Power of Distribution
Most important factor when you try to estimate how many tricks your side can take is distribution. Actually, when Jean-Rene Vernes published his famous article "The Law of Total Tricks" in The Bridge World in 1969, he wrote these wise words:
The Law of Total Tricks was developed with this in mind. Vernes' answer to the important question "How do we value the distribution ?" was "by counting the two sides' trumps." But, as we have shown in our book and on this site, that wasn't the way to go. Knowing how many trumps the two sides have does not tell us how well our trumps will work for us. Therefore, our answer "look at the short suits !" is the correct one. If you haven't been convinced yet, you will be eventually.
To show the power of distribution, we will start by looking at a simple example:
 A K 8 7
|
| Q J 10 9
|
 A K 2
| 8 7 6
| |
 A 8 7
| K 3 2
| |
 4 3 2
| A 7 6
|
Deal No. 1
East-West have 28 HCP, which all are working, a solid 8-card trump suit
in , and yet they can't make 4,
unless they can engineer some sort of endplay.
We have all been told that 26 Pts is enough for a game in a major
and here 28 Pts isn't. Yes, they can make 3NT, but for the moment,
let's concentrate on the 4-4 fit.
The reason why East-West take only 9 tricks in is that they have
the worst possible distribution, which is reflected in the highest possible SST:
6. Our formula says 28-30 WP should produce three more tricks than the SST
suggests, but here 28 WP is not enough, because of having too many
Aces. Since an average trick is won by 3 WP, if too many of
your tricks are won with Aces, you "waste" 1 WP per trick,
therefore you make bad use of your points. If you click on the link Working
Points, you can read more on this "paradox." And if we change
East-West's honors, so that their 28 WP is a mix of Aces, Kings and Queens, they
will take one more trick:
| A K 8 7
|
| Q J 10 9
|
| A K Q
| 8 7 6
| |
| A 8 7
| K Q 3
| |
| 4 3 2
| 8 7 6
|
Once more, East-West have 28 WP, but now they take 10 tricks because, instead of using 4 WP to win one trick with
A,
they use 4 WP to win two tricks (Q and Q). Since all East-West's tricks
outside of trumps are won with honors, it is surely one trick better to
have 6 such honors than 5.
| A K 8 7
|
| Q J 10 9
|
| A K Q
| 8 7 6
| |
| A 8 7 6
| K Q 3
| |
| 3 2
| 8 7 6
|
In the above deal (No. 2) we have given West a diamond instead of a club, a change which is worth exactly one trick for them. That is reflected in the SST, which goes down from 6 to 5. And the trick count goes up from 10 to 11. If we move one more club, we add one more trick to the total:
| A K 8 7
|
| Q J 10 9
|
| A K Q 4
| 8 7 6
| |
| A 8 7 6
| K Q 3
| |
| 2
| 8 7 6
|
Now, the SST is 4, and the difference between Deals No. 3 and No. 4 is exactly one trick: East-West take 12 tricks, one more than in Deal No. 3.
| A K 8 7
|
| Q J 10 9
|
| A K Q 4 3
| 8 7 6
| |
| A 8 7 6
| K Q 3
| |
| —
| 8 7 6
|
With the SST going down from 4 to 3, we add another trick. Cutting your SST by one without reducing your WP at the same time, is equivalent to adding another 3 WP without changing your SST.
When Charles Goren popularized Milton Work's method of valuing honors, he
made it easier for ordinary people to value their cards.
Our guess is that
Milton Work chose the values (4 for an Ace, 3 for a King, etc) for simplicity,
and experience has shown that it was pretty accurate.
But when Goren also
introduced a scale for distribution (3 for a Void, 2 for a Singleton and 1 for a
Doubleton) and combined it with point count, he made an error — because the two
scales are not compatible. According to Goren, a doubleton is equal to a
Jack; a Singleton is equal to a Queen and a Void is equal to a King. As these
examples show, that is way too little.
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