[Trumpace]

LHO is an intermediate and RHO is an expert (on BBO though, but first hand).

You know that RHO is 5-4 in majors and LHO 4-2 in majors (from the bidding and your and dummy's holding).

You need to bring in the whole club suit which is as follows.

---

(dummy contains the 6 carder)

When you play the ♣K, LHO follows with the 2, and RHO with the J.

When you play the ♣6, LHO follows with the 5!.

What do you do? Is there even a choice between finesse and drop?

[TimG]

This is not a restricted choice situation -- east could have played the Jack or Ten from JTx without cost (and apparently to significant gain if declarer applies restricted choice to the situation).

[Trumpace]

Sorry. RHO is 5-4 in majors and LHO is 4-2. I had it the other way round.

Modified the original.

[Stephen Tu]

Restricted choice does have an effect on the relative probability of various holdings here, but as Tim noted, the possibility of RHO playing J from JTx makes the drop overwhelmingly better. Also LHO with T9xx can insert an honor to guarantee a trick unless you have a trump coup possibility on the hand.

More interesting problem with the suit divided 5-3, and RHO having many more known major suit cards.

[helene_t]

RC may or may not apply, depending on whether you think RHO could have played the J from J9. (Strictly, RC always applies because RHO could have played the J from JT9). If he could have J9 RC does not apply because there are three ways of chosing a singleton from 3 and also three ways of chosing a doubleton, unlike when you sample from 2 (two ways of sampling a singleton but ony one way of smapling a doubleton.

But even if RHO could not have J9, I would play for the drop. The apriori chance that he has a singleton with his 7 slots vs the 4 slots by LHO is small, and besides LHO might have smelled a finese and played high in trick two if he held two high cards.

Edit: oh, RHO has the 4 slots and LHO the 7. Then it's close. I'm still more likely to play fo the drop but it will depend on my table presence.

[SoTired]

even if dummy was AQ9xxx, this is NOT restricted choice situation because of the reasons stated previously. Lack of entries or cards to resolve a situtation does not make restricted choice situation. To look at it logically, a falsecard should not be able to fool a restricted choice.

In this case, you don't even have that. You are not missing 1 card of equals, you are missing 2. So RHO could have played the J from J10, J9, or J109. And that is not counting the fact that LHO should split.

So .... by finessing the 8, you assume a lower probability layout (J-stiff is about a 3-1 underdog before considering the 9-6 major suit discovery, but I don't think that is enough to turn it positive) AND that LHO was careless in not splitting.

[Stephen Tu]

I don't understand why people keep on insisting that this is not restricted choice situation. It absolutely is, since you have missing equal cards & RHO has a choice of plays holding more than one of them, that's all you need for RC to apply. Restricted choice isn't something that only applies when there are two equal honors missing. It applies in general, when you have equal cards; for holdings where an opponent has more than one of them you have to decrease the probability of those vs. the ones where he has fewer to do proper calculation.

It's just that after you tally all the possibilities up, the drop is still better here, though closer w/ the known cards info (I think need a 4 cd imbalance to tip, by my calcs). There are many examples where restricted choice leads you to a conclusion to hook rather than drop, just not here. That doesn't mean that restricted choice didn't have an effect on the relative success of things. If we were looking at the 5-3 split case, and LHO wouldn't split from H9xx 2nd round, then restricted choice makes you conclude that stiff J or stiff T is more likely than JT doubleton. However, since RHO is supposed to be playing honor from JTx, that tips it back to going for drop.

You can & should apply restricted choice here, then conclude the drop is best. Applying restricted choice != always go for hook. Applying restricted choice is just properly weighting probabilities of JT, JTx vs. J stiff before you make decision.

[inquiry]

If LHO held T9xx he would split on the second round. Play for the drop, this is silly question, forget restricted choice, play bridge.....

Who knows, RHO might have dropped jack from JT9

[ArtK78]

There is one reason above all others why this is not a restricted choice situation.

If RHO has a singleton CJ, then LHO has the T9xx and knows that you have a doubleton club. Therefore, LHO would split.

Posing the problem in this manner is kind of silly. What you are really asking is "assuming LHO is stupid enough not to split with an original holding of T9xx, is it more likely than not that RHO's original holding was a singleton J?"

That would imply that when LHO holds T9xx, he would not split his honors 100% of the time. That is a silly assumption. Even if you turn this into some sort of game theory problem, allowing for LHO to be stupid or careless 50% of the time that he holds T9xx, you can not raise the probability of an original holding of T9xx to over 50%.

Therefore, play for the drop.

[kenrexford]

Context, context, context.

