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KQT9 opposite xxxx

The combination of KQT9 opposite xxxx occurred in the recent finals of the WBF junior online championship. In his book Expert Bridge Simplified, Jeff Rubens discusses this position and the play by a defender (p. 350.) He states that a competent defender behind the KQT9 will win his ace from ace-low "enough to make you pay" for playing for AJ doubleton. But Rubens does not answer the question, "How much is enough?"

Assuming sufficient entries and no further information regarding the deal (Those of you who want to argue that there is always additional relevant information, knock yourself out. I am just trying to lay down the basic premise as posed by Rubens.), a singleton A behind the KQT9 occurs approximately 2.8% of the time, AJ occurs 3.4%, and Ax occurs 10.2% of the time. Normalizing the percentages yields 17%, 21% and 62% respectively. So, if the deal is played 100 times and the defender always plays the singleton A and the A from AJ, and 5 times plays the A from Ax, the declarer will optimize his expected number of tricks by playing first round to an honor and second round to the ten or nine. As 5 out of 62 is about 8%, I believe that is the amount which is enough.

I am sure there is a more elegant way of expressing the problem and proving the answer but this is the way that makes sense to me. I am posting this because I have three questions. First, am I correct? Second, how does one at the table determine when to play the A from Ax to make the play 8% of the time? (Yes, the defender can win the A from Ax greater than 8% of the time (How much greater?) without altering declarer’s optimum strategy, but this sort of mixed strategy problem occurs in a number of contexts and, given that aids such as flipping coins or random number generators are not permitted, I have always been curious how one can actually employ a mixed strategy.) And third, based on your experience, what do you think your competent opponents actually do in practice?

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