Suit combination

[awm]

Maybe this is a simple analysis.

 

When we played the jack, we were planning to play RHO for the queen. If we were always going to play RHO for ace, we should lead the 8 initially so we can pick up singleton queen on our left. So it would be silly to lead the jack and then always play king when RHO follows low.

 

What should RHO's strategy be when he sees the jack? Ignoring holdings we can never pick up (like AQ9 with RHO), the only time it might help to play an honor when not forced is if RHO has Q9. If the play goes jack-queen-king-ace, we must guess whether to play for the nine to drop or to finesse. It will help RHO in this case to play queen from Q9 at least some of the time. Assuming RHO does this, when we play jack and see the 9 it is more likely A9 than Q9 and we should play king.

 

So our best play is to let the jack ride if RHO plays 7 and put up the king if RHO plays 9.

[drinbrasil]

Im surprised most dont understand the problem at all.

 

1 first of all you need 4 tricks. so all case when you are sure to lose 2 trick are pointless. and extra undertricks are also pointless.

.....

the post says you need 5 tricks.

[awm]

Ignoring positions we can never pick up for one loser, the possible holdings for RHO are:

 

Q

AQ

A9

A7

Q9

Q7

AQ7

A97

 

If we start by playing the 8 to the king, we will pick up: Q, AQ, A9, A7, A97 for five positions.

 

If our plan is to play the jack and then low when RHO plays 7, we will pick up AQ, Q7, and AQ7 but will lose to A7 and A97. This leaves singleton Q, Q9, and A9. If RHO always plays low from Q9 or A9, it's a pure guess whether to put up the king and in any case we will pick up one of the two positions and lose to the other, giving us five positions (of which two are 3-1 breaks), exactly the same odds as leading 8 to the king. If RHO varies his strategy, occasionally playing queen from Q9, he will only do this if it's to his advantage and our odds will become worse.

 

So I think leading 8 to the king might've been the best line all along...

[ArtK78]

I don't know if this adds anything to my original post, but this appears to be very simple:

 

Given that the "conditions of contest" state that the J was played from dummy and RHO followed small, then, if small is not the 9, running the J is best. It picks up all 2-2 breaks with the Q onside, and it picks up AQ7 onside. Playing the K picks up only all 2-2 breaks with the A onside.

 

If the 9 does appear, then one can eliminate AQ7 onside from consideration. It is then a straight 50-50 guess - Q9 or A9 onside.

 

Therefore, the K is certainly not always right, but it will be the winning play in a significant number of cases. Running the J is better, but not overwhelmingly better.

 

**************

 

By the way, I agree that leading the 8 to the K is best - it picks up all 2-2 breaks with the A onside, and it picks up all 3-1 breaks with a singleton Q. But, as I mentioned above, the "conditions of contest" state that the J was played from dummy.

[pclayton]

Agree with floating the J when the 7 appears as others have pointed out.

 

If I know that my opponent will always play the Q from Q-9, it makes my life a lot simpler. When RHO plays the 9 on the 1st round, I know that RHO can't have Q-9. Therefore the only relevant holding for RHO is A-9. Right?

 

If only it were that easy.

 

RHO's optimal play with Q-9 is really the 9. This makes the A-9 / Q-9 combos to be equal. As others have mentioned, if RHO has the tendency to play the Q sometimes, or even, at all, then the King becomes a better play than the Jack.

 

Perhaps the relevant question is what would you do with Q-9 as defender?

[fred]

So I think leading 8 to the king might've been the best line all along...

Actually it is tied for the best line.

 

But the line that it is tied with is a lot more interesting :)

 

Hint: the strategy that East chooses to adopt with Q9 (and AQ7) does not matter.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[ArtK78]

Since playing the J to the K is clearly inferior to playing the 8 to the K, then it must be right to float the J when the J is not covered.

[Stephen Tu]

Hint: the strategy that East chooses to adopt with Q9 does not matter.