Some have been saying that this is clearly not a restricted choice problem because LHO would split from 109xx 100% of the time. That is clearly wrong, as splitting the 10-9 is not always the right move, and it may be that it is usually the wrong play.

Proof is by an example. Suppose that Declarer has taken steps to enable an elopement. For example, perhaps he has ruffed a club in dummy before leading small to his hand. Whatever. Maybe Declarer is just up to it. If the hand could make against a 4-1 split on elopement grounds, splitting the 10-9 from 109xx will leave an end-position of 10x in LHO's hand and A8 on dummy. Not splitting will convert the end position to 109 in LHO's hand and A8 in dummy. Bad split early on, eh?

Furthermore, suppose that you are not sure whether an elopement does or does not exist. If you are 100% certain that you should split here, then are you not also 100% certain that Declarer will play for the drop if you do not split? So, why split when you may be assisting an elopement?

That being said, you then have seven a priori possibilities for LHO that are relevant (where LHO can play low twice):

xx (drop)
10xx (drop)
9xx (drop)
Jxx (drop)
J9xx (hook)
109xx (hook)
J10xx (hook)

This makes the odds seem to favor the drop. This is not "restricted choice," but it is much more interesting than some are claiming. This is especially the case is context would also allow you to rule out the xx-J109 option because you know that RHO cannot have three cards but might have one or two, or if a necessary assumption requires him to have one or two, but not three, clubs. Now we are approaching 50-50.

[inquiry]

kenrexford, on Jul 25 2007, 11:44 AM, said:

Furthermore, suppose that you are not sure whether an elopement does or does not exist. If you are 100% certain that you should split here, then are you not also 100% certain that Declarer will play for the drop if you do not split? So, why split when you may be assisting an elopement?

That being said, you then have seven a priori possibilities for LHO that are relevant (where LHO can play low twice):

xx (drop)
10xx (drop)
9xx (drop)
Jxx (drop)
J9xx (hook)
109xx (hook)
J10xx (hook)

This makes the odds seem to favor the drop. This is not "restricted choice," but it is much more interesting than some are claiming. This is especially the case is context would also allow you to rule out the xx-J109 option because you know that RHO cannot have three cards but might have one or two, or if a necessary assumption requires him to have one or two, but not three, clubs. Now we are approaching 50-50.

0You seem to be confused about probabilities. The conditions you state are not equally likely.

Chances for LHO to hold T9xx at the start is less than the chance he holds, for instance Txx or 9xx or even xx. And you included in your conditon, the chance he holds the jack, which he clearly does not.

At the point of no return (Cash king and play second and LHO plays low), the possible holding for WEST are:

xx
9xx
Txx
T9xx

The odds the way you are trying to calculate them with these four possibilities are 3 to 1 for the drop. The real odds are better for the drop, because the chance WEST holds T9xx is something like 2.8% initially, and for each of the others 3.9% initially. Assuming EAST will drop the J from JT, J9, and JT9 some reasonable percentage of time (JTx and J9x have been eliminated when LHO plays low twice), restricted choice can not come close to overtaking the odds of playing for the drop. To say that if you can prove RHO has at most 1C, then hook is right has nothing to do with restricted choice question, and the odds of RHO h aving JT9 is greater than him having singleton JACK, but odds of h im having sington 9, oar T have been eliminanted.

[kenrexford]

That's not right.

You are interested in a priori cases, not a posteriori cases. If the "jack" was replaced with an "h", with "h" being "Jack, Ten, or Nine," then the odds are much different.

[ArtK78]

To quote myself from a different thread:

"OW OW OW OW OW!"

And again, I mean that in the most complimentary way.

[Halo]

RC does not change the odds. If we give East all his singleton jacks, versus half his J9,half J10,half J10x,half J9x (and weight the results by vacant spaces) you still play for the drop.

[Stephen Tu]

Quote

That's not right.

You are interested in a priori cases, not a posteriori cases. If the "jack" was replaced with an "h", with "h" being "Jack, Ten, or Nine," then the odds are much different.

No, you're both right. There are two ways to apply restricted choice reasoning, and they are both perfectly valid ways of thinking about it, that work out equally mathematically:
1 - think of JT9 as identical cards, all "h", also take into account cases where RHO plays the T or 9.
2 - Only consider the case of RHO playing J. Reduce the weight of holdings where RHO has 2 honors by half, 3 honors by 1/3.

Restricted choice DOES apply in this analysis despite all these people trying to say it does not. It's just that after you apply these appropriate reductions in weighting for cases where RHO has > 1 honor, the drop is still indicated. RC is needed to compute the correct odds.

