In Bridge the knowledge of probabilities and chances in distribution of cards are fundamental to make decisions during the play of a hand.

After the cards have been distributed each crosshead, NS or EW, have a distribution of cards in each suit and, per example, the fit in one specific suit for NS have a complemet of cards for EW called residue.

In the play of a hand Declarer has the resideus information in the odds tables and the additional information of the bidding when opponents shows long suit or be shorter in a specific suit. Thus during hand's play the informations of residuos distribution change dynamically after each round of a trick is completed, but sometimes Declarer haven't the option to choose for DROP or FINESSE dynamically and so his decision must be take early based in the a priori table of the residuos.

Another commom situation in bridge occurs when North bids a long suit and E-W is playing a contract in Spades with AJ32 - K1098.
North    East     South West
 3      double    pass    3
pass    pass       pass 
North leads A and K and all positions serve 2 cards showing Diamonds distribution 7222. Then North plays low Hearts to AKQ on dummy.
If declarer have in Spades AJxx - K1098 who is now the most probable to have the Spades's Queen?
Of course if North shows 7 cards in Diamonds he has only 6 vacant places to have the Q while South have 11 vacant places to have the Q. In other words the chances for North is 6/17 (35,3%) and for South is 11/17 (64,7%).
Thus the information of the a prior table of probabilities are no more valid for our decision and now vacant places are the dominant information.

Example1 how to use this table:
Suppose you have a suit with AKQ10 in North and a singleton in South, so 5 cards in this suit implies a residue of 8 cards, and you want to know if is better: finesse the jack or play for the drop of the jack to make 4 tricks in this suit?
Of course you should add the chances of 8 cards resideus for the drop:
- singleton jack      =>       0,36%
- dubleton jack       => +   4,28%
- tripleton jack       => +  17,67%
  chances to drop    => = 22,31%
Considering that chances for finesse are 50% you must finesse.

Example2 how to use this table:
Suppose now you have AKQ1098 in North and singleton in South, 7 cards with a resideu of 6 cards, and you want to know what is better: finesse the jack or play for the drop of the jack?
Of course you should add the chances of 6 cards resideus for the drop:
- singleton jack     =>         2,42%
- dubleton jack      =>  +  16,15%
- tripleton jack      =>  +  35,53%
  chances to drop  =>  =  54,10%
Of course if the jack is fourth the chances for drop the jack after the finesse are half (54,10% + 32,30) => 86,40 / 2 = 43,20% to make 4 tricks because if jack is fifth you don't make all tricks in 6,05$. That means the play for drop ensures 54% to make all tricks against 37% making the finesse. So you should play for drop.

Exemple3 how to use this   
Now suppose you have AKQ109 in North and singleton in South, 6 cards with a resideu of 7 cards, and you want to know what is better: finesse the jack or play for the drop of the jack?
Make the count and you will have the answer.

==================================

Table to determine the residue repartition between opponents

Residue  cards	Distribution of residues	%	Quantity  ocurrences
2	1 - 1 2 - 0	52,00 48,00	2 2
3	2 - 1 3 - 0	78,00 22,00	6 2
4	3 - 1 2 - 2 4 - 0	49,74 40,70 09,57	8 6 2
5	3 - 2 4 - 1 5 - 0	67,83 28,26 03,91	20 10 2
6	4 - 2 3 - 3 5 - 1 6 - 0	48,45 35,53 14,53 01,49	30 20 12 2
7	4 - 3 5 - 2 6 - 1 7 - 0	62,17 30,52 06,78 00,52	70 42 14 2
8	5 - 3 4 - 4 6 - 2 7 - 1 8  -0	47,12 32,72 17,14 02,86 00.16	112 70 56 16 2
9	5 - 4 6 - 3 7 - 2 8 - 1 9 - 0	58,90 31,41 08.57 01,07 00,05	352 168 72 18 2

This table it is important to show the chances of residues distribution in oppoent cards independent of our distribution cards in this suit.
So if we have a fit of 8 cards with distribution 4-4 or 5-3 or 6-2 or 7-1 the probabilities of opponents residues are the same.

Loking the table we see that if the residue are 6 cards when we have a suit 5-2 or 4-3, the chances of the residue are 4-2 or 2-4 are great than the chances of 3-3 in a ratio of 48 to 36.

This means that if we are playing in a fit 4-3 we should assume that the residue are 4-2 and only play for it be 3-3 if there are no way to make if the residue are 4-2 or 2-4.

If our fit is 8 cards then the residue of 5 cards are near 68% of chances in be 3-2 or 2-3, so unless the bidding shows possible 4-1 distribution for the residue we should play for normal distribution in MP tournament and in IMPs we should be more careful in play for possible 4-1 distribution and make safety always it is possible.

Per example, if you are playing 6nt and have a favorable leas with jack Spades (J) that allow us make a safety play in the second trick playing low Clubs:

AKx    ===== xx         contract 6nt Lead J
Axxx  !    N    ! 10x
AKQJ  !W   E ! xxx
xx      !    S    ! AKQxxx
            =====

Of course in a tournament MP this is a hard decision but in a IMPs match this is fundamental to do because you will increase your chances of success from 68% to 96% (67,83%+28,26%);

Now lets change East hand to have only 5 cards Clubs. Now the chances of Clubs divided 3-3 are only 36% so independent of any lead we must upgrade the chances of make 4 tricks in Clubs playing low Clubs in this suit.