 

No, I disagree. If RHO is known to ALWAYS cover from Q9, then declarer could counter by playing low to the T the second round instead of low to the 8, which picks up Q9 onside but gives up on stiff Q onside for a 0.57% gain. So RHO better cover less than 91.67% of the time otherwise declarer could take advantage by changing his plan after Q K A.

 

Now, if RHO covers NEVER, it really doesn't matter what declarer does when the 9 appears, since if he rises he'll pick up A9 but ducking picks up Q9, which would be equal. But in practice RHO will cover sometimes, so A9 > Q9 playing the 9, so if the 9 appears you should rise.

 

So the right answer is for the defender to cover from Q9 a decent chunk, but < 91.67%, to guarantee the 2nd trick vs. people who haven't studied the combination, and might have chosen to finesse, but showing no gain vs. people who know it, since they were going to rise anyway.

 

Also in real life defenders may not know for sure that declarer has 6, if declarer has 5 then covering from Q9 is likely the better tactic.

 

8 to the K is actually equivalent in picking up 5 tricks but it does poorly against bad (4-0) splits so the J line is superior to get 4 more often.

[Cascade]

If RHO plays the 7 then you play low winning against Q7 or AQ7 onside (stiff 9 offside) and lose to stiff Q or Q9 offside.

 

If RHO plays the 9 then playing low only wins against Q9 onside and loses to Q or Q7 offside. Q offside is not that likely since RHO with A97 should play the 7 in case you play the King. Playing the king when the 9 appears wins against A9 onside and picks up the stiff Q on a defensive error.

 

I think this means:

 

if the 7 appears play low

 

if the 9 appears play the king against imperfect defenders and you have a 50-50 guess against perfect defenders (king or low).

[fred]

A lot of people have come up with the right answer and a few have at least come close to describing why the right answer is right (notably Lamford and Cascade - sorry if I missed someone else who deserves to be mentioned). However, I don't think anyone has expressed the right answer quite as clearly as this:

 

Several people correctly pointed out that when East plays the 7 you have the choice of playing East for:

 

A: Q7+AQ7

B: A7

 

Clearly A is more likely. You should play low when East plays the 7.

 

Similarly, if East plays the 9 you have to choose between:

 

C: Q9

D: A9

 

Clearly these are equally likely which, at first glance, would suggest you have a complete guess when East plays the 9.

 

You can resolve this guess by considering what you will do if East plays the Queen. Now you have to choose between:

 

E: Q9

F: Q

 

It is now easy to see that you should resolve the C/D guess by picking D since that allows you to get F as well. If instead you pick C, you also get E, but you already had that in the form of C :)

 

Summary: if East follows with the 7, play small. If East follows with the 9, play the King.

 

To me it is really strange that the correct play depends on whether East follows with the 7 or the 9. Who would have thought this could possibly matter?

 

This line gains against 2 3-1 breaks (stiff Queen with East and stiff 9 with West) and 2 2-2 breaks (Q7 and A9 with East).

 

The alternative of leading the 8 to the King is just as good - it also gains against 2 3-1 breaks (stiff Queen in either hand) and 2 2-2 breaks (A7 or A9 with East).

 

One final point that probably does not to be said to anyone who has read this far: if you lead the Jack, East plays the Ace, and West plays the 9, you should finesse on the 2nd round.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[ralph23]

Similarly, if East plays the 9 you have to choose between:

 

C: Q9

D: A9

 

Clearly these are equally likely which, at first glance, would suggest you have a complete guess when East plays the 9.

Hmmm.... isn't this Restricted Choice?

 

Case D (A9): East played the 9 because he really had no choice.

Case C (Q9): East could rationally have chosen to cover the Jack with the Queen. or to play low with the 9. Suppose he just plays randomly, covering or not covering on his whim.

 

Thus, Case D where his choice was restricted is more likely. He will always play the 9 from Case D, but only play it half-the-time (or so) in Case C.

 

At least, that makes sense to me. :)

[fred]

Similarly, if East plays the 9 you have to choose between:

 

C: Q9

D: A9

 

Clearly these are equally likely which, at first glance, would suggest you have a complete guess when East plays the 9.

Hmmm.... isn't this Restricted Choice?

