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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