On May 8, 4:58 pm, "raija d" <musti...@[EMAIL PROTECTED]
> wrote:
> How are squeeze possibilities (if any) calculated into the formula of
"what
> is the percentage/dds for 12 tricks", at the onset of a hand.
In principle you consider every possible (relevant) layout and course
of play, note which ones are winners, calculate the percentage chance
of each of them to occur, and add them up.
Presumably your question is mainly about "calculate the % chance of
each to occur." With the hand you have presented that's pretty hard as
a practical matter but nonetheless, precisely because there's nothing
trivial about it, it's a good example to illustrate the ideas.
I'll just discuss a single winning case. Suppose we have in mind a
line of play among whose winning cases is squeezing LHO when LHO holds
Jxxx, Jxxx, x, Jxxx.
The probability that LHO would be dealt that hand is
(# combinations equivalent to that hand) / (# hands LHO could be
dealt),
of which the numerator is simply the number of ways to assign the x's,
and the denominator is the number of available hands once NS's 26
cards are removed from the deck, hence
p=C(4,3) C(5,3) C(7,1) C(5,3) / C(26,13),
where C() is the combinatorial function e.g. C(5,3)= 5! / (3! 2!).
Thus
p=2800 / 10,400,600 = 0.027%. (Is Jxxx,Jxxx,x,Jxxx really that
unlikely or did I misanalyze?)
However, this value p must be adjusted based on the given facts of
bidding and play to date -- to wit, that LHO did lead a spade -- which
makes the computation not only tedious but subjective. The numerator
is not too tough: multiply by your estimate of the probability that
LHO, if holding this hand, would lead a spade (given the bidding).
Adjusting the denominator means that we must estimate, somehow, the
effective number of hands from which LHO would lead a spade. In
principle that means writing down all C(26,13)=10 million hands LHO
might hold, and for each estimating the % chance of a spade x lead,
then adding those %'s. Possibly the work can be sped up a bit by
listing the possible hands by category rather than one at a time. For
example for hands with no spade x or with J109x of spades the % is 0.
It would be a lot easier if the opening lead were apparently something
like Q from QJ10, in which case we could assume as a reasonable
approximation that the lead is automatic.
Charles
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