AKQ9xx vs xx

[foo]

Justin's post on Restricted Choice should be pinned.

 

Nice job Justin.

[glen]

even better he should blog it

[han]

Yes, very nice explanantion. Perhaps correct spelling before stickying it B)

[Stephen Tu]

Yep. Those advocating the play for drop against the expert but the play for split honors vs a non expert are basically saying that the expert is predictable enough that you should =trust an expert to always Falsecard=.

 

No, to prove the drop is superior, it is sufficient to trust the expert to falsecard *enough*, here playing the J at least ~1/9 of JTx and playing randomly from JT tight, *always* is not necessary. But an expert really ought to always play either the J or the T here. Why would a known expert ever not want to falsecard vs. a person like you? *It is a ZERO COST play, and he gets an extra trick from you every time he does it!* Experts don't pass up freebies. Against the people like me who know that he is an expert, and will play for the drop, he has lost nothing, because if he doesn't falsecard I was still going to pick up the suit. And he picks up massively vs. people like you who don't know how to analyze the situation, and laughs when you insult him and hook into his T when the suit was 3-2 all along.

 

*On this layout (JTx behind declarer, xx in dummy), there is never an advantage in not falsecarding by playing x*. Falsecarding anything less than the maximum only reduces the # of tricks vs. mistaken declarers. Not falsecarding never takes more tricks on this layout. If you play x, everyone from rank novice to expert is playing the drop.

 

Vs. me, it doesn't matter what he does, I pay off to stiff J.

Vs. you, who is picking up stiff J, why would he not want to falsecard all the time to counter this?

 

There do exist suit combinations where best defense involves falsecarding less than 100% of the time, where doing it too often is exploitable just like not doing it often enough. THIS IS NOT ONE OF THEM.

 

Now when the card that hits the table is one that =could= be a Falsecard, the opponents can not so easily assume that the expert has Falsecarded. After all, sometimes he does and sometimes he doesn't...

Never helps him on this layout. If it could be a falsecard with any reasonable frequency I am supposed to play for the drop, so this layout I pick up JTx & Jx, lose to stiff J. You on the other hand want to hook, so you pick up stiff J, lose to JTx/JT playing the J, so the opponent maximizes his advantage by playing the J as often as he can.

 

Even if defender is falsecarding much less than always for no particular reason, it doesn't have to be much for the drop to be superior because there are just so many more JTx combos than J stiff combos. They have to falsecard w/ J ~< 1/10 of the time, (~ < 1/5 time overall including falsecards w/ T) for hooking when an honor appears to the left to be better.

 

Since this strategy rates to work more often than either never Falsecarding or always Falsecarding, it would seem clear that an expert's defense will indeed evolve in this direction.

On some combos, yes. For example QJ doubleton tight behind declarer. Playing the Q all the time is as exploitable as playing the J all the time, it is right to mix it up.

 

BUT THIS COMBO IS NOT THE SAME. Best defense from JTx is to mix it up between playing the J & playing the T, BUT YOU NEVER HAVE TO PLAY THE x to be optimal here.

[Blofeld]

Hence the term, "mandatory falsecard".

[skjaeran]

If you want to throw in the Jx/Jxx combos

Jxx can't come into play either, since the T will then show up when you play a small from xx (originally) after laying down the ace.

[Edmunte1]

Justin isn't this a restricted choice situation as well? I must admit I have never fully understood the rationale behind restricted choice, but if lho drops the J or T doesn't restricted choice suggest you should hook?

The classic restricted choice holding is AKT9x opp xxxx. When you play the ace the queen drops behind you. The reason you should hook is that the possible options are

 

QJ tight

Q stiff

 

QJ tight is slightly more likely (since a specific 2-2 combo is more likely than a specific 3-1 combo), however the Q will be played from only half the QJ tights combos, so stiff Q is much more likely than half of the QJ tights.

 

 

More, for understanding the restricted choice, here are the probabilities for the holdings:

23 - QJ holding is 40.7% / 6cases = 6.78%

23J - Q holding is 49.7%/8 cases = 6.21%

 

So it seems that playing for the drop is more likely, and most of the people don't understand that explanation that holding QJ tight the defender has to choose from playing either the Jack or the Queen because they don't think about another case

 

23Q-J holding is 49.7%/8 cases = 6.21%

 

I'll try to explain it differently.

Suppose that your general strategy in 100 boards when you hold AKT9x opp. xxxx (or similar holdings) after cashing Ace, and the QUEEN drops - to play for the drop, you'll win in 52/100 boards (6.78/(6.78+6.21).

 

BUT this strategy doesn't take into account the case when you cash the Ace and JACK drops.

