Problem no. 32, 2008

Färgglad linje

The problem

Recently I reread some old classics: Ely Culbertson's Blue Book and Golden Book plus Charles Goren's The Standard Book of Bidding. Culbertson advocated a method of valuation, in which you counted honor tricks according to the scale below. Goren followed Culbertson in his teachings for many years, until he gave it up and popularized Milton Work's old 4-3-2-1-scale, which today is universally used.

A-K2 tricks
A-Q1 1/2 tricks
A1 trick
K-Q1 trick
K-x1/2 trick

What most people don't know, is that you counted honor tricks not to estimate how many tricks you could expect to win in your own contract, but to estimate a hand's defensive value: how many tricks you could expect to win if the opponents played a trump contract. When you bid in order to find your own side's best contract, you reverted to counting tricks with honors, ruffing tricks and long-suit tricks.

Culbertson's assumption when he counted honor tricks was that a third round of a side-suit rarely won a trick if there was a trump suit. Therefore, a side-suit like A-K-Q-x-x was marginally better defensively than A-K-x-x-x. Both were valued to 2 honor tricks.

What Culbertson wrote makes much sense, and the fact that the same scale is found in a much newer book, Marty Bergen's Hand Evaluation, but under the name of quick tricks means that the reasoning has stood the test of time.

In spite of this, I have an objection to the scale above. I think there is (at least) one error in it.


The most obvious error in the table is the last line, the one for a king with no higher honors. Traditionally, a lone king is valued to half a trick, since "the ace is as likely to be over the king as under it". But that reasoning forgets that the ace may also be with your partner.

When you look at your thirteen cards only, and have no auction to help you, the ace can be in any of three places. In order for our king to win a trick, the ace should be either opposite or to our right. That is two cases out of three. So, to value the king to 1/2 trick is not correct. It should have been 2/3 tricks instead. And if we add the possibility that the king wins a trick even though the ace is over it (say, when partner had J-x-(x) and the queen is to your right), estimating K-x to be 2/3 tricks is rather too low than too high.

Conversely, you may think that A-Q-x should be more than 1 1/2 tricks; the queen isn't worth anything only when the king is to our left. But it that suit is a side-suit and we agree to Culbertson's assumption that a queen in a side-suit seldom wins a trick in a suit contract, 1 1/2 tricks is instead too much. Only when the king is to our right (1/3), will our queen win a trick. If partner has the king, we have two tricks in the suit even without the queen (if that is the case, our queen is useless).

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