[Tramticket]

Needing three tricks from the trump suit, you lead the two towards dummy. West plays the six and you finesse the nine - losing to the jack.

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What is your strategy on the second round?

Spoiler

My instinct was to play the ace on the second round and then finesse the ten if necessary - based on the restricted choice idea that West is more likely to have the queen since east could equally have played the queen instead of the jack if holding both honours.

In this situation we will always succeed if the suits are divided 3-2 (or 5-0). So it is only the 4-1 divisions that are important. Essentially we have to decide whether it is more likely that west started with a singleton six or east started with a singleton jack.

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or

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East's play of the jack is influenced by restricted choice (he might have played the queen), but west's play of a low card is also affected by restricted choice (he might have played the five or seven)! In the cases that matter, playing the king first wins in the ratio 3:2.

This idea was new to me. Have others encountered this?

[kenberg]

lho was dealt at least one spot, rho was dealt at least one quack. We must choose:
Stiff quack on my right, two ways. Lho can play his spots randomly.
Stiff spot on my left, three ways. Rho can play his quacks randomly.

So I continue with the ace. No No Continue with the K. That's the one with 3 chances to 2. I said A before. No, the K, not the A.The K seems counter-intuitive but I think it is right.

I think the key aspect of Restricted Choice is treating blocks of cards, the quacks or the spots, as indistinguishable. Now whether or not lho, holding quack-x-y-z will really play the x-y-z spots randomly? That's another question.


It's easy to miss something, but I think this is right.

Added: This is, essentially, suit combination261 in the Bridge Encyclopedia 6th edition.

They give
AT98
opposite
Kxxx.

To maximize the chance for 3 tricks:
Run the ten. If it goes small-small honor, then cash the King.


If you think about it, it is the same problem. After the Ten loses to a quack, you can go to the ace and run the 9, or you can cash the King and lead toward the dummy. By cashing the King, you are guarding against QJxx on your left, by going to the A you are guarding against stiff Quack on your right. The advice is to guard against QJxx on your left rather than stiff quack on your left.


It actually makes sense but it took a bit for my intuition to catch up with reality.

[billw55]

 kenberg, on 2016-July-05, 06:41, said:

So I continue with small spade towards the ace. 3-2 favorite, as I get it.

I'm thinking 2-1 favorite.

Ignoring spots, there are four ways the two quacks can be located. RHO showed up with one quack, eliminating the case where LHO has both. Of the remaining three cases, LHO holds the remaining quack twice, RHO only once.

[kenberg]

 billw55, on 2016-July-05, 06:48, said:

I'm thinking 2-1 favorite.

Ignoring spots, there are four ways the two quacks can be located. RHO showed up with one quack, eliminating the case where LHO has both. Of the remaining three cases, LHO holds the remaining quack twice, RHO only once.

Yikes, I said it wrong. Play the K, not the Ace. 3-2 is still right. I corrected the earlier post.


As the OP notes, not only are the Q and J interchangeable, so are the 5,6, and 7 (these being the spots this time).

So suppose we program a robot, or set a strategy, as follows.

First we play small to the 9. If rho wins with either the J or the Q we will next cash one high honor. Which one?

We suppose that our strategy will not vary depending on which spot was played on our left.

Deals where it will be right to play the A after rho wins the first round:
LHO was dealt
Q765
J765

Deals where it will be right to play the K
LHO was dealt
stiff 7
stiff 6
stiff 5


Assuming rho plays the Q or J randomly when he holds both, and assuming that lho plays one of the 5,6,7 randomly when he holds all three and an honor, it seems that these are the five holdings of interest, with no way to tell which is the actual holding.So we go with frequency.

This is an interesting problem it seems counter intuitive, or at least counter to my intuition, to next play the K, but it is right, is it not?

[FrancesHinden]

(deleted)

[Spisu]

 billw55, on 2016-July-05, 06:48, said:

I'm thinking 2-1 favorite.

Ignoring spots, there are four ways the two quacks can be located. RHO showed up with one quack, eliminating the case where LHO has both. Of the remaining three cases, LHO holds the remaining quack twice, RHO only once.

Nicely stated...It's rare among the restricted choice fallacies that anyone knows the key to it all is a relevant honor by an East rules out both honors in West. This eliminates that 25% from the equation and, since divided honors are not mutually exclusive but inclusive, their 50% likelihood continues against the 25% a priori for both honors in East. So you've got an automatic 2:1 for a second finesse when it's an AJ10 sort of a double finesse.

Here though I would have tried to guess who might be short, then played the top honor first that allowed a finesse through the other hand.

[Stephen Tu]

 billw55, on 2016-July-05, 06:48, said:

I'm thinking 2-1 favorite.

Ignoring spots, there are four ways the two quacks can be located. RHO showed up with one quack, eliminating the case where LHO has both. Of the remaining three cases, LHO holds the remaining quack twice, RHO only once.

If you finesse losing to RHO, you are definitely supposed to cash the K next, and this is better by 3:2 margin.

You are right that if RHO wins an honor, LHO is ~2:1 to hold the other honor. If someone were asking you to simply bet who has the other honor, you should say LHO. However, the vast bulk of these are 3-2 splits which cashing either honor will pickup. The only cases that matter are the 4-1 cases, and there are simply more cases where the split was x-qjxx than qxxx-j or Jxxx-Q.

