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One problem in Edwin Kantar's "Test Your Bridge Play"

#1 User is offline   frank0 

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Posted 2011-April-17, 02:47

I was looking at volume 2 problem 106, here is the question and answer

TWO QUEENS MISSING

S Q3
H QT72
D 2
C AKJT43
---------
S A2
H J65
D AKJT98
C 62
EW vulnerable, bidding is 1D-2C-2D-2H-2NT-3NT(opp all pass)
Opening lead: S5

Question: You put up the queen-but would this hand be in here if the queen held? Of course not. East plays the king. Plan the play.

Answer:You have two powerful minor suits and if you can bring home either one for six tricks you make your contract. But which one?
Back to the basics. When you are missing a queen in each of your two suits, and bringing in EITHER suit will give you your contract, play the ace-king of longer suit(clubs) and if the queen has not dropped take a finesse in the shorter suit, (diamonds).

----------------------------

My question is I think the answer here is wrong, follow his line we only make when CQ dropped or DQ on and D3-3, so the probability is

CQ siglenton or doublenton 28%*1/5+68%*2/5=32.8% (28% is probability of 4-1, 68% is probability of 3-2)

DQ on and 3-3 36%*50%=18% (36% is probability of 3-3)

So combination percentage is 1-(1-32.8%)(1-18%)=45%(roughly)

However, if we play dropping DQ and finesse CQ, we make when DQ dropped or CQ on and C not 5-0

probability of CQ on and not 5-0 is 50%*96%=48%>45% (96%=68%+28%,3-2 or 4-1)
In addition we still have chance to drop DQ so actual probabilty of this line is much higher.
Do I miscalculate or the answer is wrong?
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#2 User is offline   CSGibson 

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Posted 2011-April-17, 03:25

View Postfrank0, on 2011-April-17, 02:47, said:

I was looking at volume 2 problem 106, here is the question and answer

TWO QUEENS MISSING

S Q3
H QT72
D 2
C AKJT43
---------
S A2
H J65
D AKJT98
C 62
EW vulnerable, bidding is 1D-2C-2D-2H-2NT-3NT(opp all pass)
Opening lead: S5

Question: You put up the queen-but would this hand be in here if the queen held? Of course not. East plays the king. Plan the play.

Answer:You have two powerful minor suits and if you can bring home either one for six tricks you make your contract. But which one?
Back to the basics. When you are missing a queen in each of your two suits, and bringing in EITHER suit will give you your contract, play the ace-king of longer suit(clubs) and if the queen has not dropped take a finesse in the shorter suit, (diamonds).

----------------------------

My question is I think the answer here is wrong, follow his line we only make when CQ dropped or DQ on and D3-3, so the probability is

CQ siglenton or doublenton 28%*1/5+68%*2/5=32.8% (28% is probability of 4-1, 68% is probability of 3-2)

DQ on and 3-3 36%*50%=18% (36% is probability of 3-3)

So combination percentage is 1-(1-32.8%)(1-18%)=45%(roughly)

However, if we play dropping DQ and finesse CQ, we make when DQ dropped or CQ on and C not 5-0

probability of CQ on and not 5-0 is 50%*96%=48%>45% (96%=68%+28%,3-2 or 4-1)
In addition we still have chance to drop DQ so actual probabilty of this line is much higher.
Do I miscalculate or the answer is wrong?



You are wrong. One obvous reason is because if you decide to cash the diamond AK, you don't have an entry back to hand to finesse a club a 2nd time, so you can only pick up Qxx onside, not Qxxx.
Chris Gibson
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#3 User is offline   frank0 

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Posted 2011-April-17, 03:45

To CSGibson: thanks for your response, I didn't pay attention on that. We can not play pick up the chance of both CQxxx and dropping DQ.

However, if I play this way, after I win SA, I finesse CQ immediately, if it works(if not works) I cross back with D, cash DAK and finesse again. For this line I make it when CQ on and not 5-0, which is 48%, and an extra chance is DQ drop and CQ on with 5-0(when I cross back with DA and cash DK if I see DQ drop I don't care about distribution of C at all).

The line given by answer is about 45% if I don't miscalculate.

Or, another alternative line is cash DAK first and hope CQxx on, probability is
DQ drop: 48%*2/6+15%*1/6=18.5% (4-2 48% and 5-1 15%)
CQxx on: 68%*50%=34% (3-2 68%)
Total chance: 1-(1-18.5%)(1-34%)=46% (roughly)
Which is also slightly better than the line given by answer.

So I still don't quite understand the answer.
Any additional comment?
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#4 User is offline   quiddity 

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Posted 2011-April-17, 08:34

View Postfrank0, on 2011-April-17, 02:47, said:

My question is I think the answer here is wrong, follow his line we only make when CQ dropped or DQ on and D3-3, so the probability is


No, we also make when DQ is onside and singleton or doubleton.
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#5 User is offline   frank0 

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Posted 2011-April-17, 15:04

To quiddity: thanks, I'll appreciate your answer, the original answer in the book is correct.
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