Re: 11 pts opposite weak NT

On Jul 14, 3:15 pm, Charles Brenner <cbren...@xxxxxxxxxxxx> wrote:
On Jul 14, 2:13 pm, "Lorne" <lorne_ander...@xxxxxxxxxxx> wrote:



"Charles Brenner" <cbren...@xxxxxxxxxxxx> wrote in message

news:1184423144.320441.269330@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

On Jul 14, 6:54 am, "Lorne" <Lorne_Ander...@xxxxxxxxxxx> wrote:
"Keith Sheppard" <keith-shepp...@xxxxxxxxxxxxx> wrote in message

news:4694cd3c$1_4@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Traditionally, with 11 opposite partner's weak (12-14) NT opener, one
would invite. At IMP, when you cannot afford to miss game if it's
there,
that seems to be sound policy. What about pairs though? Now the
magnitude of the expected gain/loss is unimportant. You just need to
do
as well, or better, than the others with your cards.

Let's assume, for the sake of argument, that partner's 1NT could be 12,
13
or 14 with roughly equal probability. You've got a flat 11 count with
nothing really extra going for it.

There is only a 1 in 3 chance that our combined count is 25.

your odds are a bit out. a 14 count is only about 28%, whereas a 12
count
is closer to 38%.

Your odds are a bit out. A 14 count is about 26%, and a 12 count 40%.

-- Charles

Not sure where you got your numbers from but if you do not constrain the
hand shape the calculated odds are:
14 = 27.6%
13 = 33.5%
12 = 38.9%

For explanation of my numbers, see my earlier message in this thread
--http://groups.google.com/group/rec.games.bridge/browse_frm/thread/6a6...

If you do constrain the shape calculating it is more complex but a 100,000
run on a dealer using normal weak NT constraints (ie 4432,4333,5332 shapes
only) gave:
14 = 28.0%
13 = 33.7%
12 = 38.3%
so I stand by my original numbers.

Your have done enough simulations that the difference between your two
sets of %'s look like real differences, not just sampling variation.
That is interesting.

However, by neglecting the fact that responder is known to have 11
points, you neglect a factor that will change the %'s about twice as
much as the distributional factor.

-- Charles

On second thought I doubt that it matters, to the HC distribution,
whether the hand is known to be balanced or not. The following tables
are from exact calculations, not simulations. Each of the 3 blocks
gives the probability of a hand having 12, 13, and 14 points given
that
(a) it is in the 12-14 point range
(b) possibly one additional condition.

any distrib balanced (5332 ok)

12 0.389001783 12 0.387950830 remaining hands
13 0.335085662 13 0.334792731 unconstrained
14 0.275912555 14 0.277256439

12 0.401914008 constraint of
13 0.334447864 11 hcp opposite
14 0.263638128

Top left is the table we all agree, apparently published. Top right
corresponds to Lorne's simulation -- constraining the distribution to
be balanced. Practically no difference.

Bottom left does not constrain the distribution but does condition on
the fact of a known 11hcp hand. This information is 10 times as
significant as the balanced hand constraint. Consequently I
extrapolate that the missing lower right block -- the most relevant
statistics from the perspective of an 11-point responder -- would be
practically like the lower left.

-- Charles

.

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