And in this 6-2 case you have to further weight by the frequency LHO splits 2nd round, further indicating the drop.

[kenrexford]

Let me try this again, just to see if I am missing something.

We know that RHO has 5-4 in the majors (9 cards), leaving him with four minor cards. We have seen one of his minor cards, and it is in the club suit and is one of the Jack, 10, or 9, which are equal cards that could be played, contextually, in any random order, and which we can call "H" as a group (H1, H2, or H3). We are considering a finesse at round two of the club suit.

We also know that the opponents have five clubs and six diamonds.

We will know that LHO has two small cards in clubs. Therefore, we know that RHO started with one of the following holdings:

H/♦♦♦
H/♦♦♦
H/♦♦♦
HH/♦♦
HH/♦♦
HH/♦♦
(HHH/♦)

We may be able to eliminate the last option, because we may either (a.) know that RHO has at least two diamonds or (b.) need to assume 2+ diamonds in order to make the contract. That leaves all of the above, except the one in paratheses.

LHO is known to have 4-2 in the majors, leaving him with seven minor cards. We know that two of the minor cards that he holds are two small clubs, and that he does not have three top clubs. So, he must have one of the following:

(xx/♦♦♦♦♦)
Hxx/♦♦♦♦
Hxx/♦♦♦♦
Hxx/♦♦♦♦
HHxx/♦♦♦
HHxx/♦♦♦
HHxx/♦♦♦

Again, we may be able to eliminate the one option, in parentheses.

So, whenever this situation occurs (honor played to the right at trick one, two small played to the left on tricks one and two) and we have this choice of what to do, these are the only relevant holdings.

One might say that a 4-1 split is less likely than a 3-2, but that is overly simplistic. As you can see, LHO's relevant patterns are either 4-3 or 3-4, with the finesse favored if LHO has the 4-3 holding where his 4-card suit is the suit that is the shorter of the 5-6 minors held by the partnership. This tends to suggest a play for the drop.

However, the margin is much tighter than would be suggested by the "LHO always splits" claims, which are incorrect claims anyway.

The chances of RHO's first card being a club H and no other card being a club seems to be 3/11 times 6/10 times 5/9 times 4/8, or 4.55% of the time. The chances of this happening for each of the four occurrences seems to be 18.2% of the time.

The chances of two cards being H and then no other seems to be six occurrences of 3/11 times 2/10 times 6/9 times 5/8, or 13.6% of the time.

This surprised me. I must be doing something mathematically wrong, it seems. But, it somehow seems to be that the odds are against the drop, strangely. It seems that the parameters of what we must assume make the ratio 18.2:13.6 for the hook. This is a roughly 57% play to succeed, on the relevant layouts.

[Stephen Tu]

Use Pavlicek's card combo analyzer, it's best for this sort of thing. With the major suit imbalance, it's indeed fairly close, and depends on the frequency LHO splits from HHxx.

I get, given RHO 9 known cards to LHO 6:
(analysis option 1, consider all honors equal):
RHO stiff honor, 18.18% * (prob LHO plays low twice from HHxx)
RHO T9/J9/JT/JT9, 15.46%

(analysis option 2, assume J played, weight accordingly)
RHO stiff J, 6.06% * (prob LHO plays low twice from HHxx)
RHO, JT/J9/JT9 and choose J (5.15%)

It's the same analysis either way, 1st is just considering all 3 quacks falling = 3x cases where specifically the J fell.

Basically the conclusion is if LHO splits 2nd round from HHxx >= 15% of the time, you should play for drop, otherwise hook.

[Trumpace]

Stephen Tu, on Jul 25 2007, 06:51 PM, said:

<snip>


Basically the conclusion is if LHO splits 2nd round from HHxx >= 15% of the time, you should play for drop, otherwise hook.

I was hoping someone (rather everyone) would come up with an analysis like this.

That is the reason I made it a point to mention the BBO skill of LHO and also made this into a poll as I expected some variation depending upon individual evaluation of "intermediate".

(In my opinion) most of the intermediates will not split more than 15% (relying on your calculation here) of the time in defense. Counting is still an issue to them. They will probably blindly follow second hand low, but that is just my evaluation.

[kenrexford]

Stephen Tu, on Jul 25 2007, 06:51 PM, said:

Basically the conclusion is if LHO splits 2nd round from HHxx >= 15% of the time, you should play for drop, otherwise hook.

Well now!

[ArtK78]

So are you saying that you are gearing your game towards competing against players who can barely distinguish one card from another?

This does not belong in a serious discussion forum.