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West led the spade 9 and I paused to count my tricks: 3 spades, 3 hearts, 2 clubs and 2 diamonds. I had some work to do. Where were my extra tricks coming from? Well, if hearts are 3-3 all is good, or if the heart jack is singleton. 3-3 hearts is only a 36% chance and the chance of a singleton jack is infinitesimally small. 3-3 diamonds only gives me one extra trick and I need two. 3-3 clubs is another possibility, yielding 3 spades, 3 hearts, 2 diamonds and 4 clubs. Things look pretty bleak.
Maybe there is a squeeze possibility. In order for a squeeze to operate for 12 tricks I need to lose a trick to rectify the count. I won the spade king in dummy and led a small diamond, intending to duck it. Lo and behold East played the jack! Now there are two possibilities, East either has played the stiff jack, or he holds QJ(x). The principle of restricted choice says he is twice as likely to have played a stiff jack as he was to be holding QJ, so I rose with the K and finessed dummy’s 10 – winning as East showed out. Darn, should have put in the 8!
Now my chances of making this hand have gone up dramatically. I could rattle off hearts from the top hoping for a 3-3 break and claim 13 tricks, and as long as hearts are not worse than 4-2 I can give up a heart and take 12 tricks. But there is another line and it is the one I chose. Given that diamonds were 5-1 with East holding only one, I fear that hearts may be splitting poorly with East having the length. So I chose to lead a low heart from dummy toward my 10 spot and make the hand anytime the heart jack is onside. As the cards lie, this was the winning play. Note I will still make 12 tricks any time hearts are 3-3 or 4-2, but this line also wins when East holds all 6 hearts or 5 hearts and the jack.
Our opponents bid all the way to 3N and took the 10 obvious tricks. Plus 1440 and minus 630 was worth 13 IMPs.
Here is the complete deal:
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