When the play of a card, that can be selected between cards of equivalents values (generally contigouos), increase the chance that the player did that because he was restricted in that choise.
The rule of Restricted Choise permits declarer have a better line of play with the evaluation of "a posteriori" probability after a significative event occurs (the drop of high card). So, this permits a best decison make to improve declarer play avoiding just of a simple guess for dubious situations.
This
principle was showed in 1936 in the book "LES IMPASSES AU BRIDGE" by Pierres
Bellanger, and in 1940 was mathematically analyzed by professor Émile Borel and
André Chéron (journalist) in
the classical book
"Theorie Mathematique du Bridge", but only
have a entry in the bridge community when Alan Truscott put this in discussion
on the "Contract
Bridge Journal" and finally when Terence Reese unify this principle in a
chapter of his book "THE EXPERT GAME" in 1958 (published in U.S. with
the title "Master Play").
We believe that only after the South African Alec Taube translated in 1954
Borel-Chéron's book, from French to English, that the principle of Restricted
Choise became mathematically understood by the bride community.
In 1985, Hugh Kelsey & Michael Glauert' book
"Bridge
Odds for practical players" shows a chapter for the Restricted Choise principle
using a new name for this principle as "Freedom of Choice". This is a high
recomended book for all advanced bridgists that need a better knowledge in a
mathematical bridge.
Before make an enunciation of Restricted Choise principle let's reproduce professor Borel probability explanation and examples:
SITUATION-1) Suppose we have a suit where the residues (cards with opponents) are QJ5432 and let's assume that nobody discard a Queen or a Jack when we play Ace or King, unlesst they are restricted to follow these honors.
The "a
priori" probability (before the play of the hand is started) of these 6 cards distribution between E-W obey the
follow frequencies:
6-0 + 0-6 => 1,4907% (02 configurations each one = 0,745%)
5-1 + 1-5 => 14,5342% (10 configurations each one = 1,454%)
4-2 + 2-4 => 48,4472% (32 configurations each one = 1,514%)
3-3 =====> 35,5280% (20 configurations each one = 1,776%)
After
NORTH play a top honor (ace or king) let's suppose West plays "4" and
EAST plays "2". What are now the new % from the remain
distributions?
Example:
|
A106 N 4 W E 2 or S K987 |
AK1098 N 4 W E 2 S 76 |
Considering that the 6-0 distribution
is eliminated, considering
also that the configuration of singleton Queen, singleton jack and QJ second
does not exist, then in the calculation of the percentage of each distribution we maintain the
35,528% from 3-3 distribution (now 2-2), we alter the 4-2 distribution
(now 3-1) deleting the QJ distribution, 48,44%-3,23%=45,22% and finally for the
distribution 5-1 (now 4-0) we alter deleting singleton Q and J, so 14,5342%-2,42%-2,42%=9,69%. Thus the new % of the distributions will be now:
2-2 ………35,52 / (35,526 + 45,22 + 9,69) x 100 = 39,29% ==
Bayes
3-1 ………45,22 / (35,526 + 45,22 + 9,69) x 100 = 50,00% => rule of 3
4-0 ……… 9,69 / (35,526 + 45,22 + 9,69) x 100 = 10,71%
== application
Now we
continue with the K and WEST plays "5" and EAST plays
jack, so the 4-0 distribution also is eliminated and raimain only the initial
4-2 and 3-3 distributibution What are now the percentages of the possible
remaining distributions
2-0 and 1-1 for the opponents?
Example:
|
A106 N 54 W E J2 S K987 |
AK1098 N 54 W E J2 S 76 |
If EAST discard was a non significant card, then the 1-1 distribution is
52,38% against 47,62% of the 2-0 distribution, but EAST shows up the jack, thus
there are two hipotheses to consider:
1 -EAST started with Jx and WEST
with Qxxx;
2
-EAST started with QJx and WEST with xxx.
Analysing the "a priori" percentagens of both possible events, we observe that the Jx has 6,46% and QJx has 7,10% probability. So, what are now the "a posteriori" probability after these events?
