hand 3

[cencio]

[hv=pc=n&s=skq2hakqdaj54cak4&w=sjt943h432dqt87c5&n=sa5h65dk932cj9832&e=s876hjt987d6cqj76&d=n&v=0&b=1&a=pp6nppp]399|300[/hv]

lead J♠

[johnu]

What book or article was this very nice safety play problem taken from?

[cencio]

amici del bridge on line

What book or article was this very nice safety play problem taken from?

It is from an italian web site. Amici del bridge on line

My link

[Cyberyeti]

Is this actually meant as a single dummy or double dummy problem ?

[johnu]

Is this actually meant as a single dummy or double dummy problem ?

 

This is a textbook quality single dummy problem.

[Cyberyeti]

This is a textbook quality single dummy problem.

 

OK, I see it now, was always obvious double dummy, but when people show 4 hands, you sometimes assume that's what's intended.

[johnu]

OK, I see it now, was always obvious double dummy, but when people show 4 hands, you sometimes assume that's what's intended.

 

Double dummy might be something like West having all 5 clubs and short diamonds with ♦Q.

 

 

 

 

You have 10 tricks in top cards, and the most likely source of 2 more tricks is in clubs with your 5-3 fit. Look at clubs and you should see a standard safety play in clubs.

 

Win ♠K, cash top club, then cross to dummy with spade to lead club 8/9. If East shows out or covers, win other top club and lead a club to guarantee 4 club tricks (plus 3 spades, 3 hearts, 2 diamonds = 12 tricks). If East plays low, finesse the 8/9. If it loses, you also have 4 club tricks.

 

If the 8/9 wins and West shows out, you have 3 club winners and no losers yet, but not enough entries to set up 5th club. You need 3 diamond tricks to make the contract.

 

Back to diamonds (which you should have already looked at) and it is another textbook safety play for 3 tricks. Cash the ♦A and lead to the ♦K9x planning to finesse the 9 if West plays low, or high if West play Q or 10 or shows out.

 

Nothing about this problem seems like a double dummy problem to me.

 

 

 

[Cyberyeti]

Double dummy might be something like West having all 5 clubs and short diamonds with ♦Q.

 

 

 

 

You have 10 tricks in top cards, and the most likely source of 2 more tricks is in clubs with your 5-3 fit. Look at clubs and you should see a standard safety play in clubs.

 

Win ♠K, cash top club, then cross to dummy with spade to lead club 8/9. If East shows out or covers, win other top club and lead a club to guarantee 4 club tricks (plus 3 spades, 3 hearts, 2 diamonds = 12 tricks). If East plays low, finesse the 8/9. If it loses, you also have 4 club tricks.

 

If the 8/9 wins and West shows out, you have 3 club winners and no losers yet, but not enough entries to set up 5th club. You need 3 diamond tricks to make the contract.

 

Back to diamonds (which you should have already looked at) and it is another textbook safety play for 3 tricks. Cash the ♦A and lead to the ♦K9x planning to finesse the 9 if West plays low, or high if West play Q or 10 or shows out.

 

Nothing about this problem seems like a double dummy problem to me.

 

 

 

 

Only the fact that 4 hands were shown not 2.

 

And yes that was the solution I got.