13−x−y },
where the constant factor
_{13}C
_{a} ×
_{13}C
_{b} ×
_{26}C
_{13−a−b}
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,
we are led to the probability formula for
P (
x, y ) above.
In Table (3), we asked,
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−a−b−c 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−a}C_{x}
_{13−b}C_{y}
_{13−c}C_{z}
_{a+b+c}C_{13−x−y−z}
P (x, y, z )
= −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−.
_{39}C_{13}
To obtain Table (3), it suffices to sum this probability
P (
x, y, z) for those sets of
(
x, y, z) which achieves an 8card (or more) fit in at least one of the
four suits.
Fitting probabilities for 9card (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.
 Start again with spade cards. Of the 13 spades, N takes a cards,
S takes x cards,
(remaining 13−a−x spade cards are held by E and W ).
So, the number of combinations for such holdings is _{13}C_{a} × _{13−a}C_{x }.
 Similarly, combinations in hearts are _{13}C_{b} × _{13−b}C_{y} .
 And for diamonds, _{13}C_{c} × _{13−c}C_{z} .
 Finally, consider clubs.
Of the 13 clubs,
N takes 13−a−b−c 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−x−y−z cards (here, also subtraction). So, in clubs,
we have _{13}C_{13−a−b−c} × _{a+b+c}C_{13−x−y−z} 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−a}C_{x}
× _{13−b}C_{y}
× _{13−c}C_{z}
× _{a+b+c}C_{13−x−y−z} .
The above probability
P (
x, y, z) is obtained by dividing
it by its total
_{x} _{y} _{z}
N (
x, y, z) =
_{39}C
_{13} .
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 _{52}C_{26} combinations.
Probability is calculated assuming that each of these _{52}C_{26} deals is evenly likely to happen.
Now, suppose that NS 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
_{13}C_{s} × _{13}C_{h} × _{13}C_{d} × _{13}C_{c}
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,
_{52}C
_{26} .
Let us consider in detail the case where NS 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, EW will have a complementary 6  6  7  7 distribution.
So, both pairs have 7 trumps.
This dealing will own
_{13}C_{7} × _{13}C_{7} × _{13}C_{6} × _{13}C_{6}
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
_{4}C
_{2} = 6 to obtain
_{4}C_{2} × ( _{13}C_{7} × _{13}C_{7} × _{13}C_{6} × _{13}C_{6} )
= 5202 59937 50016.
In the other case,
(b) 7  7  7  5,
EW pair will have a complementary 6  6  6  8 distribution.
So, they will have an 8card fit.
This dealing owns
_{13}C_{7} × _{13}C_{7} × _{13}C_{7} × _{13}C_{5}
combinations. Consider other dealings like (7, 7, 5, 7) and multiply
_{4}C
_{1} = 4,
to obtain
_{4}C_{1} × ( _{13}C_{7} × _{13}C_{7} × _{13}C_{7} × _{13}C_{5} )
= 2601 29968 75008.
The relevant probabilities are obtained by dividing them with
_{52}C_{26} = 49591 85329 48104,
as
Prob(a) = 10.4908 %,
Prob(b) = 5.2454 %.
Their sum gives the probability 15.7362% of 7card 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 7card 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.