an interesting hand, at least for me

[cencio]

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After 2 tricks lost on heart. east plays ten of club. How to play for making 5

[AL78]

The only thing I can think of is a black suit squeeze. That needs West to be guarding both black suits. For that to happen, East has to have no more than a doubleton spade and a doubleton club. The question is how to play the diamonds for no losers. What hearts were played on the first two tricks? If East is marked with ♥AKQ they probably won't have ♦Qx(x) otherwise they may have opened 1♥. In the absence of other information I don't know whether it is best to play East for 2722 or 2731 shape, but if I can pick the ♦Q up, I'd run the diamonds and hope West is forced to give up a black suit guard.

[cencio]

The only thing I can think of is a black suit squeeze. That needs West to be guarding both black suits. For that to happen, East has to have no more than a doubleton spade and a doubleton club. The question is how to play the diamonds for no losers. What hearts were played on the first two tricks? If East is marked with ♥AKQ they probably won't have ♦Qx(x) otherwise they may have opened 1♥. In the absence of other information I don't know whether it is best to play East for 2722 or 2731 shape, but if I can pick the ♦Q up, I'd run the diamonds and hope West is forced to give up a black suit guard.

West play ace and then 2 for k of east

[cencio]

You chose the right path. For an average player like me to make a Squeeze is very rare and generally I imagine it on a Slam contract. In a minor contract, I could not imagine it

[smerriman]

If you're at the point where you have to take all the rest of the tricks but you're off one, it doesn't matter whether it's a part score, game, or slam - it can never hurt to cash all of your winners and see what happens.

 

All you have to do here is pick one black suit - say spades - and count how many the opponents are throwing. At the end, if they've thrown too many, your small spade will be good. If they haven't, maybe they threw too many clubs, and your last club is good.

 

They're squeezed if they're *forced* to do one of those things, but you don't even have to know that - as long as you keep both options alive as long as possible. Even when the opponents aren't squeezed, they may throw the wrong thing!

 

(If you're not at that point of having to win all the tricks, and you can't see any way to get more, try losing tricks until you are.)

[Cyberyeti]

(If you're not at that point of having to win all the tricks, and you can't see any way to get more, try losing tricks until you are.)

 

Well there's the other very simple position that usually arises when you know the position of a particular card from the bidding where this is not true that is the other one that beginners should learn (known as a strip squeeze).

 

Assume E opened the bidding so you know he has ♠K and ♥AQ (W has 2 hearts higher than the 7), you need 3 tricks and unfortunately you're not in N to lead a heart, but in S.

 

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You cash the diamond A and now what can E discard from ♠KJ ♥AQ ? if he discards a spade you drop his K, if he discards a heart, you exit with a low heart and he has to lead into your ♠AQ.

[cencio]

the complete hand

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[bluenikki]

 

All you have to do here is pick one black suit - say spades - and count how many the opponents are throwing. At the end, if they've thrown too many, your small spade will be good. If they haven't, maybe they threw too many clubs, and your last club is good.

 

Love's book gives good advice here. Don't count anything. Just watch the ♣ discards. Has everything higher than the 6 been played? If yes, you are done. If not, what is there to do except see if the ♠s are running?

 

This won't work for all squeezes. But it shows where "automatic squeezes" got their name.

[ee57]

The only thing I can think of is a black suit squeeze. That needs West to be guarding both black suits. For that to happen, East has to have no more than a doubleton spade and a doubleton club.

Why do you say East has to have no more than a doubleton in either suit? I imagine you are right, but I can't see why.

[Cyberyeti]

Why do you say East has to have no more than a doubleton in either suit? I imagine you are right, but I can't see why.

 

Because if he has 3 of either, one of them will be higher than your 3rd card in the suit, so W can unguard that suit

[Douglas43]

I like Blunikki's advice, which is well worth repeating for players not used to trying squeezes. It's a good habit to count the suit which is in the same hand as the squeeze card (the last diamond). On this hand it doesn't matter, but if you don't have an entry back to you hand, it would matter.

[bluenikki]

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After 2 tricks lost on heart. east plays ten of club. How to play for making 5

I have

[bluenikki]

the complete hand

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I have to admit I would go with the 8:5 odds and finesse trumps into the pre-emptor.

[AL78]

So the squeeze is on and it comes down to first picking up the diamond queen. Playing small to the king and small back picks up Qx on the actual layout, but what happens if East follows small on the second round. Play him for Qxx or xx? I guess the principle of vacant spaces favours playing the ace, dropping the queen in West's hand.

[AL78]

I have to admit I would go with the odds and finesse trumps into the pre-emptor.

 

That cannot be the best line even if you are correct and West holds the queen. If East only holds a singleton which is where finessing works and trying to drop doesn't, they will hold at least three cards in one of the black suits and the necessary squeeze will not work. You have to ignore the possibility of ♦QXX with West because you are going down on that layout. The important holdings are whether East holds XX, QX or QXX assuming West follows on the first round (you have to play East for at least two diamonds), and I'm guessing the pre-empt makes it more likely East holds two rather than three diamonds, so the correct line is try and drop the queen.

[smerriman]

Don't count anything. Just watch the ♣ discards. Has everything higher than the 6 been played?

Well, this still requires counting, just that you're now looking at how many cards there were above 6 to begin with, and then counting them down one by one each time one is played.

 

But yes, my point was that you only need to be focused on a single suit.

[Lovera]

In this case, 3 squeeze endings act: one positional with the unilateral ♣ 8 with the communicating suit of ♠ working only on W and two automatic ones with divided threats (unilateral ♣ 6 in S) which instead works on both W and E and, not touching the ♠, the ending acts a trick before. It is always the unilateral that must be checked if it has become a winner after the squeeze card has been played because, otherwise, the threat in the communicating suit will have become a winner.(Lovera)