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Tripel Squeeze
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On the last hand of some casual BBO bridge, we pick up a nice one-suiter:

North
K3
AKJ8432
Q
1063
W
N
E
S
1NT
P
?

What's our plan for this hand?

Despite its limited high-card strength, it's far too nice to sign off in 4. We can make slam opposite a good number of strong 1NT openers. Many of partner's hands that produce slam will be difficult to diagnose, but the majority of slam-suitable hands have no wastage in diamonds. When partner does not have the K or J, the rest of his points should be working well enough to make slam. Our plan then is to transfer to 2 and splinter with 4. If partner has wastage in diamonds, he'll sign off, or else he'll move on, possibly with Keycard.

We bid 2, but then LHO ruins our plans:

North
K3
AKJ8432
Q
1063
W
N
E
S
1NT
P
2
3
P
P
?

Now what? We haven't discussed this auction, so we'll need to rely on meta-agreements. LHO's bid has indicated we might have some club losers. On the other hand, if partner holds something like AJx, the holding now has full value. The bid also implies there may be some bad breaks on the hand. What about partner's pass? He would likely bid with a max and 3 trump. He would also double with a trump stack.

Without discussion, our priority now should be to just avoid an accident. Slam may still be available, but it shouldn't be disastrous to miss it. How do we get out of this? If we've agreed to play retransfers, 3 should get partner to a heart contract. Or we could play it safe and just bid 4.

On the actual deal, our hand tries 4. Partner isn't sure if this is a retransfer or a splinter, but he bids 4 anyway, ending the auction. The 4 is led. Although we are in 4, how should we play this hand if we had bid 6?

North
K3
AKJ8432
Q
1063
South
QJ8
Q5
AK9754
A9
W
N
E
S
1NT
P
2
3
P
P
4
P
4
P
P
P

North
K3
AKJ8432
Q
1063
South
QJ8
Q5
AK9754
A9
W
N
E
S
1NT
P
2
3
P
P
4
P
4
P
P
P

It looks like our best chance for 12 tricks is to win the A in hand, play a heart to the A, unblock Q, and then play a heart to the Q. We now try to pitch two clubs on the diamonds. We make whenever hearts are 2-2, diamonds are 3-3, or the long heart is with the long diamond. It's a reasonable slam, though the enemy bidding does indicate things are unlikely to split as we need them.

In 4, we may choose to play the same way, but at least at IMPs, it seems prudent to just draw trump, overtake the diamond to pitch one club, and guarantee making 5. On the actual lie of the cards, though, we can take all 13 tricks after this start. Can you see how it may be possible?

The full deal:

West
109654
1096
J10
854
North
K3
AKJ8432
Q
1063
East
A72
7
8632
KQJ72
South
QJ8
Q5
AK9754
A9
W
N
E
S
1NT
P
2
3
P
P
4
P
4
P
P
P
D
4 South
NS: 0 EW: 0

The doubleton JT gives you 12 tricks (and 11 top) without much difficulty. Where does the 13th come from? Well, East guards spades and clubs, and his 8 guards diamonds. Since we have the beer card(7) as a menace, we have what I like to call a tripel squeeze: a triplesqueeze where the beer card is a threat.

Here is the end position with North on lead:

West
109
J10
85
North
K3
2
Q
106
East
A
8632
K
South
AK9754
D

On the last heart, if East parts with a diamond, the diamonds run and we score all the tricks. But East does no better if he discards a black winner. This creates a winner in dummy, which can be used to squeeze East again for the 13th trick: a repeating squeeze.

I've already defined the tripel squeeze. Any suggestions on a nice name for the repeating three-suit beer squeeze?

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