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4H play problem


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First look::

I would try for ten tricks via four hearts, two spade, four diamonds.

So the first thought is:

Spade ace

AK of diamonds.

If either the Q or T falls, I only need hearts to be 3-2:

K of spades throwing a club, AK of hearts,

J of diamonds. If it was the diamond ten on my left that fell earlier, from Tx, I don't ruff, I establish the suit by losing to the Q. I lose one heart, one club, one diamond.

 

Now suppose instead that as I play off the AK of diamonds, everyone follows small. The chances for a 3-3 split are now, with the evidence, about 50%. This is because the diamond spots are indistinguishable from each other, but the two honor card have the special feature that neither could be pitched. So the relevant probability refers to the chances that the two honor cards are in the same hand, given that the spots are 2-2. So I think I still cast my fate with the diamonds. Spade, tossing club. AK of hearts, diamond.

 

 

I imagine a careful analysis can do better, but this is plausible. Probably I should cash one high heart at trick 2, just to see what happens.

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My speedball line is SA, DA, DK, SK discarding a club, club.

 

I like this. Also might be worth ruffing a spade before exiting a club, aiming to score some more low trumps in hand by ruffing clubs later. if LHO is forced to win the club A we're pretty well placed on any return.

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