 

Case D (A9): East played the 9 because he really had no choice.

Case C (Q9): East could rationally have chosen to cover the Jack with the Queen. or to play low with the 9. Suppose he just plays randomly, covering or not covering on his whim.

 

Thus, Case D where his choice was restricted is more likely. He will always play the 9 from Case D, but only play it half-the-time (or so) in Case C.

 

At least, that makes sense to me. :)

What you say is true (that East's play is forced with A9 and that he had a choice with Q9), but this is not the type of situation that "restricted choice" normally refers to.

 

Restricted choice situations can normally be expanded out so that the "right" play is more likely to work than the "wrong" play (on an a priori basis). In all examples I have seen restricted choice applies when the choice is between equal cards (or equal suits). That is not the case here because the Queen and the 9 are not equals.

 

The point you make is valid and it might well be possible to use this sort of reasoning as a basis for coming up with the right answer. But I don't think this would be considered a restricted choice situation by people who know enough about math to say such things with authority (I am not one of those people).

 

Fred Gitelman

Brdige Base Inc.

www.bridgebase.com

[Stephen Tu]

Similarly, if East plays the 9 you have to choose between:

 

C: Q9

D: A9

 

Clearly these are equally likely which, at first glance, would suggest you have a complete guess when East plays the 9.

 

You can resolve this guess by considering what you will do if East plays the Queen. Now you have to choose between:

 

E: Q9

F: Q

 

It is now easy to see that you should resolve the C/D guess by picking D since that allows you to get F as well. If instead you pick C, you also get E, but you already had that in the form of C

 

I don't like this method of explanation. You should not treat C & E as completely separate cases, because it's the same combo. If RHO with Q9 covers, the frequency of case E increases while the frequency of C decreases. If this frequency is too high, the optimal declarer line changes. The defender has to not cover too often to make the stated line correct, it DOES matter what defender does with this, it is possible for him to choose an exploitable strategy. Now if declarer doesn't know this, he simply plays the book line & it doesn't matter if the defender is playing suboptimally.

 

I would phrase as the following:

When RHO plays 9, declarer can rise or duck.

When RHO plays Q, 2nd rd declarer can hook or play the T.

 

There are 4 relevant combos for RHO:

C1: Q9, RHO chooses to cover

C2: Q9, RHO chooses to duck

D: A9

F: Q

 

There are 4 lines for declarer to choose:

1. K if 9 appears, hook if Q covers (as stated, best vs. optimal defense)

Picks up D+F.

 

2. K if 9 appears, T if Q covers (best if RHO covers >= 91.67% from Q9)

Picks up D+C1, will be > D+F if C1 > F

 

3. duck if 9 appears, hook if Q covers

Picks up C2+F. equals D+F only if C2 = D. (so only equal if RHO never covers

from Q9)

 

4. duck if 9 appears, T if Q covers

This is by far worst, only picks up C1+C2, = D.

 

So 1 is best, but switch to 2 if RHO always covers from Q9. Now in real life, how often do defenders know declarer has 6 not 5? So the answer to best play in this combo seems entwined with how to handle the JT8 vs. Kxxxx combo, shouldn't RHO w/ Q9 cover?

 

So definitely one should play K if 9 appears on the right, but if Q appears, I am not so sure what the right play is. I think in real life most of time should play for drop, vs. 5 in declarer I think RHO should cover Q9 100%, many auctions will be impossible to know declarer has 6, and 5 cd suit more likely than 6. If have shown 6 in auction then maybe this doesn't apply, but I think a lot of people will not be familiar with this combo & possibility of declarer popping K on the 9 so will cover too often.

[fred]

Similarly, if East plays the 9 you have to choose between:

 

C: Q9

D: A9

 

Clearly these are equally likely which, at first glance, would suggest you have a complete guess when East plays the 9.

