Durango Bill's
Bridge Probabilities and Combinatorics



How to Calculate Bridge Suit Split Combinatorics/Probability

If you and dummy hold "N" number of cards, what is the probability of various possible splits?

In particular, what is the basis for the "With 8 ever, with 9 never" rule?

We will cover 3 topics:
Suit Splits in General
"With 8 Ever"
"With 9 Never"

Suit Splits in General

   For all of the following combinations, we note that the opponents hold 26 cards. One opponent will hold 13 of these while the other opponent will hold the remaining 13. Thus, the number of combinations for any one opponent is COMBIN(26, 13) = 10,400,600. In all cases, once we define what one opponent has, we do not have to make calculations for what the other opponent has. He simply gets what is left over. With 3 hands known (yours, dummy's, and one opponent) the other opponent's hand is explicitly known, hence his "combinatorics" is always equal to 1.000.

   All of these combinatoric calculations are based on distributions resulting from a random deal. As soon as knowledge is gained from bidding or play of a hand, these calculations are no longer valid. If an opponent bids, the hand is no longer random. We will give an example showing how this changes random distribution.

   When we calculate the probability of any particular possible split of some suit, there are two COMBIN() functions that are used. The first COMBIN() is used for the suit in question while the second COMBIN() is used for everything else that is left over.

   The general equation given that "N" cards are outstanding in a particular suit and you are calculating how many ways "K" of these might be in a particular opponent's hand (Choose either left opponent or right opponent) is:

Total combinations = COMBIN(N, K) * COMBIN(26 - N, 13 - K)

   For example: Assume there are 5 cards outstanding in a suit and you want to calculate how many ways any two of these could be in your left opponent's hand. (Note: the "26" represents the number of cards left in the deck after you and dummy have 13 each. The 13 is the total number of cards in each Bridge hand.)

First we have N = 5 (There are 5 cards outstanding in the suit of question.) and K = 2. (How many different combinations are there such that your left hand opponent has any 2 of them.)

The calculation becomes:
Total combinations = COMBIN(5, 2) * COMBIN(21, 11) = 10 * 352,716 = 3,527,160

(Part 1) For this particular distribution the opponent can have any 2 of the 5 cards outstanding in a suit. Then
(Part 2), for each of these 10 combinations, he can have any 11 of the 21 remaining cards that are in the other 3 suits.

Now let's calculate all the combinatorics given that you and dummy have 11 cards in a suit.

"Split"
  2,0    Your left opponent has both missing cards (right opponent has none)
  1,1    Each opponent has one card
  0,2    Your left opponent is void in the suit, right opponent has both

For the left opponent:

For  "2,0" we have COMBIN(2,2) * COMBIN(24,11) = 2,496,144 combinations
For  "1,1" we have COMBIN(2,1) * COMBIN(24,12) = 5,408,312 combinations
For  "0,2" we have COMBIN(2,0) * COMBIN(24,13) = 2,496,144 combinations

We note that the sum of the number of combinations equals the 10,400,600 total that we calculated earlier (COMBIN(26,13)).

To get the probability of any particular split we just divide the number of combinations for the split by the total number of hands. Thus:

Split             Probability
2,0      2,496,144 / 10,400,600 = .2400
1,1      5,408,312 / 10,400,600 = .5200
0,2      2,496,144 / 10,400,600 = .2400

   Assume the King of the suit is outstanding. If you play the Ace (either directly from your hand or lead to it), you will drop the missing King 52% of the time. If you try the finesse, the suit will split 2,0 favorably 24% of the time, plus the 1,1 split (prob. = .52) will have the King in front of the Ace one half of the time. Thus the finesse will work .24 + 1/2*.52 = .50 or 50% of the time. Thus, the play for the drop has a slightly better chance of working, but it is close.

   Now we will try a slightly more complicated problem. Dummy has put down his hand, and you and dummy have 11 cards in your long suit, but are missing the king. However, during the bidding, the opponent to your left bid 1 of some other suit and the opponent to your right gave a simple raise. Also assume that you and dummy have a total of 5 cards in the opponent's suit. Thus the opponents have 8 cards in their suit. You have a choice of leading the Ace (in your hand) of your long suit, or you could try a finesse. Which is the better strategy?

   First we assume that the opponent to the left has 5 cards in his long suit. We also assume the missing king could be in either hand. The left opponent has more strength, but this could be distributional strength.

