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A Suit Combination. Who is right?

Playing online Wednesday night, the following suit combination comes up:

A106

Q987

The contract is 6NT. There are plenty of entries to both hands. Declarer need three tricks from this suit. After the hand was over, one player commented to me that the correct play in the suit was to lead to the Ace and back to the Queen. I said I did not see how that could be right, (exact phrase  "Can't be right" and  I annoyed the person). I said I thought the correct play was to take two finesses in the heart suit. However, I was not certain. So I decided to try to figure it out.

First, I listed the possible distribution of the heart suit in the E/W hands. Took a while, I found 20 distinct cases. They are listed later.  Then I compared three lines against each of the cases. A. Finesse twice.  (lead 9 first)   B. Lead to the Ace and finesse on the way back.     C. Lead to the Ace and back to the Queen. In the table that follows WWW means all three lines win, WLW means the A and C win, B loses, LWW means A loses, B and C win,  etc. I hope this notation is clear. There are also some large numbers in the last two columns. I will explain those later. (Btw, does anyone know how to get columns to line up?)

 1) K   Jxxxx      1 case    WWW   C(20,12) = 125,970        125,970

 2) KJ  xxxx       1 case    WWW   C(20,11) = 167,960        167,960

 3) Kx  Jxxx       4 cases   WWL    C(20,11) = 167,960        671,840

 4) KJx xxx        4 cases   WLL     C(20,10) = 184,756        739,024

 5) Kxx Jxx        6 cases   WWL    C(20,10) = 184,756      1,108,536

 6) KJxx xx        6 cases   WLL     C(20,9) = 167,960        1,007,760

 7) Kxxx Jx        4 cases   LWW    C(20,9) = 167,960           671,840

 8) KJxxx x        4 cases   WLL     C(20,8) = 125,970           671,840

 9) Kxxxx J        1 case     LWW    C(20,8) = 125,970          125,970

10) KJxxxx --     1 case     WLL     C(20,7) = 77,520              77,520

11) J  Kxxxx       1 case     WWW  C(20,12) = 125,970        125,970

12) Jx Kxxx        4 cases   WLW    C(20,11) = 167,960         671,840

13) Jxx  Kxx       6 cases   WLW    C(20,10) = 184,756      1,108,536

14) Jxxx Kx        4  cases  WLL     C(20,9) = 167,960           671,840

15) Jxxxx K        1 cases   WWW  C(20,8) = 125,970           125,970

16) x  KJxxx       4 cases   LWL     C(20,12) = 125,970         503,880

17) xx KJxx        6 cases   LWL     C(20,11) = 167,960       1,007,760

18) xxx KJx        4 cases   LWW   C(20,10) = 184,756          739,024

19) xxxx KJ        1 case    LWW    C(20,9) = 167,960           167,960

20) -  KJxxxx      1 case   WWW   C(20,13) = 77,520             77,520

It is easy to leave a cases out, I did at first, but there are six card to be distributed between two hands, so the total number of cases is 2^6 = 64. Which is indeed the sum of the number of cases listed. If my wins/loses in the third column is correct and my addition is correct, I get Line A wins in 44 cases, Line B wins in 35 cases, and Line C wins in 25 cases. This strongly indicates that finessing twice is correct. However, some cases occur more frequently due to more numerous ways of the non-hearts being distributed in those cases. The fourth column gives the total number of ways the non-heart cards can be distributed in each case.  And the fifth and last column the the total number of hands. It is the number of cases in column two times the number of ways the non-heart cards can be distributed in column four. As a reasonableness check, the total number of E/W hands is C(26,13) = 10,440,600. That is indeed the sum of the numbers in the fifth column. It took me several tries to get the addition to come out right.

Now to sum up the total number of winning layouts for each line. For Line A, I get 19*167,960+16*184,756+7*125,970+2*77,520 = 7,184,166 winning layouts. For Line B, I get 16*167,960+10*184,756+8*125,970+1*77,520 = 5,620,200. And for Line C, 10*167,960+10*184,756+4*125,970+1*77,520 = 4,108,560.

Thus the probability for Line A is 7,184,166/10,400,600 = 69%

Thus the probability for Line B is 5,620,200/10,400,600 = 54%

Thus the probability for Line C is 4,108,560/10,400,600 = 39.5%

 

Anyone willing to confirm this or tell me where I went wrong?

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