 

You can resolve this guess by considering what you will do if East plays the Queen. Now you have to choose between:

 

E: Q9

F: Q

 

It is now easy to see that you should resolve the C/D guess by picking D since that allows you to get F as well. If instead you pick C, you also get E, but you already had that in the form of C

 

I don't like this method of explanation. You should not treat C & E as completely separate cases, because it's the same combo. If RHO with Q9 covers, the frequency of case E increases while the frequency of C decreases. If this frequency is too high, the optimal declarer line changes. The defender has to not cover too often to make the stated line correct, it DOES matter what defender does with this, it is possible for him to choose an exploitable strategy. Now if declarer doesn't know this, he simply plays the book line & it doesn't matter if the defender is playing suboptimally.

 

I would phrase as the following:

When RHO plays 9, declarer can rise or duck.

When RHO plays Q, 2nd rd declarer can hook or play the T.

 

There are 4 relevant combos for RHO:

C1: Q9, RHO chooses to cover

C2: Q9, RHO chooses to duck

D: A9

F: Q

 

There are 4 lines for declarer to choose:

1. K if 9 appears, hook if Q covers (as stated, best vs. optimal defense)

Picks up D+F.

 

2. K if 9 appears, T if Q covers (best if RHO covers >= 91.67% from Q9)

Picks up D+C1, will be > D+F if C1 > F

 

3. duck if 9 appears, hook if Q covers

Picks up C2+F. equals D+F only if C2 = D. (so only equal if RHO never covers

from Q9)

 

4. duck if 9 appears, T if Q covers

This is by far worst, only picks up C1+C2, = D.

 

So 1 is best, but switch to 2 if RHO always covers from Q9. Now in real life, how often do defenders know declarer has 6 not 5? So the answer to best play in this combo seems entwined with how to handle the JT8 vs. Kxxxx combo, shouldn't RHO w/ Q9 cover?

I don't want to get into an argument about it, but I still think my reasoning is valid (and a lot simpler than yours). For whatever its worth, some bridge players with excellent credentials have agreed with my analysis of this combination.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[Stephen Tu]

Just because your reasoning leads you to correct conclusion about what to do when the 9 appears, and is "simpler", doesn't mean that it is also mathematically accurate & complete.

 

I think your reasoning doesn't adequately show what the proper thing to do is when the Q covers the J, for example. My reasoning does; it shows clearly that you play T if the covering percentage is >= 91.67%, and hook otherwise.

[han]

Certainly agree with Fred's analysis. Also frustrated with my own analysis, I missed AQ7 for my strategy 3.

[fred]

Just because your reasoning leads you to correct conclusion about what to do when the 9 appears, and is "simpler", doesn't mean that it is also mathematically accurate & complete.

 

I think your reasoning doesn't adequately show what the proper thing to do is when the Q covers the J, for example. My reasoning does; it shows clearly that you play T if the covering percentage is >= 91.67%, and hook otherwise.

Just because your reasoning leads you to the correct conclusion about what to do when the 9 appears and is "more complicated" doesn't mean that it my reasoning is not mathemetically accurate and complete.

 

I think your reasoning tries to answer a question that was not asked and fails to take into account the assumption of best defense (which I admit I forgot to include in the original specification of the problem but added to a later post in this thread).

 

I think my reasoning demonstrates that you have to choose between A9+Q and Q9.

 

IMO that is sufficient for answering the question that was asked.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[cherdano]

I would just add one comment to Fred's reasoning: One might wonder whether there is a better line if RHO told us in advance how often he will cover from Q9 (but RHO will choose this percentage optimally, of course). It is quite easy to see that if RHO chooses to always plays the 9, we cannot improve on the correct line.

 

 

[Game theorists would say that Fred's line, plus RHO never covering from Q9, is a Nash equilibrium.]

[kenrexford]

My brain is about to explode here.

 

Lots of us worked out the you ride with the 7 showing but rise if the 9 appears. This is not all that magical. There are various ways of saying essentially the same thing. So, what difference does it make how you arrived at this conclusion?

 

I also have no idea what the approved mathematical defintion of "restricted choice" might be. However, in my layman's mind, I view the issue as one where the enemy in one situation has options but in the other does not, at least when coming up with the meaning for the term.