The number of ways your suit could split 2,0 (left opponent has both missing cards in your suit) is:

   Your                        Their                    What is
   Suit             times      Suit         times    Left over
COMBIN(2,2) * COMBIN(8,5) * COMBIN(16,6) = 448,448 total combinations

Translated: This calculation involves the number of ways the left opponent could have both of the 2 outstanding cards in your suit times the number of ways he could have any 5 of the 8 outstanding cards in their suit times the number of ways he could have any 6 (his hand must have a total of 13 cards) from the 16 remaining cards in the other 2 suits. (Note: You and dummy have 11 + 5 = 16 cards in the first two suits. Thus, the combined hands (declarer and dummy) will also have 10 cards in the two remaining suits. This leaves 26 - 10 = 16 cards in these other two suits for the opponents.)

Similarly for 1,1 and 0,2 splits we have:
For "1,1":  COMBIN(2,1) * COMBIN(8,5) * COMBIN(16,7) = 1,281,280
For "0,2":  COMBIN(2,0) * COMBIN(8,5) * COMBIN(16,8) =   720,720

   If we add these together, we get 448,448, + 1,281,280 + 720,720 = 2,450,448 total hands. This is less than the 10,400,600 we got before because we reduced the number of combinations when we required the left opponent to hold 5 cards in the opponent's suit. (But it does equal COMBIN(8,5) * COMBIN(18,8) = 56 * 43,758 = 2,450,448). When we divide each of the three possibilities by 2,450,448 we get the following:

Split                Probability
"2,0"        448,448 / 2,450,448  =  .1830
"1,1"     1,281,280 / 2,450,448  =  .5229
"0,2"        720,720 / 2,450,448  =  .2941

(Slight rounding errors may be present for all calculations.)

Thus, if your long suit is going to split unevenly, the opponent to your right (He has only 3 cards of the opponent's long suit) is more likely to have both of the missing cards.

If you play the Ace of your suit (play for the drop), it will drop the missing King 52.29 % of the time.

If you lead to your Ace through the opponent that is shorter in their suit (try the finesse), it will work all the time the cards split "0,2" and half the time if they split "1,1". This gives a success rate of: .2941 + (1/2 * .5229) = .5555 or 55.55% of the time. Again it is close, but now the odds slightly favor the finesse.

Because both combinations are close, in practice, you might want to look at both opponents to see if either looks nervous.


With 8 ever, with 9 never

With 8 ever: (always finesse?)


   This rule of thumb is an (overly) simple way to remember if the best strategy is to finesse, or play for the drop when you are missing a Queen, and you and dummy have 8 or 9 cards of a suit. First we will look at the table, given that you have 8 cards of a suit. Again for each row, we divide the total number of combinations by the total ways an opponent's hand can be dealt COMBIN(26,13) = 10,400,600.

Split            Number of Combinations                     Probability
-----------------------------------------------------------------------
"5,0"  COMBIN(5,5)*COMBIN(21, 8) =   203,490    203490/10400600 = .0196
"4,1"  COMBIN(5,4)*COMBIN(21, 9) = 1,469,650   1469650/10400600 = .1413
"3,2"  COMBIN(5,3)*COMBIN(21,10) = 3,527,160   3527160/10400600 = .3391
"2,3"  COMBIN(5,2)*COMBIN(21,11) = 3,527,160   3527160/10400600 = .3391
"1,4"  COMBIN(5,1)*COMBIN(21,12) = 1,469,650   1469650/10400600 = .1413
"0,5"  COMBIN(5,0)*COMBIN(21,13) =   203,490    203490/10400600 = .0196


   Assuming you have entrees for any finesses, the best way to pursue the finesse strategy is to take the Ace (or King) and then try the finesse. The best way to pursue the drop strategy is to initially take the Ace (or King). If this reveals a favorable "5,0" split, then switch to the marked finesse, otherwise continue with your other high card.

   First, we will calculate the success rate using the finesse strategy. We will assume that you can finesse through the right hand opponent of the above "splits".

