[eyhung]

A friend of mine asked me this question about Restricted Choice and I thought it was a doozy that should be shared.

You need to play the following suit combination for 3 tricks:

---

Assume that there are no other indicators from the bidding or other suits as to how the suit lies. Say you lead low to the 8, which forces the jack from LHO. LHO returns another suit, you win, cash the king and one of the remaining honors drops on the right. When you lead up to the A9, LHO plays low. Finesse or drop?

Does the answer change if LHO wins with a different honor besides the jack?

[jdonn]

It seems to me restricted choice applies to each honor play, so essentially cancels and you go with the odds. 3-3 is more likely than 4-2 in a specific direction, so drop.

Here is another way to look at it. The two possible holdings (if rho plays the ten on the second round) are:
QJxx Tx
Jxx QTx

The a priori odds of the first holding are 48% * .5 * 4/6 * 3/5 * 2/4 = 4.8%
The a priori odds of the second holding are 36% * 3/6 * 3/5 * 2/4 = 5.4%

So before any plays, the specific 3-3 break is the 5.4 to 4.8, or 9 to 8 favorite over the specific the 4-2 break.

Half of the time LHO has QJ he would have won the queen, so when he plays the jack then the odds he holds QJxx are 4.8% * .5 = 2.4%. And half of the time RHO has QT he would play the queen on the second round after the jack is good, so when he plays the ten then the odds he holds QTx are 5.4% * .5 = 2.7%.

So after their plays, the specific 3-3 break is the 2.7 to 2.4, or 9 to 8 favorite over the specific 4-2 break. Nothing has changed.

[whereagles]

The restriced evil choice seems to be what josh says, if he's right

What I mean is that "the principle of restricted choice" is a simple (though twisted) way to say that the odds usually favour playing for the dropping honor to be singleton. Usually, not always.

[quiddity]

eyhung, on Apr 3 2009, 02:27 PM, said:

Does the answer change if LHO wins with a different honor besides the jack?

I'm not sure, but I don't think it changes. Say LHO wins the T on the first round and RHO drops the J on the next. The possible relevant holdings are
QTxx Jx
Txx QJx

LHO might always win the T on the first round because he didn't realize the two honors were equivalent, while RHO would have a free choice on the second round if he started with QJx. By this reasoning, it would be right to play RHO for the doubleton J. But if you were to alter your play based on this reasoning, then RHO should always play the J from QJ on the second round to make you go wrong; so his choice is also restricted.

Sorry if this makes no sense at all; I find the subject very confusing.

[hanp]

Obviously there are an equal number of HHxx and HHx holdings, and given that each 3-3 split is slight more likely you should play for the drop if you think the opponents are capable of playing their honors randomly.

In practice against 99% of bridge players you should finesse if LHO plays the queen and RHO plays the jack and play for the drop otherwise.

[jdonn]

hanp, on Apr 3 2009, 04:21 PM, said:

In practice against 99% of bridge players you should finesse if LHO plays the queen and RHO plays the jack and play for the drop otherwise.

Wait, you are playing LHO to be winning the queen from QTxx?

[hanp]

Against 99% of all bridge players... no the other way around. This is why I don't win national swisses.

[quiddity]

quiddity, on Apr 3 2009, 04:12 PM, said:

But if you were to alter your play based on this reasoning, then RHO should always play the J from QJ on the second round to make you go wrong; so his choice is also restricted.

On the other hand, I guess the vast majority of players (and perhaps even bridge-playing robots) would not realize the advantage of playing the J, and would still play either honor at random from this holding. So perhaps this means the finesse is right after all.

[jdonn]

hanp, on Apr 3 2009, 04:26 PM, said:

Against 99% of all bridge players... no the other way around. This is why I don't win national swisses.

Well you actually came up with a good point I hadn't considered, which leads to a really neat chain of reasoning.

If LHO wins Q and RHO plays J on the second round, the drop is (in practice, essentially) 100%.

Therefore RHO should always play the T on the second round from JTx.

Therefore When it goes Q from LHO T from RHO it's 9 to 4 to play for the drop since restricted choice doesn't apply if RHO is defending correctly (sticking with the assumption that LHO defending "correctly" by sometimes winning Q from QTxx will never happen.)

Therefore LHO should win the Q from QJxx since if you are playing correctly you will go for the drop on the third round.

But what would mean when LHO wins the jack you will know to always play for the drop on the third round if RHO follows with the ten.

Therefore if LHO wins the jack and RHO has QTx he should always play the queen on the second round. That kills the restricted choice in THAT situation so from JTx LHO should always win the ten.

Etc etc etc

My head hurts. Maybe this changes everything?

Thanks a lot Eugene, just when I thought I knew the answer to a suit combination...

[hanp]

RIP jdonn.

[JLOL]

Just think about it like Han for theory, and in practice it is obvious what the right thing to do will be.