🚀 go-pugleaf

In article <118@aplcen.UUCP> prm@aplcen.UUCP (Paul McMullin APL x7823) writes:
>
>	    (Dummy), W: Ax
>			Qx
>			AJ9
>S exposed:		KJ8643		N exposed:
>7c, 9c					Qc, Ac
>		You, E: KJT98x
>			Jx
>			Q8xx
>			2

   The situation seems clear; North has three clubs and hence is 4423, 4333,
or 3433.  It is possible that you will be forced to lose a diamond, a spade,
and a club, so you must avoid heart losers.  So you cash KA of spades to draw
South's trumps, and start pitching hearts on clubs.  If North does not ruff,
or ruffs with the queen, you pitch your second heart, and lose at most a club,
a diamond, and a trump.  If he ruffs small you overruff, cross to the board
with the AD, and play another club pitching your heart.
   Note that if North refuses to ruff you plan to simply cash AD and continue
the clubs.  If North ruffs you have no problem with control, and if he lets
you pitch two diamonds you are down to JT9 of trump and QD, and can hardly
lose more than two tricks.

   -- David desJardins

In article <118@aplcen.UUCP> prm@aplcen.UUCP (Paul McMullin APL x7823) writes:
>
>	    (Dummy), W: Ax
>			Qx
>			AJ9
>S exposed:		KJ8643		N exposed:
>7c, 9c					Qc, Ac
>		You, E: KJT98x
>			Jx
>			Q8xx
>			2
>
>
   One point in the play of this hand may be of further interest, as it
illustrates a simple but valuable application of bridge probabilities.
After the club ruff, KA of spades, LHO showing out, J of clubs pitching a
heart, club ruffed small by RHO and overruffed, diamond to the ace, club
ruffed with the QS while you pitch a heart, heart return ruffed in your
hand, the position will be as follows:

		     W: --
			Q
			J9
			64

		     E: JT
			--
			Q8x
			--

The question now is how to play the diamond suit for one loser.  Obviously
there are two possibilities, small to the J or small to the 9.
   Against good players the following inferences can be made:
1) RHO has a heart honor (or LHO would have led one).
2) LHO has a heart honor (or RHO would have played one at trick two).
3) RHO has the king of diamonds (if LHO had this, with five or six hearts to
	the ace or king, he would have made a negative double (or NFB))

Given that RHO holds the KD, the two lines may seem to have roughly the same
chance of success.  One looks for the TD to be onside; the other looks for
the diamonds to break evenly.  Let's compute the actual probabilities:

   Suppose no diamond honor dropped under the ace, or appears when you lead a
small diamond toward the J9.  Then RHO was dealt one of the following hands
(where H represents either the ace or king of hearts):

     A: Qxxx	     B: Qxxx	     C: Qxxx	     D: Qxxx
	Hxxx	or	Hxx		Hxx	or	Hx
	Kx		KTx		Kxx		KTxx
	AQx		AQx		AQx		AQx

If he has (C) either play will work; if he has (D) neither play will work.
The relevant hands here are (A) and (B).  The number of possible hands of
the two types (and thus their relative probabilities) may be computed as
follows (assume spades and clubs are already determined):

   # of hands = (# ways to choose heart honor) x
	(# ways to choose heart spots) x (# ways to choose diamond spots)

   Pattern A: (2C1)(7C3)(3C1) = (2)(35)(3) = 210 hands
   Pattern B: (2C1)(7C2)(3C1) = (2)(21)(3) = 126 hands

Thus hand A is 67% more likely than hand B, and so the 9 should be played
from the board.

   At the table, it should be easy to see that RHO has only four unknown
cards (apart from Qxxx/H/K/AQx) and LHO has eight unknown cards (he is known
to hold only x/H/-/T9x), so of course the TD is much more likely to be with
LHO.

   -- David desJardins

In article <11492@ucbvax.BERKELEY.EDU> I write:
>   One point in the play of this hand may be of further interest, as it
>illustrates a simple but valuable application of bridge probabilities.

Actually what it illustrates is that too much calculating can prevent
you from looking closely enough at the hand...

>After the club ruff, KA of spades, LHO showing out, J of clubs pitching a
>heart, club ruffed small by RHO and overruffed, diamond to the ace, club
>ruffed with the QS while you pitch a heart, heart return ruffed in your
>hand, the position will be as follows:
>
>		     W: --
>			Q
>			J9
>			64
>
>		     E: JT
>			--
>			Q8x
>			--
>
>The question now is how to play the diamond suit for one loser.  Obviously
>there are two possibilities, small to the J or small to the 9.
> (long analysis of probabilities)

   Unfortunately I seem to have gone astray at the "Obviously"; the 100%
line is to lay down the QD.  This loses only if one hand holds KTxx (and
ducks), which is not possible since we are assuming that RHO holds the KD.
   Does the fact that no one pointed this out to me mean that nobody saw
it, or that nobody is reading this stuff anyway?

   -- David desJardins