AKx   ===== xxx    contract 6nt lead J,
AKx   !    N    ! 10x
AKQx ! O   E  ! xxx.
xx      !    S    ! AKQxx
           =====

So with a low Clubs we upgrade our chances from 35,53% to 48,45 + 35,53 = 83,98%. Of course in a tournament some player could play Clubs for 3-3 but this type of player few times will won a tournament.

2) About the Restrited Choise and the orientation for finesse or drop

a) If you have a suit configuration A1092 - K8765 nine card fit missing the Queen and Jack and more 2 low cards. Let's say you play the 2 and opponent serve an honor (J or Q). You make the King and play low to the ace and the other opponent serve a low card.
But
Thus now you must make a decision Finesse or Drop?
We recommend to lead Restricted Choise here.
But the probability between H43 - H (singleton honor) against 43 - HH is 66% againt 33% de chance. So you should make the finesse, unless you have a more strong motive to not do the finesse, like need ruff another suit.

b) Now you have fit of 8 cards A1092 - K876 also missing Queen and Jack, and you play the 2 and opponent serve an honor and you make the King. You play low to the Ace and the other opponent serve low.
Thus againt you must make a decision Finesse more 2 times or play for a normal 3-2 division serving the Ace?
Again the probability between H543 - H (singleton honor) is 66% against 33% or the double chances for finesse than play for the drop!

Thus understanding the Restriced Choise Principle you will be more like to have success in the play than a play that just play for normal division 3-2.

3) About the Vacant Places and the play for Finesse or Drop

a) Example1
South is playing 4 where opponents only pass.
West leads J and dummy play the Ace and East serves the K
 976
 AQ32
 A2
 QJ76
======    Contract 6 
!     N     !    lead: J   NORTH plays A and EAST serves K !!!
! W    E  !
!     S     !
======
 AKJ1084
 4
 K3
 K1054

Now declarer play Hearts and ruffs while East discard low Diamonds.
Declarer play Ace Spades and both opponents serve low Spades.
Declarer plays Diamonds to Ace in dummy and play another Spades and East play again low.
What should Declarer do now? Play Finesse or for the Queen Drop?

Analyses: after the play of Ace Spades the a priori probability table recommend that the play for drop better in 52,5% against 47,5% but the after the drop of K we know that West has 7 cards in Hearts so West has only 6 vacant places to have the Q while East have 12 vacant places to have the Q. In other words West has 6/18 chances in have the Queen Spades while East have 12/18 chances in have the Queen Spades. This means that the probability of West is 33% and East is 66%. So forget the information of a priori table and use now the information of Vacant Places and make the finesse.  

Note: the rule for avaliation probabilities using vacant places demands that only suit with known distribution can be counted.

b) Example2 - Suppose E-W is playing 7 and the lead was trump where South discard low Diamonds.
KJ9         N     A108         
QJ987 W     E AK1032          
Axx         S     Kx                 
AK                  865            

trick01 - trump lead and South discard Diamonds - so North have 10 vacant places for have the Q and South have 13 vacant places;
trick02 - trump and South discard another Diamonds and a Clubs;
trick03-04 - AK Clubs are played and all serve;
trick05-06 - AK Diamonds are played and all serve;
trick07 - last Diamonds are ruffed and North discard Clubs;
So Diamonds was 6-2 and there are now an inversion in the vacant places because North shows 3 + 2 so has 8 vacant places to have the Q while south has 0 + 6 then only 7 vacant places to have Q.
The ratio 6 to 7 in percentagens is 46,15% agaubst 53,85 something that justify do not use a simple guess but take this value to make the finesse against as North the most probable player to have the Queen of Spades.

Thus when we find the player that have more cards in the suit where we the Queen is missing that player is the most inclinable to have the Queen.

So when we have AJx - K10x if we find that a specific player have 4 cards in this suit and the other have only 3 cards in this suit, the probabilities are 4/7 (57,14%) against 3/7 (42,86%). If a specific player has 5 cards and the other only 2 then the probabilities are 5/7 (71,43%) against 2/7 (28,57%). Even if during the play of the hand the opponent with 5 cards have discard 2 cards of this suit the ratio are unchanged.

c) Example3 - North are playing 7nt and East leads 10
KJ97     E      AQ108    bidding: N   E    S     W    
AK2   N     S  QJ3                      2   -    2    -     
AK3      W     QJ2                     2ST  -  7ST   -  
KJ9               A108        Lead by East 10 

In this twin distribution we must find a indication of the player that has more Clubs to be the candidate to have Q.
After Hearts lead Declarer play 3 rounds of Hearts and West discard a Clubs is the third Hearts round.
Then Declarer play 4 round of Spades and East shows 4 cards Spades while  West discard 2 Diamonds and 1 Clubs.
Then Declarer play now the Diamonds and East serves 2 diam and discard one Hearts. So East has only 2 Clubs.
The count is: East has 5 cards Hearts + 4 cards Spades + 2 cards Diamonds. Conclusion West has 5 cards Clubs.
Declarer must play West with the Q with 71% of probability.

/ / /