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3 odd/ 1 even or 1 odd/ 3 even suits


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EDIT: Flawed analysis... thx to posts #2,3 and 4 .

 

[hv=pc=n&s=skq8htdaqj6cjt742&w=st43hj652d52cq653&n=sj752hak743dkcak9&e=sa96hq98dt98743c8&d=s&v=n&b=15&a=1dp1hp1np2c(NMF)p3cp6nppp]399|300[/hv]

Sp 10 lead

 

I'd never find this at the table.

Opps have their "book" ... the Sp Ace at trick one....

I win the Spade return w/Q.

I have 11 top tricks.

To finesse or not to finesse ( Clubs ) for the 12th after East plays the Club 8 under my Ace lead.

 

Take the A and K of Hts.... both Opps follow ( key point as you shall see later )

Clear Diam K.

Back to hand w/Sp Q ....all follow , so Opps were 3-3 in Spades.

 

Now play the rest of the Diams.

- - West shows out on the 3rd Rnd, so East started with 6 cards ( another key point ).

- - So West now known to have 1 even and 1 odd ( number of cards in a ) suit.

 

This means East must have started with a stiff Cl.

- - Why? Because it is impossible for East to have 1 odd/ 3 even suits..

- - so, East is 3 odd/ 1 even....

- - and since East has already shown 2 cards Hts, s/he must have started w/ 3 cards ( no room for 5 cards ).

- - ie: 3 3 6 1

 

The HOOK ( for the Cl Q ) is 100% ON !

Edited by TWO4BRIDGE
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[hv=pc=n&s=skq8htdaqj6cjt742&w=st43hqj652d52c653&n=sj752hak743dkcak9&e=sa96h98dt98743cq8]399|300[/hv]

I'm confused, why would the above layout not be possible?

You are right.

( I knew I posted this in the right forum ).

 

I think I could have done it if I were able to cash the 4th Spade earlier, but

there were transportation problems.

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If this is supposed to be one of those learned "proofs" where you make everything so complicated that people miss the tiny critical step that invalids the whole argument then I fear it hasn't worked since we are all baulking at this statement:

 

- - Why? Because it is impossible for East to have 1 odd/ 3 even suits..

-

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