Prof. Borel teached us that in the computation of probabilty we need also to consider the behavios of the players. Lets look EAST possible discards:
- If EAST
have QJ in the second trick, one first hipothesis is he always plays
the jack to
avoid tricking his partner;
- but in a second hiphotesis EAST can discard a false card to trick declarer
and so plays the
Queen;
- finally a third hipothesis is he half time plays the jack and half time plays
the Queen.
So,
in hipothesis 1 and in hipothesis 2, the "a priori" probability of Jx is 6,46%
and the composed "a priori" probability for
EAST started with Jx is:
6,46% x 1 = 6,46 where 1 is the assurance that EAST have the Jack.
In
hipothesis 3, the "a priori" probability of QJx is 7,10% and the composed "a
priori" probability
that EAST started with QJx and
played the jack
in the second trick is:
7,10 x 1/2 = 3,55%, where 1/2 represents the "a priori" probability that he
discarded the jack when he had Queen and Jack.
Now, for
the applycation of the formula of the English priest Thomas Bayes (1702 -
1761) we need find the "a posteriori" probability after EAST played the Jack, and this is made
by applying a rule of 3 to adjust the %
of the sum of all remaining probabilities, so:
Jx => is 6,4596 / (6,4596 + 3,5528) x 100 = 64,516%
QJx => is 3,5528 / (6,4596 + 3,5528) x 100 = 35,484%
Suprisingly, the third hipothesis shows that EAST probability in have a honor second is near 2 times x 1 time, the probability of have QJX.
Borel also suggest, for a more easy understanding, that we paint the two honors contiguous in green so that it will be higher of 10 and lower of king, whitout distinguishing each other. The 52 deck pack will now have 2 cards green, thus the a priori probability for EAST to have started with 2 green cards and one low card is 7,1056%, but the a priori probability for EAST to have started with only one green card and only one low card is 12,9192% (adding Qx% + Jx%), so the "a posteriori" probability is:
Green+low card...........= [12,9192 / (12,9192 + 7,1056)] x 100 = 64 ,516%
Green+green+low card= [ 7,1056 / (12,9192 +
7,1056)] x 100 = 35,484%
SITUATION-2) Suppose E-W has the following cards QJ10432 and after we
play a high honour (A/K) WEST discard the 2 and EAST the 10. What is now the
distribution probability of the remain 4 cards?
Example:
|
A987 N 2 W E 10 S K65 |
AK987 N 2 W E 10 S 65 |
To answer this question, in the same way as in SITUATION1, we need postulate a behavior for the E-W discard. Let's admit that the cards 4/3/2 are discarded in any way by E-W always before the Q/J/10; and with only Q10 E-W will play first the 10; and with only J10 or QJ E-W will serve any of this cards with the same probability.
Let's
considere now the a priori probability for:
singleton
10 =
1,2112% ; dubleton J10 or Q10 = 1,6149% ; tripleton QJ10 = 1,7764% and the "a
priori" probability composed for EAST to have:
- 10.... and serve the 10 = 1,2112% x
1 = 1,2112% ;
- Q10.. and serve the 10 = 1,6149% x
1 = 1,6149% ;
- J10.. and serve the 10 = 1,6149% x 1/2=
0,8075% ;
- QJ10 and serve the 10 = 1,7764% x 1/3= 0,5921% ;
- 10xx and serve the
10 =
1,7764% x 0 = 0,0 (hipothesis never
10).
____________________________ total = 4,2257%
So after EAST plays the 10, the "a posteriori" probabilities will be calculated using Bayes formula as follow:
singleton 10... [1,2112 /(1,2112 +1,6149+0,8075+0,5921)] x100 =28,26%
dubleton Q10. [1,6149 /(1,2112 +1,6149+0,8075+0,5921)] x100 =38,22%
dubleton J10.. [0,8075 /(1,2112 +1,6149+0,8075+0,5921)] x100 =19,10%
tripleton QJ10 [0,5921 /(1,2112 +1,6149+0,8075+0,5921)] x100 =14,01%
the follow table for QJ10 with the "a posteriori" probability summarizes the drop of the 10 using the hipothesis for QJ10432 cards.