If you look at the original percentages corrected for restricted choice in choosing between Q & J, it basically looks like:
Qxx-Jx 10.17
Qx-Jxx 10.17
xxx-QJ 1.70
xx-QJx 5.09
Qxxx-J 2.83
x-QJxx 4.24

As you can see LHO is more likely to hold the Q, but it is right to arrange to finesse against RHO since the 3-2 cases can all be dropped.

Essentially, playing the losing finesse to have been to stiff honor picks up 2 breaks: Jxxx-Q and Qxxx-J. But there are 3 x-QJxx breaks. That's more. 3>2. Restricted choice means you are supposed to halve the chances of qjxx (8.48% to begin with) vs. stiff J if you are comparing a priori odds. But 3 cases divided by 2 still > 1 case.

[billw55]

 Stephen Tu, on 2016-July-08, 22:28, said:

As you can see LHO is more likely to hold the Q, but it is right to arrange to finesse against RHO since the 3-2 cases can all be dropped.

Interesting, I had not thought of it this way.

[cherdano]

Yes. I do.

[WellSpyder]

I agree with the approach that says there are 3 small singletons and only 2 singleton honours, so it is better to play for the small card to be the singleton. In fact, if you are going to play for the honour to be singleton then I think you should start by cashing a high honour rather than taking a finesse.

[1eyedjack]

Not had time to look into it in detail (slipping this through office firewall), but out of interest I tried it out with Suit Play.

It reports 2 equally good lines to get 3 tricks both of which involve taking a first round finesse.

Well, actually, it reports 4 equally good lines, but only because you have a 2-way finesse available on the first round which are equally good either way but listed as a separate "line" in Suit Play.

Then, when the first round finesse loses to an honour, the two lines are distinguished by which top honour you cash next, Suit Play ranking them equal in effect, both lines providing a 94.3478% success rate.

As I recall, Suit Play does apply restricted choice when appropriate, but maybe it does not get it right all the time? Some of you lot will know.

[EDIT - load of tosh - scroll down]

[Stephen Tu]

Don't know what you are seeing but my copy of Suitplay gets this right. If the finesse loses it is catering to always picking up the Qjxx cases as this thread has been discussing, rather than playing for the honor to have been stiff. The lines for 3 tricks are distinguished by which way you take the finesse at trick 1, and what your response is if an honor pops at trick 1 by 2nd hand and you are trying for 4 tricks (so are guaranteed 3 at this point, 2 lines play for QJ tight, other two play for restricted choice honor stiff which is better). If 4th hand wins it always plays for the honor not to be stiff.

[1eyedjack]

You are right, I misread the output. Suitplay recommends playing for the small card to be singleton rather than the honour. Not equal at all.

[helene_t]

Cashing the ace is wrong if W had a small singleton which could happen in 3 ways.

Cashing the king is wrong if E had a singleton honour which could happen in 2 ways.

So I cash the king.

[billw55]

Ah, yesterday evening I had a chance to apply what I learned in this thread:

example


Perhaps I should have bid 1♠. I had hoped for some form of NMF or checkback, but such are the perils of pickup partnerships. In fairness, it would also be a risk for my partner to try that with a random.

Anyway, my line in diamonds was not a success. But at least I went with the odds ... I think.

[Stephen Tu]

 billw55, on 2016-August-01, 06:54, said:

Anyway, my line in diamonds was not a success. But at least I went with the odds ... I think.

I don't think so, not if opps playing 4th best leads. Person with only 4 clubs way more likely to have 4 diamonds than the person with 6. Esp after testing the spades which you should have done.

The line discussed in this thread is only best in a vacuum where either opp is equally likely to have 4 and you don't have other clues to work with.

[kenberg]

Another way of looking at it: If you play on spades and find E with two, then giving him a stiff diamond gives him, if you trust the lead, a 2=6=1=4 shape. He had a chance to bid some number of hearts over your 1D opening. All in all, a stiff D with E would be highly unlikely.

Myself, I rebid 1S over 1H but I realize many prefer 1NT. I am not prepared to give a solid argument either way.


My view of bridge is that at least three quarters of the good or bad results I get come from such decisions. Bid 1S or bid 1NT. How to play out the hand to take into account that suits re not always 3-2. That sort of thing. It would be good to have a thread, or a whole forum, devoted to it. Often the issue is not advanced versus expert or intermediate, it's just some simple thing where afterward you say "Of course I might have thought of that".

[billw55]

Assuming 4th best by BBO randoms is not always reliable. But maybe it was better than nothing. And yes, playing spades first would be better. I think I got too excited about seeing precisely the suit layout we had been discussing.

[kenberg]

Here is a hand I mis-played this afternoon.

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Yes I would like to be defending 3D, preferably doubled, but I have to make 3H. The opening lead is the diamond T to the Q and A. I will lose a diamond and two spades, so if I can hold the heart losers to one, I make it, trusting spades to be 3-3 as seems likely on the auction ( and is correct).

The is to illustrate what I said earlier. I don't have to be an expert, I have to give it a little thought. More thought than I gave it. Kx or Qx in hearts on my right, again likely on the auction, not certain but likely, and I am home. Oh well, next time.