Source: [ilink url=»http://www.maths.usyd.edu.au/u/richardc/bridge.html»]http://www.maths.usyd.edu.au/u/richardc/bridge.html[/ilink]

** Honour values with balanced hands played in NT contracts: In 1987, after about 12 years of work (albeit only the odd few weeks here and there), I published the following paper: Cowan, R. **Hand evaluation in the game of Contract Bridge.

*J. Royal Statistical Society: Series C; Applied Statistics*

**36**58-71 (1987).

It has become common practice in the game of Bridge to evaluate the Aces, Kings, Queens and Jacks in a 13-card hand in proportions 4:3:2:1. This paper considers the trick-taking potential of combinations of these cards, with Tens also. On this basis we find optimal weightings for the top five cards. The optimal weightings are very close to the proportions 5:4:3:2:1.

The paper was reviewed in the US-based journal «Bridge World», in «Science News», in «The Australian» newspaper and in the Arts & Science pages of «The Economist». After publication, interesting correspondence on the topic came from many in the bridge scene, with major encouragement and constructive criticism from Jeff Rubens, Danny Kleinman, George Havas, David Askew, Alec Traub, Alan Truscott and Tim Seres.

My paper concluded that, if one must use an additive valuation scheme for the honours A, K, Q, J and 10, for the suits one encounters when you and your partner each have a balanced hand and play in NoTrumps, then the best weightings for the honours are not 4, 3, 2, 1, 0 (the basis of the High-Card Point count system that we all play).

The paper shows that the trick-generating potential of two balanced hands is best predicted by honour weightings which are approximately proportional to 5, 4, 3, 2 and 1. Some people find this hard to believe. Surely an Ace is not equal to a Q + J?

When both players have a balanced hand, 4-3-3-3 or 4-4-3-2, the suits they have to play for tricks are either 4-4, 4-3, 4-2, 3-3, 3-2 or 2-2. For example you might have to play the 4-3 combination Axxx/KJx where x is an insignificant card. For each suit combination, there is an optimal way of playing to maximise the expected number of tricks produced. In this example, we play the A, finesse against the Q, then K and the final x if the split has been 3-3. Overall the expected number of tricks from this play, an entity that I call the «trick-taking potential» (ttp) is 2.87 and intuitively this sets up an equation A + K + J = 2.87. I worked out this ttp for each suit combination and so set up a vast number of such equations in the 5 variables A, K, Q, J and T. As is standard procedure when there are more linear equations than variables, I found the best values for the honour cards by a weighted least squares analysis. The mathematical values given to the 5 variables A, K, Q, J and T (ie. the least-squares solution of the equations) were:

or (if A is scaled up to be 5) in proportions 5 : 3.97 : 3.06 : 1.93 : 0.95.

There are issues of control (for which aces are best, in the sense of immediacy) and these are discussed in the pdf document along with comments from other people. Also the finding does not apply to suit-contracts, except in regard to the honour valuations within a trump suit and any second suit which must be played for tricks. (Generally in suit-contracts the lower honours have little value, compared to aces and kings, in the suits that declarer ruffs or discards. Aces and kings also dominate in the trump suit if the fit is extreme, say 10+ trumps.)

If the subject has been of interest for you, read the comments: [ilink url=»http://csbnews.org/?p=45573″]Notas sobre la Evaluacion de la Mano por Richard Cowan[/ilink] October 1987. It’s Worth doing it…