best play for maximum matchpoints

[Mbodell]

[hv=pc=n&s=sak7h32d8cat86542&n=s632ha7dakt972ckq&d=s&v=n&b=15&a=1c1h2dp3c3h6cppp]266|200[/hv]

 

Trick 1 is:

♥K - A - 4 - 3

 

Matchpoints, your play.

[gnasher]

I'd just play safely to make it - with a combined 27-count, the field won't be in slam. Draw trumps (overtaking if ♣J appears early enough, and otherwise using a spade to cross to hand), then cash the rest of the trumps in case there's a squeeze.

 

Then if West has all the hearts above the 7, there may be a compound squeeze. If I'm able to preserve ♠AK, I can get to:

 

[hv=pc=n&s=sAk7h3d8c42&w=sq98hqdq43c&n=s63h7dakt9c&e=sjt5h5dj65c]399|300[/hv]

 

When I cash the penultimate club, West is squeezed. If he throws a spade, we have a double squeeze with diamonds as the pivot; if he throws a diamond, we have a non-simultaneous double squeeze with spades as the pivot.

 

If I've had to use a spade winner to get to hand, I don't think the compound squeeze plan works - West unguards diamonds, and dummy is squeezed on the last trump. (I'm looking forward to Rainer's explanation of why that's wrong :) .) Hence I'll have to hope for a more pedestrian double or single squeeze.

 

How come these come up so often in the forums but never at the table? In my entire life, I've seen one compound squeeze sucessfully executed at the table.

[whereagles]

Careful, you're beginning to get obsessed by compound squeezes :)

 

Anyway, you could try ♦AK (heart discard) and a third, eventually ruffing with the ten, depending on you read the suit. If you get overruffed, remaining trumps probably are split 2-1 and diamonds still set up for 12 tricks. But yeah, you will be going down on some layouts and the contract is probably a great score in all but world-class fields. Since I'm stupid, I'd probably try for 13 tricks anyway lol.

[VM1973]

As long as clubs run you have 12 tricks off the top. The field is probably not reaching this slam and so you should play it a bit safe. The only risk is a 4-0 club break so I'd start with the ♣K to see how things look.

[semeai]

Careful, you're beginning to get obsessed by compound squeezes :)

 

Anyway, you could try ♦AK (heart discard) and a third, eventually ruffing with the ten, depending on you read the suit. If you get overruffed, remaining trumps probably are split 2-1 and diamonds still set up for 12 tricks. But yeah, you will be going down on some layouts and the contract is probably a great score in all but world-class fields. Since I'm stupid, I'd probably try for 13 tricks anyway lol.

 

Let's say West has 7 hearts. Then clubs 0=4 is 14%, but the compound squeeze line reverts to your line when finding no clubs in West (though you do better when diamonds 4-2), so I'll ignore this. My numbers will be off a bit because I'm now really assuming clubs not 0=4, but it shouldn't change them too much.

 

Given 7 hearts in West, a split of 0=6 or 1=5 in diamonds is 26%, 2=4 in diamonds is 40%, 3=3 or 4=2 in diamonds is 34%. Given 2=4 diamonds, West having ♣J is 1 in 3, and stiff ♣J is roughly 1 in 9. So you're down roughly 30% of the time and making seven roughly 34% of the time.

 

I won't work out how often the compound squeeze line makes seven. Let's just do the spade-diamond simple squeeze against East: West needs at most 2 spades and at most 2 diamonds (or longer diamonds without ♦QJ, but I'll leave this out). This is almost 23%. It will go down due to clubs 0=4 more often than your line, because it needs to pick up diamonds with one ruff, but even 23-14 is more than 34-30.

 

So even if you're playing BAM and expect your opposition to get to 6♣, you want to play safely with the squeeze chances.

[VM1973]

Let's say West has 7 hearts. Then clubs 0=4 is 14%, but the compound squeeze line reverts to your line when finding no clubs in West (though you do better when diamonds 4-2), so I'll ignore this. My numbers will be off a bit because I'm now really assuming clubs not 0=4, but it shouldn't change them too much.

 

Given 7 hearts in West, a split of 0=6 or 1=5 in diamonds is 26%, 2=4 in diamonds is 40%, 3=3 or 4=2 in diamonds is 34%. Given 2=4 diamonds, West having ♣J is 1 in 3, and stiff ♣J is roughly 1 in 9. So you're down roughly 30% of the time and making seven roughly 34% of the time.

 

I won't work out how often the compound squeeze line makes seven. Let's just do the spade-diamond simple squeeze against East: West needs at most 2 spades and at most 2 diamonds (or longer diamonds without ♦QJ, but I'll leave this out). This is almost 23%. It will go down due to clubs 0=4 more often than your line, because it needs to pick up diamonds with one ruff, but even 23-14 is more than 34-30.

 

So even if you're playing BAM and expect your opposition to get to 6♣, you want to play safely with the squeeze chances.

I don't follow your figures as my back-of-the-envelope calculation put the chances of clubs 0-4 at 16.18 percent.

[gnasher]

With hearts 7=2, West has 6 vacant spaces and East has 11, so clubs 0=4 is

11/17 * 10/16 * 9/15 * 8/14 = 0.139

That's the same answer as I get from Richard Pavlicek's calculator

[Mbodell]

If you play AK of diamonds pitching a heart, and then a third round of diamonds which RHO follows to, should you ruff with the A, T, or 8? It turns out to be moot on this hand, but I was thinking 8 when I was considering that line.

 

I ended up playing 1 round of clubs first (worried about 1 club and 2 diamonds with LHO), but this proved fatal when LHO had no clubs.

 

The complete hand is:

 

[hv=pc=n&s=sak7h32d8cat86542&w=sqjthkqt965dq543c&n=s632ha7dakt972ckq&e=s9854hj84dj6cj973]399|300[/hv]