Join Bridge Winners
Part VI - Moving Honors Around
(Page of 5)

We will jump right in and return to my first example hand discussed in Part II of the Theory of Total Tricks:

North
KJ32
KJ5
1087
954
South
AQ654
AQ87
932
6

We have nine winners, and can expect to cash 3½ tricks on defense, on average, so this hand yields 18½ total tricks.  Now let’s remove an ace, to give, say

North
KJ32
KJ5
1087
954
South
AQ654
Q876
932
6

Now, we can only win eight tricks, most of the time.  Occasionally, hearts split 5-1 and this missing ace allows for several ruffs, so our winner count actually drops by a little more than one.  What about our defensive prospects?  Obviously, we have one less potential winner.  The missing ace also costs a tempo.  Sometimes, the opponents will draw trumps and throw heart losers on the diamonds, so their expected trick total increases by more than one.  This is complicated, but the expected trick total is still close to 18½, but may be a little higher. 

Suppose we swap aces? 

 

North
KJ32
KJ5
1087
A54
South
AQ654
Q876
932
6

This evens out the tempo issue, and returns us to around 18½ total tricks.  Swapping around aces doesn’t change our basic total trick analysis by very much.  To put this another way, aces are pure cards, and moving aces around doesn’t detract from the purity of the hand by much.

 Now, let’s lose a king:

North
J1092
KJ5
1087
954
South
AQ654
AQ87
932
6

We are now on a finesse for three spades – the missing king costs us half-a-winner.  What happens on defense?  If the finesse is onside, we still have 3½ expected defensive winners.  When the finesse is offside, and spades are 3-1 or 4-0, the king won’t help them.  The losing finesse costs us a full trick, but is only worth a trick in a club contract when spades are 2-2, 40% of the time.  Thus, the expected total number of tricks drops by .3, to 18.2. 

Here is another interesting example:

North
J932
KJ5
1087
954
South
AQ874
AQ87
932
6

Double dummy, this won’t be any different[1], but, at the table, we will lead a low trump to the queen, and lose a trump trick when East has K10x or K10xx.  Missing both the king and ten of spades drops the total trick estimate to around 18. 



[1] This is one of the big flaws of double-dummy simulations that I and others have used to approach this problem.

 What if we swap kings:

North
J1092
KJ5
1087
K54
South
AQ654
AQ87
932
6

The K is worthless on offense, but half-a-trick on defense, so the trick total falls to 17.7.  Or, how about:

North
J1092
KJ5
K87
954
South
AQ654
AQ87
932
6

Here, the K is mostly neutral, but, if both finesses are onside, it gains us a tempo, and we will take eleven tricks.  This king actually increases the trick total slightly, to 18.4. 

Again, real life is complicated, but, as a general rule of thumb,

          Each king moved costs around ¼ of a trick from the total trick estimate.

Obviously, kings facing shortness are very poor, and cost nearly half-a-winner, while kings facing length are mostly neutral.  Likewise, intermediates facing shortness are purely defensive.  Let’s say the opposition is competing in clubs where you hold K32.  Subtract ¼ to ½ a trick from your total trick estimate.  Change this to K102, and you should subtract ¾ to a full trick.

Now, let’s consider what happens when queens are moved about.  Try this layout: