if { ([west has QS] && [spades west] < 3) || ([east has QS] && [spades east] < 3) || ([east has QD] && [diamonds east] > 1)} {lineB add 1}
should be:
if { ([west has QS] && [spades west] < 3) || ([east has QS] && [spades east] < 3) || ([east has QD] && [diamonds east] > 0)} {lineB add 1}
The way it is coded, East must hold 2 or more diamonds, if I read it correctly, and that is not the case.
(Unless the language is smart enough to know that since East has to hold the diamond Q, he must hold at least 1 diamond).
I don't use this language so am not really sure.
finessing 3nt.
#21
Posted 2009-October-12, 14:08
Is the word "pass" not in your vocabulary?
So many experts, not enough X cards.
So many experts, not enough X cards.
#22
Posted 2009-October-12, 20:41
I don't think you can play diamonds for 4 tricks with a singleton Q with east if you start with the ♦J

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