 

If we consider a general strategy when we cash the Ace and a Honour drops (from our point of view Jack and Queen are equals, we can call them Q1 and Q2) then:

 

- playing for the finesse will win in (6.21 (Q1 drops)+6.21(Q2 drops))/(6.78+6.21+6.21)=64.7%

- playing for the drop will win in 6.78/19.2 =35.3%

[lilboyman]

In "Bridge Odds fot Practical Players", by Kelsey & Glauert, it is written on page 24 that "If two honour cards in a suit are missing,there is a 52% chance that they will be split between the opponents, a 48% chance that the same opponent will have them both." Contining to to page 95 where the authors discuss restricted choice it is written, "When a defender follows suit or wins a trick with one of two equivalent cards, the probability of his having both cards is halved." Hence, it would seem that there is only a 24% chance that LHO has the ten and the finesse should win 76% of the time.

[Stephen Tu]

Oh wonderful, another person who can't count ...

 

It is true, *before we play any cards*, that LHO will have J+T only 24% of all possible times. But once we play a round and RHO follows & LHO plays the J, the sample space collapses. All the hands where RHO might have J disappear. The possibility RHO is void disappears. The possibility RHO has stiff T disappears. Also, your reasoning presumes that LHO will play the J whenever he has it, even when he started with irrelevant holdings like JTxx, Jxx, Jx, where it won't matter what we do and the opp wouldn't do it anyway.

 

The ONLY holdings where it matters are JTx, JT, J, to do the analysis you halve the first two because of restricted choice principles. And you multiply the first one by an estimate of whether your opponent knows this is a mandatory falsecard situation, if an expert this would be 100% or close to it, if a novice you might be close to 0%.

 

The odds of LHO having J and T having no prior knowledge, is not the same as asking how often LHO has the T already knowing that he has the J and either chose or was forced to play it.

[gwnn]

"Even though while I read through the thread I saw many many words I didn't really understand, let me just write down what I think is right. After all, all that incomprehensible talk about falsecards and Latin words and "%" signs might just be nonsensical"

[ASkolnick]

Interesting post. I am not sure either is right per se. If expert always false cards, then you can ignore any false card. If a person never false cards, you are still ignorning the false cards.

 

Jlall, A priori, you are right its 5 time as likely, but once a 2nd spot is led from dummy, you are now limited to much less in combinations. So, there are not 3 JTx's at all, only 1 JTx since you have already seen the two of the remaining spot cards.

 

You have 5 cards, J T A B C (ABC are spot cards)

 

On round 1, card A and the Jack are played

On round 2, card B is played

 

Now the only possible holdings which are relevant for LHO are

 

JTC or (JTA and JTB are not possible anymore)

JT

J

 

so, it is twice as likely not 5 times as stated by many before this. You should still play for the drop, but it is much closer than indicated in any one else's proof.

 

Similar to the situation with restricted choice, the reason the hook becomes a better play is because once left hand opponent shows up with a 2nd card, the cards cannot be 1-3 offsides.

 

This problem seems very similar to playing Axxx opposite QT98x for 1 loser. The correct play is the Ace and low, only because of single case of KJ offside.

[ASkolnick]

Technically, proof not quite correct. It is slightly over 2-1 since there will be 12 card places that LHO holds versus the 11 card places that RHO has.

[dburn]

Sadly enough, it is the mathematical abuse by people who don't have a clue that gives mathematicians a bad name. I hope dburn isn't reading foo's "mathematics".

dburn has been reading foo's "mathematics" with much amusement - he was unaware of the existence of this thread until Frances referred to it in another thread, and too lazy to search for it until it appeared on the first page of threads in this forum.

 

He has little to add to the excellent analyses by Stephen Tu, Jlall and others. He would make this general observation.

 

The Principle of Restricted Choice may be summarised in a few simple words: always assume that a player has done something because he had to, not because he chose to.

 

In the simple case of

 

AK10xx

 

xxxx

 

one assumes when an honour drops to the right of the ace-king (from South's viewpoint) that it was played from necessity (because it was a singleton) rather than from choice (because it was selected from two honours doubleton). Thus, one finesses on the second round.

 

In this case of

 

AKQ9xx

 

xx

 

when an honour appears to the right of the ace-king-queen, one again assumes that it was played from necessity rather than from choice. But the positions are different because in the second case, LHO "must" play an honour from J10x in order to create a chance to take a trick in the suit. Thus, one plays for the drop because the holdings from which an honour "must" be played are J10x, J10 and singleton honour. Obviously, the a priori probability of J10x and J10 combined divided by two (because either honour will be played half the time) exceeds the probability of a singleton honour.