 

If you want weird analyses, or terms, I'd call this a sideways "V" center of gravity problem. I am considering two scenarios with one common point but two unrelated points. If I draw it out sideways, I have a point up high and left, crossing diagonally to a point low and right. crossing back to another point lower yet and left, creating a sideways V. I draw a line down the middle and notice that the same amount of lines are one both sides. However, if I fill in the space between the lines, the filled-in area left of the center line weighs more than the area to the right. So, I lean toward the left. Those two points to the left are then selected.

 

I can even analyze it quickly:

 

"I"ll play the low if the 7 is played but high if the 9 is played, because of the Sideways V principle."

[fred]

Lots of us worked out the you ride with the 7 showing but rise if the 9 appears.  This is not all that magical.  There are various ways of saying essentially the same thing.  So, what difference does it make how you arrived at this conclusion?

Some people find simplicity to be elegant.

 

Also simpler solutions tend to be easier to comprehend and learn from than unnecessarily complex solutions.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[Cascade]

I am not sure that "restricted choice" is a mathematical term.

 

My understanding and recollection is that Terrence Reese coined the term restricted choice to explain the bridge situation.

 

"Restricted Choice" applies when we are considering a player who might have two different combinations - e.g. stiff queen (or jack) or queen-jack tight - and we reason that the combination in which there was a 'restricted choice' is the one that is most likely since with the other option the player had a choice and therefore the probabilities can be discounted since sometimes we will see the one choice and sometimes the other.

 

To me, as one who overlooked the defender having a choice of plays with Q9 I think this is a "restricted choice" problem. The defender with A9 has a "restricted choice" and with Q9 has a choice.

 

As Fred said normally "restricted choice" is from among equals. Here while the queen and the nine are not equals in the normal sense they are equal in the sense that either might take a trick if the other is played. In fact whatever you play with Q9 should get you a trick. As we have seen if you play the nine declarer should play the king thus promoting your queen. While if you play the queen declarer should be inclined to finesse for the nine on the second round.

[Apollo81]

Some people find simplicity to be elegant.

Best thing Ive heard all day

[matmat]

As Fred said normally "restricted choice" is from among equals. Here while the queen and the nine are not equals in the normal sense they are equal in the sense that either might take a trick if the other is played. In fact whatever you play with Q9 should get you a trick. As we have seen if you play the nine declarer should play the king thus promoting your queen. While if you play the queen declarer should be inclined to finesse for the nine on the second round.

that hurts my head.

[kenrexford]

Similarly, if East plays the 9 you have to choose between:

 

C: Q9

D: A9

 

Clearly these are equally likely which, at first glance, would suggest you have a complete guess when East plays the 9.

 

You can resolve this guess by considering what you will do if East plays the Queen. Now you have to choose between:

 

E: Q9

F: Q

 

It is now easy to see that you should resolve the C/D guess by picking D since that allows you to get F as well. If instead you pick C, you also get E, but you already had that in the form of C :)

Not to nit-pick, but I don't understand how this way of putting it is simplicity that is elegant.

 

Paraphrasing what you said:

 

I'm resolving what to do if one thing happens by deciding what I will do if another thing happens and doing the same thing in the first situation that I did in the second situation because this allows me to get something in that other situation. Furthermore, if I opt to do something else in the first situation that would work to help me get something else in the second situation, but I'd already have that because that's the same thing.

 

That's simpler than what I and others said?

 

Paraphrasing what I and others said:

 

How do I decide whether the 9 is from A9 or Q9? With A9, he has to play the 9 to give me a guess. With Q9, he wins no matter what he plays, because the 9 gives me a guess and the Queen gives me a guess. So, I decide that he played the 9 because he had to.

 

I must not understand what "simplistic" means. :)

[fred]

Maybe "simple" is partly a function of perception. I thought my solution was simpler than most of the others because:

 

- I used less space

- I did not use any % symbols

- I did not consider irrelevant holdings

- Everything I said was on the path between the question and the answer

- I did not need to mention "restricted choice" (a concept that is not well-understood and may not even apply here)

- I explictly stated the key point (if you play for A9 you can also play for stiff Q)

 

If you like your solution better I am happy for you :)

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com