West leads the KS and you plan the hand before touching the dummy.
rth
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D
NV
West
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Board24
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NV
East
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NV
South
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16
3
6
15
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You see that you have 11 top tricks, and, if the JD drops, you can make all 13.
But you realise that most pairs will not have reached 6NT, so you must maximise your chances of making the contract.
You have potential threats in three suits, so you prepare for a squeeze by ducking the opening lead. This is the key play to make your contract. It is known as "rectifying the count".
You win the spade continuation, cash the Ace and Queen of clubs, cross to hand with AD, and cash the clubs.
This is the position before the last club is led:
N: ♠ 7 ♥ J ♦ K Q 10 7 E: ♥ Q 8 6 ♦ J 8 4 S: ♥ A K 5 2 ♦ 2 ♣ 6 W: ♠ J ♥ 10 9 4 3 ♦ 9
On your last club, West discards a heart, you throw a diamond from dummy, and East must discard a heart in order to keep his diamond guard.
You then cross to the KD and cash QD.
Now West has a problem. He must keep JS, so he discards another heart.
You return to hand with a heart, and make the last three tricks with AK5 of hearts.
Well played!
Note: If you win the opening lead, East will be able to keep his hearts and the squeeze will not work.
Well played indeed. After you cash the clubs, you must cross to South with the AD, and not with the heart. Otherwise when you cash the long clubs dummy is squeezed first.
ReplyDeleteWe bid, uncontested, 1D-2C-2S-4NT, natural, and I passed as North as I thought I had a minimum reverse. Nigel made six on the double squeeze too.
Good point about the need to keep three hearts Paul. What happens if West switches to a heart at trick 2?
DeleteNow I as declarer have to play 3 rounds of diamnonds first and come back to hand by overtaking the QC on the second round of clubs. With four cards left, dummy has SA7, HJ, DT and declarer has ST, HA5, C8. On leading my last club, I'll assume West has kept two spades and a heart, so I'll pitch the 7S, squeezing East when I follow up with a S to the Ace.
DeleteDid cunning West bare his QS? Did East give the game away by giving count on the very first round of spades or was that 9S a singleton? We'll never know.
Suppose that the 3S and 3C are swapped so E has three spades to the 9. Will E now keep his 9S ditching QH, thereby encouraging W ditch his JS and keep hearts, an so defeating the squeeze? I suppose E might given that N will have come down to singleton JH. Would you?
ReplyDeleteI think you are right Mike - the double squeeze is depenedent upon West being the sole guardian of the spades.
ReplyDelete