|
QJ10432 |
singleton
honor |
dubleton Q10 |
dubleton QJ |
dubleton J10 |
tripleton QJ10 |
|
served 10 |
28,66% |
38,22% |
------- |
19,11% |
14,01% |
|
served J |
35,43% |
-------- |
23,52% |
23,62% |
17,33% |
|
served Q |
46,39% |
-------- |
30,93% |
------- |
22,68% |
Corollary: When opponents have these above residues, the drop of the 10 after the play of the Ace indicates a high chance of singleton 10 or Q10, rarely QJ10, and the drop of an honor, Q or J, signals high possibility of a singleton honor, despite the 5-1 distribution have low chance in occuring.
SITUATION-3) A common and practical case for analyze the
principle of Restrited Choise occurs in the follow configuration:
............NORTH (dummy)
..............QJ9
.........======
WEST ! !
EAST
? ! !
?
.........
======
.............SOUTH
..............432
SOUTH needs to make one trick in this suit and has all communication. Suppose
that the "2" was played and West plays a low card and dummy the Q that is covered by
a high honor (A or K) in EAST. Let's say K. After another round the "3" was played and
West again plays low card. What should dummy play?
________________________ WEST______EAST
Jack will make the trick if
A x
K10
9 will be the correct play if 10
x
A K
What is the
better card to be played: J or 9?
Declarer must decide and
as we know
the Restricted Choise principle says that East covers the Queen with the K
because he is restricted in use of the King, having K10, but if he has AK then it
could cover the Queen 50% with A or 50% with K.
So, if SOUTH plays 200 times in this same configuration of cards, let's suppose
that in 100 hands
that EAST have K10, he will make the trick with the K and in 100 hands that EAST
has AK he will make the trick 50 times with the King and 50 times with the
Ace. Applying what was explained before by Prof. Borel we can conclude that the
play of Jack is
superior than the play of 9 in 2 x 1 times.
In other worlds if AK are green cards then will be 2 configurations that EAST
has a green card with a low card and only one configuration that EAST
have both green cards.
**************************************************************
In 1958 an application of Restricted Choise was missed by the U. S. team against
the Italian team in the Bermuda Bowl. The contract was 3nt and the lead in both tables was little spades:
x
Qxxx
AKxx
Contract by South: 3nt
KJxx
A8xx N
KJ9xx
lead by WEST
x
J10xx O E
xx
xxx S xx
xx
Q10xx
Q10x
AKx
QJxx
Axx
In the table where Italian was declarer, the U.S. player in EAST make the
K and return a little
x and the Italian played
10 covered by the Ace in WEST, so the contract
was made, but in the table where the American was declarer
Q was played and the contract finish in down 1.
Was the option of the 10 or the Q equal and the Italian just was lucky?
The answer should be given by the Restricted Choise that, of course, was not used
by the american but was used by the Italian player. The Italian played for A
and K divided that is 66% against AK in the same hand that is 33%. The American
played for AK in the same hand was a guess that is 33%.
In our days no one makes any more this mistake in a World Champion and if it
happens with success that will be considered higher suspected!
Corollary: if you are playing the Regional and a similiar contract occurs and
you have
AK742 against 3nt and
your partner leads low spades with singleton on table. What you should do?
Suggestion: play the Ace and return little spades, your partner could have
J9xx and declarer
Q10x. Good players do
not bid 3nt with only Qx.
*************************************************************
Anothers examples to use the
Restricted Choise are:
a)
AJ10987 – 432 after your finesse of jack lose for K or Q in EAST, a new
finesse must be done, because percentage of 3 - 1 distribution was upgraded
considering that if EAST has KQ he will cover the jack 50% with the K and 50%
with the Q:
3-1=6,218% and the "a posteriori" composed 6,218% x 1 =6,218%
2-2=6,783% and tbe "a posteriori" composed 6,786% x ½=3,391%
So the
probabilities "a posteriori" adjusted by Bayes formula:
3-1 => [ 6,218 / ( 6,218 + 3,391 ) ] x 100 = 64,7%
2-2 => [ 3,391 / ( 6,218 + 3,391 ) ] x 100 = 35,3%
and the conclusition is - making the finesse again is 2
times higher than play for the
drop of the other honor.
