The concept of Working Points (WP) is an important part of the formula presented in the Lawrence/Wirgren book I Fought The Law of Total Tricks. If we had concentrated solely on distribution, we would have come up with something just as vague and inaccurate as the Law of Total Tricks, but with the inclusion of Working Points, hand evaluation becomes much more clear. Incidently, these principles are an extension of what we all learned when we started playing bridge: that we should value our honors and our distribution.
What exactly do we mean by working points? Or, rather, which honors are said to be working? If you have, say, 9 HCP in spades, hearts and diamonds, and you know your partner has at most one club, does that mean you have 9 WP?
The correct answer is "maybe". Quite often you will indeed have 9 WP, as you may expect, but it is also possible that you have less than that – or even more than that. To understand why, let's start from the beginning with a trivial combination like this one:
|A K Q||7 6 4|
East-West have three spades in both hands, including the ace, the king and the queen. All these honors take tricks, which means they are all working. As you see, one trick is taken by an ace (4 HCP), another by a king (3 HCP) and a third by a queen (2 HCP). On average, a trick taken by a working honor is done with 3 HCP. That is also why an extra 3 WP always translates into one extra trick.
|A K 3||7 6 4|
If we remove the queen, we remove 2 HCP. But since the remaining 7 HCP represent two tricks, those HCP are working. If we remember that a trick uses 3 WP, we can see that East-West in this example use a little "too much" to take two tricks; their 7 WP don't take 2.33 tricks, only 2 tricks.
|A Q 3||7 6 4|
What about this one? How many WP do East-West have here? The answer is that it depends on where the king of spades is. If North has it, the queen isn't worth anything; but if South has it, it is. In the first case East-West have 4 WP (and take one trick), in the second they have 6 WP (and take two tricks). When both honors are working, the 6 HCP we assign them is right on target.
|K Q 3||7 6 4|
Once again the value of this combination is determined by where a key card is. In the previous example it was the king of spades. Now, it is the ace. Depending on where the ace is, East-West take one or two tricks. When the ace is offside, it's useful to think of the lower honor as a means to promote the higher one to a trick, i.e. the queen is not working while the king is. Here, East-West have either 5 WP or 3 WP.
Compared to the two previous situations you note that when the ace is onside, East-West make excellent use of their honors. With 5 WP they take two tricks, which is a third of a trick more than the expected 1.67 tricks.
|A 8 3||7 6 4|
When we have only one honor in the suit, the reasoning will be the same. When we have the ace, we take one trick. Assigning the lonely ace 4 HCP or 4 WP was therefore slightly too much, since it only took one trick (not 1.33 tricks). Its correct value would be 3 points.
|K 8 3||7 6 4|
When we have the king, as in this example, we need the ace to be onside. If it is, we have 3 WP, if it isn't we have 0 WP. Counting the king as 3 HCP or 3 WP will be way too much half of the time, exactly right the other half. The king's average value when it is accompanied by no other honors in the combined hands is therefore 1.5 points.
|Q 8 3||7 6 4|
When our honor is the queen, we need both the ace and the king to be onside. That happens roughly 25% of the time. When it does, the queen is valued slightly too low (you take one trick with 2 WP, not 0.67), when it doesn't, the queen is overvalued by 2 points. Since it is worth one trick one fourth of the time, and nothing in the remaining cases, the queen's average value when she is alone is 0.75 points.
Together or separated?
When East-West had two or three honors in the suit in all previous examples the honors were in the same hand. What if they weren't?
If we start with three honors, it's clear that nothing changes if the honors are split between the hands. With ace-king-queen in the suit you will take three tricks. And the same is true for the best two honors case: When you have the ace and the king, they are worth two tricks no matter how they are distributed.
|A 8 3||Q 6 4|
When you have the ace in one hand and the queen in the other, the situation is similar to the one where you had ace-queen third opposite three low. The only difference is that now the king will be favorably placed for East-West if North has it. Then, this combination is worth 6 WP, and two tricks. Otherwise, it is 4 WP and one trick only.
|Q 8 3||K 6 4|
When you have the king in one hand and the queen in the other, things are worse for your side. Unless one of the defenders has the ace singleton or doubleton, and you can figure out which, your two honors will only take one trick. When that happens, we use the same reasoning as in example 4 above: the lower honor, the queen, is used to promote the higher one, the king. So when we take one trick we have 3 WP in the suit.
|Q J 3||K 6 4|
When we add the jack, we add almost a full trick. And now it's clear that we have 6 WP in the suit, equalling two tricks.
|Q 10 3||K 6 4|
Even the ten makes things rosier than when the defenders have it (as in example 9). If we lead towards the king, later finessing the ten, we will take two tricks slightly more than half of the time. But how many WP do we have when that happens?
Suppose the king is taken by the ace, and you later take a successful finesse with the ten. Then you take one trick with a queen and one trick with a ten. But to say that you had 2 WP in spades is obviously wrong, given that an average trick is 3 WP. Instead, we suggest you think along these lines: "With the ten lying over the defense's jack, West's spade holding can be said to be Q-J-3, and just like in example 10, we say that the lower honor is used to promote the higher honors. Therefore, East-West have 6 WP in spades, even though they only had 5 HCP in the suit.