Suit combination

[kenrexford]

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

You used less space because you skipped a hell of a lot of explanation.

 

I have no idea why anyone would use % symbols. I did not.

 

Although you claim to have not considered irrelevant holdings, you actually did. You considered them, recognized why they were ireelevant, and discarded them. You just did not include your reasons in your explanation. (See "saved space" above.)

 

I have no idea what you mean by a path.

 

You are correct that you did not need to mention the term "restricted choice." However, that concept is actually quite simple, and elegantly so, for many of us. Plus, it does absolutely apply and is the reason why you play the King.

 

Although you did state what you believe to be the "key point," I still don't get the manner of thought here. You are bouncing back and forth between a posteriori's and a priori's in a manner that works out right, but I cannot follow what your argument is enough to determine if it makes sense or if your just get to the answer by lucky happenstance.

 

Finally, the reason why I mention all of this, and mentioned similar things earlier, is that you dangled before many of us who figured this thing out a while ago a "hint" of some great mysterious manner of analyzing this, gave a bizarre way of analyzing it, and claim that weird ACDC but for's and otherwhich's is a simpler way to look at something than what others suggested.

 

What I was expecting is something like this:

 

Fred: "Very well done! I'm sure many would agree that this is quite a sublime and rare anomaly, eh?"

 

Us: "Quite!"

 

Some: "Still don't get it... :) "

 

Fred: "Well, try this way of looking at it..."

 

Instead, it read more like...

 

Fred: "Y'all got it, but you did it a weird way. My way is much simpler and more elegant."

 

Us: "What do you mean?"

 

Fred: "Well, take an albatross, let it fly backwards ten yards, count the wing strokes, subtract that from the wingspan, and you will know that grass is clearly green."

 

Us: "Uh, OK."

[Stephen Tu]

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.

 

Fred, your reasoning may be "simpler", but frankly it's just incomplete.

 

Using your terminology for C, D, E, F,

(C = Q9, 9 played,

D = A9

E = Q9, Q played

F = Q)

, you essentially argued that the choice was between

C+E and D+F.

 

But those aren't the only choices one can make!

 

One can also play for combos D+E, and C+F.

 

Since you didn't demonstrate that D+F is also greater than those other options, then your reasoning must be considered incomplete.

 

And my more complete explanation showed the conditions where D+E > D+F, and that this only applied vs. perfect defense where RHO knows you have 6 cds rather than 5, and knows that you will choose the right play. Do you not agree that a non-omniscient defender with Q9 will want to cover if 5 cd suit is a possibility?

 

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

 

That's not the key point. If the guy with Q9 never covers, you can also play for stiff Q, and hook on the 9 for the exact same percentage result. The key point is that in practice Q9 will cover sometime, so Q9 is less likely than A9 when the 9 is played.

[fred]

Stephen - the assumption was best defense.

 

Ken - apparently we do not live in the same universe.

 

Sorry guys. I have had enough of this.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[helene_t]

Don't worry Fred, we will set up a committee with a mandate to define the meaning of the term "simple". You don't have to bother about it.

[fred]

Don't worry Fred, we will set up a committee with a mandate to define the meaning of the term "simple". You don't have to bother about it.

Thank you my dear, but I was actually hoping you would weigh in on the math ;)

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[benlessard]

3 playing LOW (the 8) to the king is inferior because of (9----AQ7)

 

Oups my mistake since low to the k win when (Q---A97) its equal. In practice leading the 8 to the K is probably better because RHO might rise with AQ7 (not likely but its possible)

 

I liked Fred very simple explication.

 

For me the interesting part of the problem is that an option by the defense on another holding (A97---Q) or (A7----Q9 with the play of the Q or the 9) is the key to understand why going up with the K is better.

 

But i still think solving it with the restricted choice is still a simple and sound solution.

 

When similar problems will come up it will be simpler for me to understand it in RC term then a "duplication of holding" method.

 

Anyone here who doesnt understand RC should devote some time to it because its quite important and its not that hard to understand.

 

The 'classic' definition of RC is between card of equal values (that are assumed to be played with even frequency) but the RC method also applies when equal arent played with even frequency or when they are not quite equals.

 

 

ATxxx

 

K98x

 

playing low from hand and LHO play the J.

 

its a (QJ) VS (J---Qxx) case

 

Even if LHO (a beginner who doesnt mixed his play) with (QJ) play the J 80% and the Q 20% its still a restricted choice. Just that finessing for the Q become a small winner instead or a big winner.

 

However the day LHO will drop the Q on first round now the finesse for the J is going to be a big winner. So the fact that with Q9 the 9 and the Q arent played with the same frequency doesnt say its isnt a RC.