If the split is "5,0", the strategy fails.
If the split is "4,1", there is a (1/5)*.1413 = .0283 success probability the first high card will drop the singleton Queen (Else the finesse fails).
If the split is "3,2", there is a (4/10)*.3391 = .1357 success probability the Queen will drop on the 2nd round when you lead toward your high card.
If the split is "2,3" there is a (6/10)*.3391 = .2035 success probability the finesse will work and you capture the Queen when you lead your remaining high card on the next round
If the split is "1,4", you will capture the Queen on all combinations. Success probability is thus .1413.
If the split is "0,5", you will again capture the Queen. Add another .0196 to the success probability.

When we add all these results together, we find the success rate for the finesse strategy is: .0283 + .1357 + .2035 + .1413 + .0196 = .5283 (Numbers don't add exactly due to rounding error.)

Now, we will make a similar table for the success rate using the "play for the drop" strategy.

If the split is "5,0", the strategy fails.
If the split is "4,1", there is a (1/5)*.1413 = .0283 probability the first high card will drop the Queen (Else the strategy fails).
If the split is "3,2", there is a (4/10)*.3391 = .1357 success probability the Queen will drop on the second lead.
If the split is "2,3", there is a (4/10) *.3391 = .1357 success probability the Queen will drop on the second lead.
If the split is "1,4", there is a (1/5)*.1413 = .0283 success probability the first high card will drop the Queen (Else the strategy fails).
If the split is "0,5", the first high card will reveal the split and the marked finesse will add another .0196 to the success probability.

When we add all these results together, we find the success rate for the "play for the drop" strategy is .0283 + .1357 + .1357 + .0283 + .0196 = .3474  (Note: Components were rounded but the .3474 is accurate)

The finesse strategy has a clear advantage, hence "With 8 ever" is a valid rule.


With 9 never (never finesse?)

   Now let's do a similar set of calculations assuming you and dummy have 9 cards in a suit and are missing the Queen. Again there are COMBIN(26,13) = 10,400,600 possible hands an opponent could have. After we calculate the total possible hands for each split, we will divide by this number to get the probability.

Split           Number of Combinations                          Probability
---------------------------------------------------------------------------
"4,0"   COMBIN(4,4) * COMBIN(22, 9) =   497,420     497420/10400600 = .0478
"3,1"   COMBIN(4,3) * COMBIN(22,10) = 2,586,584    2586584/10400600 = .2487
"2,2"   COMBIN(4,2) * COMBIN(22,11) = 4,232,592    4232592/10400600 = .4070
"1,3"   COMBIN(4,1) * COMBIN(22,12) = 2,586,584    2586584/10400600 = .2487
"0,4"   COMBIN(4,0) * COMBIN(22,13) =   497,420     497420/10400600 = .0478


   Again assuming you have entrees for any finesses, the best way to pursue the finesse strategy is to take the Ace (or King) and then try the finesse. The best way to pursue the drop strategy is to initially take the Ace (or King). If this reveals a favorable "4,0" split, then switch to the marked finesse, otherwise continue with your other high card.

   First we will calculate the finesse success strategy. We will assume that you can finesse through the right hand opponent of the above "splits".

If the split is "4,0", the strategy fails.
If the split is "3,1", there is a (1/4) * .2487 = .0622 success probability the Queen will drop on the first high card. (Else the strategy fails.)
If the split is "2,2", there is a (1/2) * .4070 = .2035 success probability the Queen will drop when you start to take the finesse.
If the split is "1,3", you will always capture the Queen. Success probability = .2487
If the split is "0,4", you will always capture the Queen. Success probability = .0478

If we add these together we get: .0622 + .2035 + .2487 + .0478 = .5622
There is a probability of 56.22 % you will capture the Queen using the finesse strategy.

Now we will calculate what happens if you "play for the drop"

If the split is "4,0" the strategy fails.
If the split is "3,1", there is a (1/4) * .2487 = .0622 success probability the Queen will drop on the first high card.
If the split is "2,2", you will always capture the Queen. Success probability = .4070
If the split is "1,3", there is a (1/4) * .2487 = .0622 probability the Queen will drop on the first high card.
If the split is "0,4", you will discover the marked finesse and always capture the Queen. Success probability = .0478

If we add these together we get: .0622 + .4070 + .0622 + .0478 = .5792
There is a probability of 57.92 % you will capture the Queen using the "play for the drop" strategy.

The "play for the drop" strategy has a very slight advantage, but other factors such as not letting the dangerous opponent gain the lead are far more important.

Thus the maxim of "With 9 never" should be changed to "With 9 sometimes depending on where you would like the lead if it doesn't work"



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