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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:

North

♠

KJ32

♥

KJ5

♦

Q87

♣

Q54

South

♠

A10654

♥

A876

♦

932

♣

6

Each missing queen in our long suits means, essentially, another half-a-loser. Likewise, the queens in their suits are worth half-a-defensive trick. When the ♦A and ♦K are onside, our ♦Q may take a trick for us on offense, so we can add back in ¼ of a trick or so. Still, on this layout, the total trick estimate has dropped to approximately 16 ¾.

Complications arise, as usual.

Qxx

facing

xx

is worthless on offense, but will take a trick half the time on defense. Yet, make this

Qxx

Kx

and the result is completely neutral – one trick on offense and one trick on defense. So, deducting for a short suit queen sometimes overstates the result. In general, though, we should subtract ¼ to ½ a trick from the total trick estimate for each missing queen, and for each queen held in our short suits.

The more kings, queens, jacks, tens, and nines we swap around, the more the total trick estimate slips. Our simulations came up with SF+2 as a starting guide to estimating total tricks, but you can start with the very pure SF+3 and work downward, based on your hand. If partner shows the majors, and you hold

♠KQ86 ♥KJ5 ♦865 ♣1083,

bid more aggressively than holding

♠K863 ♥KJ5 ♦865 ♣Q108.

That is pretty obvious advice. Likewise, the trick estimate so far has been based on the expected number of defensive winners likely to cash. You can’t know their short suit distribution, but you can see you own. ♠AQxxx ♥AQxx ♦xxx ♣x is better than ♠AQxxx ♥AQxx ♦xx ♣xx. When SF=17, say, we expect, on average, three short suit losers. You can look at your own short-suit holdings and adjust accordingly.

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