 

 

[hv=n=saqxx&w=sktx&e=sjx&s=sxxxx]399|300|[/hv]

 

you play the A and RHO pitch the J

 

here we have a falsecard that is a RC problem for declarer (between KJ, Jx) (assuming declared is dead anyway if the J is stiff).

 

IMHO its a RC without equals cards

 

but some will say that since the J is worthless then the J and the spot are equal.

[Stephen Tu]

But i still think solving it with the restricted choice is still a simple and sound solution.

 

I don't think I would really call the original problem a restricted choice problem since it doesn't involve equals. Restricted choice is a specific application of Bayesian reasoning, the complete reasoning behind playing the K is similar, also Bayesian, but I wouldn't necessarily call it RC.

 

The principle to learn is that if a defender has a choice of plays from a holding, you should assign a probability to each choice, and calculate accordingly. Figure out why a defender ought to do something, and how often. Often you find a range of values for the choice where declarer should do one thing, but if the defender strays outside those ranges declarer should do something else. This can affect practical play of combinations such as the original, e.g. best play after the Q covers, which is different between omniscient "perfect" defenders who know declarer has 6, and non-omniscient "perfect" defenders who also have to cater to declarer having 5 (where you should cover with Q9 doubleton).

 

[hv=n=saqxx&w=sktx&e=sjx&s=sxxxx]399|300|[/hv]

you play the A and RHO pitch the J

 

This is the wrong example. 2nd round either the K or T is played, you don't really have any guessing to do. The one you are looking for is:

 

[hv=n=saqxx&w=sktx&e=sjx&s=sxxxx]399|300|[/hv]

 

And the idea is you have a choice between playing RHO for JT or KJ after the J falls. RC will show that you should cater to KJ.

[Codo]

Stephen - the assumption was best defense.

 

Ken - apparently we do not live in the same universe.

 

Sorry guys. I have had enough of this.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

Sometimes you open a can of worms and you find? Worms.

 

I had waved my "I did not vote the worms" sign to warn you, but it was forbidden.

 

Thx for this threat, it made me a better bridge player. Just a little bit better, but it helped.

[cherdano]

Sorry guys. I have had enough of this.

Me too ;)

Thanks for a very funny post and a suit combination that reminds me why I play bridge...

[helene_t]

I was actually hoping you would weigh in on the math ;)

It's a little embarasing for me as a mathematician, with specialization in probability theory and operational analysis, to admit this, but I feel rather helpless when confronted with this kind of problems. The complexity just makes me dizzy, since the only way for me to analyze it would be to spell out the entire game matrix, decision tree or whatever. Shortcuts such as probability tables (I don't know them by heart) and restricted choice can help, but often I'm unable to decide if those shortcuts apply.

 

Intuitively, it is expected that playing the king has more merrit when RHO plays the 9, since if he played the 7 there is a case for saving the king as a guard against the nine. Also, it feels like RHO would be more likely to play the 9 without the queen since the 9 and the queen are equally expensive, so the 9 would tend to deny the queen and thus making it necesary to play the king, hoping for Q or Q7 by LHO. This is not a proof to me because I still wonder if RHO may nullify said effect by sometimes false-carding the 9 when he wants us to cover. Maybe this false-carding argument is irrelevant for this analysis, but that is the kind of things that are not obvious to me.

[sceptic]

if the small card is the 9, then east does not have 4 cards or he would play the 7 as he can see the 810J, so the splits you can have are 2/2 or 3/1, if west has AQ or AQx or AQxx you are stuffed anyway

 

I think to make 5 tricks you need to finese the jack, beacuse if east has 79Q you have no second shot at the finese also if east has xQA it works

[Gerben42]

I would just like to say I enjoyed this suit combination.

 

;)

[benlessard]

You play the A and RHO pitch the J

 

This is the wrong example. 2nd round either the K or T is played, you don't really have any guessing to do. The one you are looking for is:

 

I just didnt put the high spot in LHO hand but its almost the same example as the 1 you show.

 

[hv=n=saq32&w=skt9&e=sj8&s=s7654]399|300|[/hv]

 

After the J from (J8) or (JT) or (J9) or (KJ) .Declarer will play have a guess between playing low to the Q or ducking all around. I think the correct solution is to play low to the Q unless RHO rarely falsecard.

 

 

 

But i agree for the rest of Stephen post.

[Stephen Tu]

It's a little embarasing for me as a mathematician, with specialization in probability theory and operational analysis, to admit this, but I feel rather helpless when confronted with this kind of problems. The complexity just makes me dizzy,

 

You know, it's really not that complex a problem, if looking only at the question of whether to play K if RHO plays the 9, ignoring the question of what to do if RHO plays the Q (which is more complex IMO).

