Questions and Answers

Question 13

I am a mathematician, which impels me to express your method of trick estimation as compactly as possible. Two contributors to your website have already independently discovered the pieces of what I'm about to say. You replied that their methods were fine, but explained why you still preferred the original formulation. I hope you find some merit in the 'complete compactification' of your method, your preferences notwithstanding. Perhaps you will even present it somewhere in your future writings as a worthwhile alternative. Here it is:

THE WIRGREN ESTIMATE of the number of tricks declarer will take (given an adequate trump fit) is the sum of two numbers. The first number is WP/3, which must be 'rounded up' if it is not a whole number. The second number is the number of 'shortness tricks.' For this second number, count 3 tricks for each nonduplicated void, 2 for singletons, 1 for doubletons.

# tricks = [WP/3] + ST

Explanation: The square brackets are a reminder to 'round up' after dividing WP by 3, and ST is the number of shortness tricks.
   Notice how short this is, once WP is understood. Tables like those on pages 139, 146, and 149-152 are not needed. Also, there is no place in this description for subtracting from 13, or for any arithmetic except dividing WP by 3. Next, let's mention SST. I believe that the concept of SST is more complicated (especially with 3 short suits) and less intuitive than the concept of shortness tricks. Although I am happy to eliminate the SST concept, I guess that you justifiably have the opposite feeling. First of all, SST led you to your exciting breakthrough. Secondly, SST has already started to enter the world's bridge lexicon.

Best regards,
Louis Brickman


Thank you for your short and beautiful formula.
   A short comment: the word 'nonduplicated' refers to doubletons and singletons also. Then, a doubleton opposite a singleton or void counts for nothing, just like a singleton opposite a void.

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