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Counting opponent's hand exact distribution on first trick

#1 User is offline   Fluffy 

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Posted 2005-July-11, 05:33

It's often easy to deducee opponent's distribution from the bidding early on the play (sometimes on the bidding is also possible), here is a quick example just happened tomy father:


Scoring: MP

W - E
1-1NT*
2-3
4-ps


Partner leads the 9 (3rd-5th).

Not its easy to deduce declarer has 3, and since has rebid 2 to the forcing 1NT he has to have more than , so declarer has 5143 or similar with void in .

After winning Q you decide a passive deefence is in order and switch to either a or a .
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#2 User is offline   Fluffy 

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Posted 2005-July-11, 05:39

Now it is probably a good time to haave a look at the full deal:



You can often misscount for 1 card for opponents making a decision on the bidding such as bidding as best minor with 3-3, but misscounting THREE cards, well daddy said it was his first time ever.
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#3 User is offline   Gerben42 

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Posted 2005-July-11, 06:39

Sorry I disagree with your analysis. As is apparent from the full hand you are playing 1st 3rd 5th, i.e. leading HIGH from 3 small. In this case partner can have 3 and declarer then has 2, making possible a wide range of distributions, for example 5242 or 5332. Now I wouldn't have guessed 4432 either but to conclude declarer is short in is premature.
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#4 User is offline   Fluffy 

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Posted 2005-July-11, 11:11

Nobody knows the ways of the sponsor ;)
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