[fred]

Hints follows:

It is possible to come up with the answer by purely by logical deduction. No math is required. Restricted choice does not come into play. The easiest path that I have found is very concise - less than 10 short lines of text. This is not a trick question - it is just a normal suit combination problem (but an unusually interesting one).

The above hint is unlikely to help very many people actually solve the problem, but it might save some people from typing more very long analyses.

Most likely I will provide some more hints in a few hours

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[ralph23]

benlessard, on Oct 18 2007, 08:47 AM, said:

Im surprised most dont understand the problem at all.

1 first of all you need 4 tricks.

Don't you need five (5) tricks per the problem statement?

Maybe I didn't understand the problem at all...

[ArtK78]

I am sure that the 4 tricks statement was just a typo, as was explained in the second half of the sentence "so all case when you are sure to lose 2 trick are pointless."

You have to win 5 tricks. So you are not going to take any line that guards against 3 losers, since even 2 losers is too much.

[Apollo81]

Now that I read Fred's hint, I don't think I'm right anymore, oh well =)

---
Sometimes.

The only holdings that matter are Q7, A7, AQ7, Q9, and A9.

If RHO plays the seven then it is right to play low two in two out of three cases clearly making low the percentage play.

If RHO plays the 9 (he could not do so from A97 or AQ9 without possibly blowing a guaranteed trick) it appears that we have a 50-50 guess, but do we? Let's consider RHO's decision.

If RHO always plays the 9 from Q9 then declarer has a 50-50 guess when he sees the 9 but has a 100% play on the next round when RHO covers with the queen (which must be singleton). If RHO always plays the Q from Q9 then declarer has a 50-50 guess on the next round when RHO covers with the queen, but has a 100% play when he seems the 9 on the first round (which must be from A9). In each case, a deterministic play from RHO will ensure 75% success rate from declarer.

If RHO plays either the Q or 9 randomly, then it is right for declarer to play the K in the problem given by restricted choice (the 9 is twice as likely to be forced from A9 as chosen from Q9). Also if the Q is covered then it is right for declarer to finesse for the 9 on the next round (stiff Q twice as likely as Q9). Declarer will succeed 66% of the time in this case.

Note that if RHO assumes declarer has used the above reasoning, his choice of Q or 9 from Q9 is irrelevant; declarer will fail in both cases.

There is one more factor to consider. Perhaps RHO is unaware of how many cards in the suit are held by each player. In this case he would be hesitant to play the Q for fear of crashing a singleton honor in partner's hand. Let's say RHO plays the Q some percentage of the time P, 0<P<50. It is still right for declarer to play the K, although his chance of success will be somewhere between 50% and 66% depending on the value of P. Additionally, his chance of success when RHO does cover with the Q goes up to somewhere between 66% and 100%, depending on P. As long RHO sometimes plays the Q, the K is the percentage play.

[hotShot]

Interesting use of "appropriate".

Since it is impossible to get more informations from bidding and playing the side suits, this problem is very restricted.

It is a tiny bit better never to play the king.
This is, because the 5% of the boards where RHO is void in ♠ can be excluded. So the average ♠ length for RHO is a little above 2. So the Queen is a little more often with RHO. RHO holding AQ9 is a little more likely due to this fact, but we would have seen the Q on the J than.

Basically it would not matter much, if humans could decide this truly random. Playing the King based on subjective randomness, will lead to worse results most of the time.

[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]

benlessard, on Oct 18 2007, 12:47 PM, said:

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]

awm, on Oct 18 2007, 04:17 PM, said:

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]

Quote

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]

fred, on Oct 18 2007, 05:34 PM, said:

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]

ralph23, on Oct 18 2007, 09:53 PM, said:

fred, on Oct 18 2007, 05:34 PM, said:

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]

Quote

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]

Stephen Tu, on Oct 18 2007, 10:07 PM, said:

Quote

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.