[kayin801]

Consider this suit in a NT contract in which you need 4 tricks:

AK87

Q106

Due to problems with entries to dummy (but starting in dummy), your best options seem to be a first round hook of the 10 or to unblock the 10 en route to leading up to the K8.

Suppose you choose the latter line (or someone starts out by playing the first trick and leaves you to clean up) and it goes:

ACE, five, ten, two
seven, nine, QUEEN, three
six, four, ?

Assume the opponents are playing their spots completely randomly, what card do you play if

i) no other information is known
ii) another side suit is known to be 4=2

[barmar]

1) is a simple restricted choice situation, so I think you're supposed to finesse.

2) Now there's an additional 2 vacant spaces in East, which probably cancels out the restricted choice probability, so play for 3-3.

[Phil]

barmar, on 2010-November-18, 22:18, said:

1) is a simple restricted choice situation, so I think you're supposed to finesse.

Why do you say this? There's no choice between the 9 and any other spot.

This is like saying that with:

Txxx

AKJ8x

after cashing the A we get the 9 on our left, so its right to finesse.

[Siegmund]

Quote

Why do you say this? There's no choice between the 9 and any other spot.

He means that because the ten was seen on the first trick, the nine and jack are equals when RHO makes his play to the second.
It is something other than a pure restricted choice situation, since players with J2, 92, and J92 had a choice to make on the first round too.

[Phil]

Siegmund, on 2010-November-19, 01:18, said:

He means that because the ten was seen on the first trick, the nine and jack are equals when RHO makes his play to the second.
It is something other than a pure restricted choice situation, since players with J2, 92, and J92 had a choice to make on the first round too.

Oh we dropped the 10? Sorry. Agree this is R/C with an alert opponent.

[gnasher]

There are 8 relevant 4=2 splits (Jx/9x) and 4 relevant 3=3 splits (J9x).

(1) With no knowledge of any side-suit
After you've given LHO 3 specific cards and RHO 2 specific cards, LHO has 10 vacant places to RHO's 11.
Specific 4=2 : Specific 3=3 = 10:11
Any relevant 4=2 : Any relevant 3=3 = 20:11

(2) Knowing that another suit is 4=2
After you've given LHO 3 specific cards and RHO 2 specific cards, LHO has 6 vacant places to RHO's 9.
Specific 4=2 : Specific 3=3 = 6:9
Any relevant 4=2 : Any relevant 3=3 = 12:9 = 4:3

Quote

2) Now there's an additional 2 vacant spaces in East, which probably cancels out the restricted choice probability, so play for 3-3.

It's easy to overestimate the effect of this sort of knowledge. In fact, the break-even point is:

(3) Knowing that another suit is 5=1
After you've given LHO 3 specific cards and RHO 2 specific cards, LHO has 5 vacant places to RHO's 10.
Specific 4=2 : Specific 3=3 = 5:10
Any relevant 4=2 : Any relevant 3=3 = 10:10

Quote

It is something other than a pure restricted choice situation, since players with J2, 92, and J92 had a choice to make on the first round too.

It may or may not be safe to play J from J92 or J2 on the first round, depending on what is known about declarer's hand.

If it's not safe to play J from one of these holdings, that means you shouldn't play it from the other either. That, in turn, means that you shouldn't play 9 from 92 either. In this situation it's best to play your small card first. But in practice I'm sure that you're more likely to get the 9 from 9x than the jack from J9x.

[han]

barmar, on 2010-November-18, 22:18, said:

2) Now there's an additional 2 vacant spaces in East, which probably cancels out the restricted choice probability, so play for 3-3.

Way way way off. You need some serious vacant space difference to overcome almost 2:1 odds. (I see now that gnasher already computed this, 5:1, and I trust that Andy is correct)

Vacant space considerations usually only have a small effect on the probabilities. If you need to guess who has the heart queen and you have no other information to go about, sure, finesse through the person with 2 spades instead of the person with 4 spades. In the long run this gives you a small edge. The only situation that comes up very often in which vacant spaces play a major role is when somebody has preempted. Fortunately we all consider it intuitive to play the other player for length, so most people will do well enough without ever having heard of vacant spaces. And even in the case of the preemptor, you will often do even better by considering the exact shapes the preemptor can hold and which of those is most likely.

[mfa1010]

I agree with gnasher and just want to point out a shortcut in the calculations.

A- If we have a normal vacant spaces problem with no restricted choice, we finesse west at the decision point if he has more vacant spaces than east.

B- If it is a restricted choice situation then the impact of this is smply, that we should halve the number of vacant spaces we count with east. Very easy.

---
So here:

1) West has 10 vacant spaces, east has 11 vacant spaces. Restricted choice => east should be counted as only 5.5 vacant spaces. Odds of finesse = 11:5.5 or 20:11.

2) West has 6 vacant spaces, east has 9 that should be counted as 4.5. Odds for finesse= 6:4.5 or 4:3.

3) Cut-off is knowing a suit is 5-1, since that would leave west with 5 vacant spaces and east with 10 counted as 5.

[Fluffy]

But the fact that East could had played the 9 from J92 at trick one is completelly irrelevant?

[gnasher]

Fluffy, on 2010-November-20, 11:48, said:

But the fact that East could had played the 9 from J92 at trick one is completelly irrelevant?

If he'd play 9 with equal frequency from J92 and 92, it's irrelevant. If he wouldn't, it's not.

[kayin801]

I'm, of course, more interested in the general case, but against the actual opponent I would judge him to:

- Always play the 2 from J92 to the first trick
- Play the 9 from J92 at least 80% of the time to the 2nd trick
- Always play the 2 and then 9 from 92 doubleton

Not sure how that affects calculations and how to play...

At the table I hooked and went down, but I guess in the general situation hooking is right?

[mfa1010]

kayin801, on 2010-November-20, 20:52, said:

I'm, of course, more interested in the general case, but against the actual opponent I would judge him to:

- Always play the 2 from J92 to the first trick
- Play the 9 from J92 at least 80% of the time to the 2nd trick
- Always play the 2 and then 9 from 92 doubleton

Not sure how that affects calculations and how to play...

It's quite easy to calculate.
In the usual restricted choice situation, we divide the odds for finessing by 50% (= multiply by 2), but now we instead just need to divide by 80%.

So with your 80-20 assumption it goes /when we see the 9/:
1)
Empty spaces: 10 to 11
Odds with 80-20RC: 10 to 11x0.8 = 10 to 8.8
Percentage for finesse = 53,2%

2) (Here we know about the 4-2 in the other suit)
Empty spaces: 6 to 9
Odds with 80-20RC: 6 to 9x0.8 = 6 to 7.2
Percentage for finesse = 45,5%

So knowing about the 4-2 in the other suit is enough to tip the scales in favour of playing for the drop under the assumption that RHO is 80% to play the 9 on the second round from J9x.

[mfa1010]

(double post)