 

If the 9 shows, there are only 2 relevant possibilities where you can pick up the suit, A9 and Q9 with RHO, because with all other combos you can see that the defender can always force the loss of 2 tricks. These are equal only if RHO plays 9 from Q9 100% of the time as he would with A9. But he won't (since covering can generate a 2nd trick vs. declarers who don't know the optimal play and might take a 1st rd hook, and can't ever hurt him unless he does it too often), so rising with the K is better, Q.E.D.

 

All you really have to do is convince yourself that it's stupid to play anything but the 9 from A9 (if pop ace, declarer can't go wrong), and that RHO will want to play Q from Q9 some non-zero amount.

[fred]

It would be nice if one of the math stars out there would say something about the restricted choice issue.

 

Besides Helene, I seem to recall that both Cherdano and Hannie are graduate students in math. I have no idea about Stephen's credentials, but at least he seems to take himself seriously - for all I know he has a right to.

 

It sounds to me like a lot of people are confused about this in the sense that I believe that this problem has nothing to do with what bridge players normally mean when they refer to "restricted choice". This has managed to confuse me too - I thought I knew what restricted choice meant and I thought it was a concept that could be defined/explained using mathematics (Bayes theorem if I recall correctly).

 

But it is not like I think I really know what I am talking about. Maybe one of you who does seem to know what you are talking about can enlighten the rest of us :P

 

If so please try to keep it simple!

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[kenrexford]

The issue with the refusal to accept this as a restricted choice problem is that folks want to define "equal" with reference to some arbitrary standard.

 

A Queen and a Jack are "equal" in the sense that, when together, they both are less than all outstanding honors above them and both are greater than all outstanding cards below them. They are not truly equal, they are simply equally positioned for the person holding them. If one person did not hold both, they would not be equal.

 

In this situation, the Queen and 9 are "equal" in the sense that the play of either card at the critical moment will yeild what appears to be a 50-50 guess for Declarer. If Declarer guesses wrong in either situation, the other will take a trick. If the Declarer guesses right, neither will take a trick. So, each has equal trick-taking power.

 

The other problem is that either card creates its own artificial image of alternative layouts, and those layouts are not the same. Playing the 9 creates an artificial image of having A9, whereas playing the Q creates an artificial image of just the Queen.

 

So, you end up with three possibilities, when RHO plays the [Q/9]:

 

1. [Q/9][Q/9]

2. A9

3. Q

 

Two of three times, the [Q/9] is resolved as one or the other. So, restricted choice applies.

 

This sort of looks like a cat-in-the-box situation. In a manner of speaking, the three options are:

 

1. [Q/9][Q/9]

2. A[Q/9]

3. [Q/9]

 

However, 2 and 3 will manifest the nature of the [Q/9], whereas 1 will not. That does not change the odds a priori.

[helene_t]

Restricted choice applies whenever one assumes that a defender, if he held two equivalent cards, would chose either with 50% probability. One could generalize this to cases where the probabilities are something else, or one could generalize it to three equivalent cards with each 33.3% probability etc.

 

This is indeed a special case of Bayes' theorem.

 

I don't think it's correct to apply restricted choice here. Even if 9 and 7 are equivalent DD and RHO knows that, we cannot a priori assume that RHO will chose either with 50% probability, because he knows that we do not know that 9 and 7 are equivalent. He knows that we will make different inference from 9 and 7 and he may exploit that.

 

(Or substitute 9+Q or whatever for 9+7).

 

Ultimatively, Bayes' theorem should be applied when responding to any mixed strategy, even if the parameters are not known apriori but change during itteration while one is looking for the Nash equilibrium. I don't think the term "restricted choice" is used that broadly, but I might be wrong.

[hotShot]

We don't know much about defenders hands, but what do the defenders know about our cards?

Does defender know that Q and 9 are equivalent?

He won't know that his partner holds the Ace. His partner could hold the single K or only small cards.

 

I don't think that this situation is equivalent to holding touching honors in your own hand.

 

But knowing that he would never play A from A9 and sometimes Q from Q9 makes playing the K a little more attractive.

[fred]

We don't know much about defenders hands, but what do the defenders know about our cards?

Does defender know that Q and 9 are equivalent?

He won't know that his partner holds the Ace. His partner could hold the single K or only small cards.

 

I don't think that this situation is equivalent to holding touching honors in your own hand.

 

But knowing that he would never play A from A9 and sometimes Q from Q9 makes playing the K a little more attractive.

There is a big difference between real life problems and problems like the one that was presented here.

 

Many suit combination problems are "solvable" only if you assume that the defenders can see your cards and never make a mistake.

 

Of course this is not realistic, but it makes it possible to solve these problems. If you do not make this assumption you have to make other (subjective) assumptions like how often a defender who is in the dark will think you have what you actually have, how often he will think you have various other things...

 

Obviously such things vary greatly from defender to defender and from hand to hand.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[ralph23]

Sorry guys. I have had enough of this.

