Questions and Answers


Question 6

I just finished reading I Fought the Law of Total Tricks and loved it. I believe it will improve my competetive bidding.
   I have found it easier to work with Odd Tricks as opposed to tricks we can take, where OT = WP/3 - SST. This gives exactly the same answer provided that you remember to always round up after dividing by 3. OT = Odd tricks or Tricks - 6.
   So in fact I calculate OT = (WP + Adj) / 3 - SST. First adding either 0, 1, or 2 so that the adjusted WP is exactly divisible by 3. I find this easier and faster than trying to remember the plus or minus adjustment tables. Anyway it seems to work better for me.

Example: with 16 WP and 3 SST. (16+2) / 3 = (6-3) = 3 OT, or we should be able to make a 3-bid. Or if you prefer: 16/3 = 5 1/3 rounded up to 6. Less 3 SST = 3 OT.

THX for a Great Book
Ben Hooyer


Answer

Once you realize that how many tricks one side can take is a function of their distribution and working honors, you are on the right track. All roads lead to Rome, and it is possible to come to the right conclusion in many ways. Your suggested method sounds excellent to us, and if you prefer it to ours, by all means go ahead and use it.
   The base for our method is "If we have half the deck in working points, we will lose as many tricks as our SST. For every full 3 WP more than average, we deduct one loser; for every full 3 WP less than average, we add one loser". In your example (16 WP and an SST of 3) we would think "16 WP is a king below average, so we will lose one more trick than our SST. Four losers = nine tricks". We find this simple.
   If you take a look at our next question (No. 7), you will find another proposed method of counting your tricks. That one is excellent too.


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