Note: This
not apply if the hand is AQ10765-432. If the Queen finesse loses to the king the
distribution 2-2 maintains the same value because we cannot apply the Restricted
Choise for non contiguous cards like K-J. So, play for the drop of the jack in
this case.
The initial play of finesse of 10 do not apply when we have 9 cards in the suit,
unless there are information in the bidding that justify this finesse.
b) to make one trick in the configuration J94-Q32 the correct play is low to the Queen (because AK can be together), and if it loses to a high honor then you play for divide A K and so finesse the 9.
c) to make one trick in the configuration NORTH K109 – SOUTH 432 we should play a little card to the 10 and if EAST covers with J or Q we must apply the Restricted Choise playing for divided Q J and so make another finesse with the 9 that have a ratio of 2 x 1 of success.
d) In the
configuration NORTH K109876 – SOUTH A32 after the play of the Ace by South, if a honor (Q or
J) drop in EAST the finesse (play of the 10) is 66% against 33%.
e) a non
example - needing 4 tricks with the configuration A2 – K9865 after NORTH play
Ace, even with the drop of the 10 (or jack) in WEST we should not play for
singleton 10 because there are possible J10 or Q10 second in WEST and so we
should play the King playing for 3-3 or 4-2 and not for 5-1 that will not allow
us to make 4 tricks. So this is not a application of the Restrict Choise.
f)
a non example of Restricted Choise: In the same A2–K9865 configuration needing make 4 tricks, after
North plays Ace and WEST drops the Queen we cannot finesse based in the before
table QJ10, because despites singleton Queen having 46% of chance, the chances of QJ (31%)
and QJ10
(23%) results in 54% and besides that the play for 5-1 does not allow make 4
tricks. So we need play the King hoping for the drop of the
Jack or 10.
Corollary: When a singleton honor drops, if there
is no way to make the contract, with singleton honor, we must play for low
chance hoping find the other honor together, unless the contract is doubled and we
need minimize the losses.
g) In the
configuration AKQ9–432 after NORTH play the A and K, EAST serve the a little and
the 10. Now the hipothesis of J10x-xxx is lower than Jxxx-10x in the ratio of 2
x 1 (20 x 11 considering the vacant places adjust after WEST plays a card). So
we need finesse the Jack.
Corollary: When we have an option in play for the drop or
play for the finesse the finesse has a higher chance.
h) In the configuration AKQ8–432, after NORTH plays the Ace and the King, EAST plays two significatives cards from J/10/9. In this case we should make the finesse more quickly because the ratio is 3 x 1 favorable to finesse (in fact the ratio is 30 x 11 when we make the adjust considering the vacant places after WEST plays a third card).
i) In the
configuration 2-QJ87654 we need make 5 tricks. NORTH play 2 to the queen and
EAST plays 9 or 10 and WEST covers with a high honor (A or K)
The question is: what is more likely: 9 10 dubleton or 9 with the other high
honer?
The principle of Restrited Choise determines that the A10 or K10 has a chance
in a ratio of 2 x 1 than 9-10 together. So the correct play now is little card
and not the play of the jack to try cover the 10.
j)
Let's supose we have a suit with 10 cards missing KQ2 and after the play of the
Ace EAST serve the the Q and WEST the 2. Based on Restricted Choise principle
that says if EAST has the KQ it could play 50% the K and 50% the Q we can
assume that in a ratio of 2 x 1 the other honor is with WEST and then we can
make the play of the hand by count of the cards or to prepare a possible endplay
using this important information. This is a important
detail used by experts players that know Restricted Choise principle.
k) Similary with 10543-AK62 after the play of AK EAST has the Jx and WEST has xx, so who is more likely to have the Q. In a ratio of 2 x 1 WEST is more
likely to have the Q and you now can play the hand assuming the information to
cound the hand distribution with opponents or to prepare a endplay when is necessary.
