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?
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?