You know that an ordinary simple squeeze requires that you have all the tricks but one. You probably also know that if you have three threats against one player, or a suit-establishment threat, the squeeze can function with all but two tricks. But did you know that if both conditions hold, n-3 can be enough? Or that you don’t need any entries?
In a rubber bridge game, this 6-card ending happened, with spades as trumps:
North: xx xx Jx ----
West: T ---- xxxx x East: ---- QJ KQ Jx
South: Q x x KT8
As you can see, South has 3 of the last 6. When he plays a trump to his Q, a red-suit pitch by East lets him establish and score a 4th trick, and a club pitch gives two tricks. All this with no entry in any threat suit! Yes, the last trump is a sort of entry-surrogate, but this doesn’t at all resemble a typical ruffing squeeze.
Since this took place at a bar, I hope declarer, Dan Wilderman, doesn’t mind my saying he may have stumbled on this position a bit by accident. Nevertheless, this should definitely be the Wilderman (or, the Wild Man?) squeeze.
6 comments:
Cute ending.
Nice. Because of tenace position in clubs it seems that Wild Man squeeze would work even if can beat one of dummy's red cards.
For example, diamond position could be
West: KT9x N Jx E AQ S x
and the squeeze will still be on?
Regards,
Alex
I meant that West can protect something and it won't help....
Alex
You mean East can be thrown in if he pitches a diamond? I don't see it; he just cashes hearts, the extra club trick crashes with dummy's trump at trick 13. Did I miss something?
My mistake, you are right Jon.
Alex
Post a Comment