Note: In many situations you play the trumps and one of the opponents discard a
card, there are a good chance that card is from a suit with 5 cards and that can
also be used for the count of the hand.
l) We need make 2 tricks with the configuration NORTH A1098 – 432 SOUTH and we don't have an entry in the dummy unless in this Ace, so we must play for a 3-3 division, but after we let two times opponents make the trick and EAST makes 2 tricks of K/Q/J. Now with this information and applying the Restriced Choise principle you have a ratio of 3 x 1 that EAST is dubleton. Thus the finesse is a must to do and not a continuation to play for 3-3.
m) In
configuration Q32–K98765 SOUTH play the "5" to NORTH Queen and WEST serve the Jack
(or 10) and EAST covers the Queen with Ace. After declarer is again the hand, he
goes to dummy and play the "2" and EAST serve the "4". The question is: Should
declarer play for the drop of the "10" or should make the finesse playing the "9"?
Based in Restricted Choise in a ratio of 2 x1 the finesse has probabilitu of 66%
against 33% of the drop.
Corollary: the chance of two contiguous cards are together
in a set of 4+ cards is lower, so the higher distribution is 3+ and 1 when a
contiguous card drop in first round.
n) Extending the use of Restricted Choise principle in the case of a honor drop
in the following situations bellow:
The question is you need make 3 tricks in those configurations and after the
play of the King a honor (Q or J) drop in EAST. In what cases can we use the
Restriced Choises principle?
|
a) A1065432 N x W E H S K7 |
b) A105432 N x W E H S K7 |
c) A10432 N x W E H S K7 |
d) A1032 N x W E H S K7 |
e) A102 N x W E H S K7 |
Our answer
is: in all this cases we can use the mathematical considerations of the Restricted Choise
when King is played and an honor drop in EAST, the probability of finesse is
better than play for the drop. BELIEVE OR NOT.
Note: Here we are not considering the joke of a player with QJ98 that discard
the Q because in this case there are no solutions, so we are only analysing possible cases where the H (Q or J) is singleton or dubleton (QJ).
To examplify let's prove mathematically the caso (e)
A102
After the play of the King drop a honor (Q/J) in EAST
N
the problem have solution in 2 cases:
x W E H
1) WEST has honor 7th (H xxx xxx) and EAST has
singleton
S
2) WEST has 6 low cards (xxx xxx) and EAST has 2 honors
K7
For a residual 8 cards the distribution 6-2 or 2-6 has 17,14% of
chance and occurs in 56 diferent configurations each one equal to 0,306%, and
the distribution 7-1 or 1-7 have 2,86% of chance and occurs in 16 diferent
configurations each equal to 0,178%.
Thus the "a priori" distribution for 6-2 is higher than the 7-1 configuration
but when we use the logic of Restricted Choise principle in the 6-2 distribution applying the
behavior of 50%
of times defender play Q and 50% times defender play J the calculation
adjusted by Bayes formula show us that the composed probability is:
6 - 2 (two honors) is 0,306 "a priori" and composed is 0,306 x ½ = 0,153
7 - 1 (singleton honor) is 0,178 a priori and composed is 0,178 x
1 = 0,178
Thus the
"a posteriori" probabilty after the event of EAST drop an honor is:
honor singleton => [ 0,178 / (0,178 + 0,153) ] x 100% = 53,77%
honor dubleton => [ 0,153 / (0,178 + 0,153) ] x 100% = 46,22%
So in the case (e) we have a ratio of 53,7% to finesse against 46,2% to play for
the drop.
IMPORTANT CONCLUSION: this technique of Restriced Choise used for having
the best line of play is not all valid when we have competitive bidding where
one opponent shows many cards in a suit. In this case we should use the vacant
places technique to count the percentagens in the play of the hand, or to make
an adjust in both techniques. This is explained in Kelsey-Glauerts" book.
/ / / that is all folks / /
/