🚀 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
>
>
   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