Restricted choice on locating honor in grand slam

[frank0]

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A hand from BM2000, lead is ♦4, after you cash all non ♣ winners you find W has 4=2=4=3 or 4=2=3=4, E has 1=5=4=3 or 1=5=3=4, without information on vacant space, who do you play for ♣Q?

 

The answer is W, the explanation is if W has it W's opening lead is restricted to non ♣ so given W lead non♣ it's more likely W has ♣Q.

 

For this hand does anyone know how to calculate(estimate) how much W is favored to have ♣Q?

 

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This is a problem from a quiz book. Lead is ♦5, the author asks you beside finesse two major kings can you find a better line?

 

The answer is if W has two major Ks and ♣ stopper W will be caught by triple squeeze(after all ♦, ♣Kxxx in dummy ♠J♥J♣Ax in your hand, if W discard a K then you cross to your hand with ♣A and cash that J to make the second squeeze), and the author says because W makes passive lead this line is better.

 

I am wondering if anyone can make probability argument on this hand. Why the line seems to require another extra condition(besides 2K you require ♣ stopper locate at right place) can be better, just because W makes a passive lead? If it does better, can you estimate by how much?

[xcurt]

Also, if you finesse a major and catch Kx on, you don't need the other major finesse.

[awm]

It depends on exactly how you model LHO's leads. On the first hand, it seems reasonable to suppose that if he has nothing, he will pick a suit randomly... and if he has the club queen he will pick a non-club suit randomly. In this case he leads diamonds on 1/4 of hands without the club queen and 1/3 of hands with the club queen. Since there are equal number of hands with/without the club queen, he will lead diamonds on 1/8 + 1/6 of all hands, of which the 1/6 have the club queen. So his odds of holding the club queen are (1/6) / (1/8 + 1/6) = 4/7.

 

On the second hand, things are much more complicated because there is a great deal of information from the auction. I would tend to lead a diamond over a major on this auction anyway (at least from moderate diamond length to no honor), to force declarer to take his diamonds early and prevent finessing partner. Anyway, if we assume that LHO will never lead from a major king but is equally likely to lead suits where he has nothing, we get the following. He leads diamonds on all the hands with both major kings, 1/2 the hands with one major king, and 1/3 of the hands with no major king. Thus given the diamond lead, he has both major kings on 1/4 of all hands, one major king on 1/4 of all hands, and no major king on 1/12 of all hands. So the overall chance of him holding both kings given the lead is (1/4) / (1/4 + 1/4 + 1/12) = 3/7. Of course, we would only make the contract half (really less) of these times because we need the club guard also, for an overall chance of somewhat less than 3/14. If we take both major suit finesses, our chances are more like (1/12) / (1/4 + 1/4 + 1/12) = 2/14, which is worse.

 

However, there are other lines to make this hand. In particular, you can try to guess one major suit king and then squeeze RHO in that suit and the clubs. The chance that RHO has the particular king we guess will be (1/8 + 1/12) / (1/4 + 1/4 + 1/12) = 5/14, after which our chance of making is better than 2.5/14 (so quite close to the original line). Further, we can delay the decision of which finesse to take until we are cashing the sixth round of diamonds (on which declarer must discard a major suit card) which may improve our chances of deciding which king to guess.

 

Note that this ignores the possibility of a club lead. Surely a club lead is unusual with clubs bid/raised, but if LHO holds something like QJT(x) or JT9(8) it would be fairly normal to lead a club from sequence as a safe option. These holdings obviously include a club control, which is part of why LHO having the club stopper is less than 50/50. Factoring this in, I would prefer running some diamonds, deciding which major to finesse, and then squeezing RHO in that major and clubs if my finesse wins.

[gnasher]

On the first deal, a further complication is the choice between xxx, xxxx and 10xxx. West will probably prefer xxxx to xxx, and he'll probably prefer xxxx to 10xxx. With two 10xxx holdings and an xxx, it's not clear what he'd lead. To make use of such inferences, we need to know whether LHO had ♠10 or not, whether ♦10 has appeared, whether anyone is up to playing ♦10 on the third round from an original 10xxx, and whether they're capable of working out when it's correct to do so.