Me too :)

Thanks for a very funny post and a suit combination that reminds me why I play bridge...

Thanks Fred, I thought it was a very interesting combination also. I didn't mean to start a firestorm of controversy. But --Remember, no good deed goes unpunished. C'est la vie.

 

It is certainly not a "textbook" case of RC -- i.e. one that you would show to a beginner to explain the concept of RC. There, I would use a singleton Jack or Queen, and compare it to a doubleton QJ. In that textbook case, the play of the singleton J is "restricted" by law; any other play is a revoke.

 

In the A9 case, the choice is "restricted" because the defender knows how to play. He could legally play the Ace; but he won't. His choice is "restricted" it seems as much as the defender holding the singleton Jack in the paradigm case.

 

Likewise, the Q9 are certainly not "equal" in the sense of QJ's being equal. But the play of either one could be "right"!

 

Playing the 9 from Q9 may persuade South to rise with the King, and lose to the Ace. This strategy will, in our case, defeat the "wise" declarer.

 

Playing the Queen may persuade South to finesse the 8 on the next round of the suit, having induced the false belief that East originally held a stiff Queen, and West started with A97. I must admit, the play of the Queen from Q9 would certainly fool me!!

 

I don't know but it seems like at least an analogue to RC, if not an actual example of the same mathematical principle that RC is based upon. Whether one CALLS it "RC" or not -- well, that is a matter of semantics, I think.

 

Whether the same logical or mathematical principle that underlies RC, also underlies this situation, is IMHO an interesting question -- and frankly one that I have never seen before in the literature.

 

Thanks again for a very thought-provoking problem.

Edited October 19, 2007 by ralph23

[ralph23]

All you really have to do is convince yourself that . . . . RHO will want to play Q from Q9 some non-zero amount.

That really sums it up for me. If all defenders always play 9 from Q9, 100% of the time, then (and only then) ex hypothesi A9 and Q9 are equivalent. Since all defenders don't play that way, 9 is more likely to be from A9 than Q9, because the Q9 defender had a "choice" -- he was not "restricted" by law or by simple common sense -- of plays.

 

Whether it's an apt opportunity to apply the nomenclature "Restricted Choice" to this or not, I don't know. It seems to exemplify the same principle as does RC to me, that's all.

Edited October 19, 2007 by ralph23

[ralph23]

So the fact that with Q9 the 9 and the Q arent played with the same frequency doesnt say its isnt a RC.

This seems exactly right. Is it stated somewhere in the definition of "Restricted Choice" (whoever the hell defines these things!!) that it must be exactly a 50-50 chance that defender will discard either Q (50%) or J (50%) from QJ?

 

All that seems to matter (to paraphrase Stephen) is that it is not 100-0, i.e. that one term is non-zero.

[whereagles]

Many suit combination problems are "solvable" only if you assume that the defenders can see your cards and never make a mistake.

Which is why I look at their faces instead of the book :)

[Stephen Tu]

It would be nice if one of the math stars out there would say something about the restricted choice issue.

 

Whether to call it "restricted choice" IMO is a bridge terminology semantics issue, not a math question, since RC only appears as a term in bridge articles AFAIK. This problem certainly falls under the general category where you have to apply Bayes theorem so it is related subset if not the same subset.

 

In any case, I think it should be clear that the crux of the specific problem of what to play on the 9 is that defender has a choice of plays from Q9 and doesn't really have much to do with "picking up stiff Q" as your original argument stated. A declarer could choose a line of always picking up the stiff Q, yet still ducking if the 9 appeared. This would just fail more & more often vs. playing the K as the defender increased the frequency of choosing to cover with the Q from Q9, thus decreasing the probability that the 9 represented a holding of Q9 instead of A9.

 

Besides Helene, I seem to recall that both Cherdano and Hannie are graduate students in math. I have no idea about Stephen's credentials

 

I don't see what credentials should have anything to do with it. To me, for a real mathematician, math/logic arguments should stand on their own intrinsic merit, not what credentials the person holds. Credential evaluation is only a (poor) substitute for evaluating the argument itself if you have insufficient knowledge/training to do so.

 

I personally put only a little faith in credentials since I have both met college dropouts who are simply brilliant, & also people with graduate degrees who are mostly incompetent. I will give people with the paper credentials the presumption of competence at first, but once they start their discourse their arguments must make sense to me.

 

In any case, my degree is in computer science not math so perhaps that disqualifies me in your view, but to me this is not an advanced math problem, and as someone who started taking university math at 13 I like to think I am at least better than most. If one of the math majors wants to start a topic on topology or something like that I would keep my yap shut as I know nothing, but this was a simple problem.

[fred]

Stephen - let's just agree to disagree!

 

The last word is all yours...

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com