| ♠ | A K Q 4 3 |
| ♥ | K 10 4 |
| ♦ | K 7 6 |
| ♣ | 6 4 |
| ♠ | 6 5 |
| ♥ | A 9 3 |
| ♦ | A 9 3 |
| ♣ | A K Q J 10 |
You're in 7N facing the ♠J lead. How do you play?
This might not be right, but let's say that you start with 3 rounds of spades and RHO shows out on the 3rd round. Take over from there (don't forget to pitch something on this trick!).
Yesterday I suggested pitching a diamond, then running 4 clubs (pitching a spade and a heart) for a compound squeeze. The problem is that this is a restricted squeeze and you have to make a key guess before cashing the last club. There's a way to make it unrestricted, though, which gives you much better odds of reading the position. Answer below.
Pitch a diamond on the 3rd spade, run 4 clubs pitching a spade and the HT dummy, then cash DK, DA, and play the last club in this position (planning to pitch a spade unless it's good):
| ♠ | 4 |
| ♥ | K 4 |
| ♦ | 7 |
| ♣ | |
| ♠ | |
| ♥ | A 9 3 |
| ♦ | |
| ♣ | 10 |
This is a compound guard squeeze that can work if LHO has either missing heart honor. You still have to read the position, but you'll almost always be able to. Your next play is the HK, and, unless the diamond is good, a low heart towards your hand. When RHO follows low, only then, halfway through trick 12, do you face your guess (and even then, only if you've seen a heart honor from LHO. Either RHO's last card is the other heart quack and you finesse (this is why you unblocked HT), or LHO has it along with a spade and you drop it.
For starters, if heart honors are split and you play for it, you'll always make: a bit over 52% of the time. You'll also pick up a bunch of scattered things, such as when RHO starts with the sole diamond guard (or LHO unguards diamonds unwisely) and both heart honors.
On the other hand, once you play this way, they can always beat you if LHO guards diamonds and RHO has both hearts.
Overall, not that big an advantage, but it's there and the play is cooler.
I noticed that when you play for the regular compound, accurate defense prevents you from guessing right more than 50-50, but this may require a mixed strategy. Suppose West turns up with 4=?=?=3. He must pitch so that he unguards the pitched suit exactly half the time. Conditional on public information, he is 4=3=3=3 40.8% of the time, and in those cases always has unguarded the pitched suit, so: when his reds are unequal he must pitch from his short suit 9.2/59.2 or 16% of the time to give declarer an equal guess.
ReplyDeleteWhen West is 4=?=?=2 defense is much easier -- he can always pitch one of each and this gives declarer an equal guess.