 

The same sort of reasoning applies to this case:

 

AK103

 

Q872

 

The ace is cashed and the nine appears to the left of the ace. If this is a singleton nine one must cash the king next; if from J9xx one must play to the queen. But the nine "must" be played from J9xx in order to create the possibility of taking a trick in the suit, and because the a priori probability of J9xx is three times the probability of singleton nine, one plays to the queen.

 

dburn awaits with interest the view of foo on

 

Q432

 

AJ85

 

When a small card is led to the jack, the nine appears from LHO. What should one do on the second round to avoid losing a trick in the suit?

[Halo]

As far as touching cards go, you can also get to the correct answers just by imagining your eyesight isn't too good, and you can't tell which was played.

 

This approach also saves you dividing things by two, a bonus typically for the sort or people whose eyesight is poor.

[MFA]

Jlall, A priori, you are right its 5 time as likely, but once a 2nd spot is led from dummy, you are now limited to much less in combinations.  So, there are not 3 JTx's at all, only 1 JTx since you have already seen the two of the remaining spot cards.

 

You have 5 cards, J T A B C  (ABC are spot cards)

 

On round 1, card A and the Jack are played

On round 2, card B is played

 

Now the only possible holdings which are relevant for LHO are

 

JTC or (JTA and JTB are not possible anymore)

JT

J

 

so, it is twice as likely not 5 times as stated by many before this.  You should still play for the drop, but it is much closer than indicated in any one else's proof.

 

...

Your argument doesn't hold, I'm afraid.

 

JTx still is three times as likely as each of the other holdings. Righty will always part with two x's, so this doesn't change anything.

 

The situation can be viewed in two ways:

 

1) I don't care (=notice) which x's, I have seen. They are all of equal rank, so it cannot matter to me in any way, which were played.

So I'm right where I started, with 3 JTx.

 

2) Ok, I look at the x's. Righty has played 2 out of 3 of them, so this constitutes a restricted choice situation among the x's! Because there are three equal cards (x's), it's not 2 but 3 times as likely that lefty will have the last spot card.

[dburn]

Interesting post.  I am not sure either is right per se. If expert always false cards, then you can ignore any false card.  If a person never false cards, you are still ignorning the false cards.

 

Jlall, A priori, you are right its 5 time as likely, but once a 2nd spot is led from dummy, you are now limited to much less in combinations.  So, there are not 3 JTx's at all, only 1 JTx since you have already seen the two of the remaining spot cards.

 

You have 5 cards, J T A B C  (ABC are spot cards)

 

On round 1, card A and the Jack are played

On round 2, card B is played

 

Now the only possible holdings which are relevant for LHO are

 

JTC or (JTA and JTB are not possible anymore)

JT

J

 

so, it is twice as likely not 5 times as stated by many before this.  You should still play for the drop, but it is much closer than indicated in any one else's proof.

 

Similar to the situation with restricted choice,  the reason the hook becomes a better play is because once left hand opponent shows up with a 2nd card, the cards cannot be 1-3 offsides.

 

This problem seems very similar to playing Axxx opposite QT98x for 1 loser.  The correct play is the Ace and low, only because of single case of KJ offside.

I am not sure what Hannie would have to say about this - it is not the kind of reasoning that gets mathematicians a bad name, but it is deeply flawed reasoning nonetheless.

 

To see this, imagine that you have this suit:

 

AQ987

 

J10653

 

I will make you this solemn assurance: the suit is not divided 3-0 (nor 0-3). You lead the jack, and West plays the two. You... well, what do you do?

 

If you are ASkolnick, you might argue that once West has played the two, there are only two possible cases: West began with 42 and you should play for the drop; or West began with K2 and you should finesse. Since both cases are equally likely a priori, you face a guess at this point.

 

You don't, of course, since from 42 West might have played the four, whereas from K2 he had to play the two. The same sort of reasoning applies to the As, Bs and Cs in ASkolnick's post - if East plays the three and the four (in either order) to consecutive tricks, one assumes that he had to do this from 43 doubleton, rather than that he chose to do it from 543 trebleton.

 

Since Hannie will by now be wondering what happened to the poetry, I offer:

 

There once was a fellow called Bayes

Who led us all into a maze.

But once we learn rules

Are not masters but tools,

We'll quickly emerge from the haze.

[Stephen Tu]

Askolnick your reasoning doesn't work for the reasons dburn & MFA stated, that RHO is allowed to play randomly from the spot cards. If they are say 234, and RHO plays 2 & 3, and you say you only want to count JT4, reducing by 1/3, you also have to reduce JT by 1/3 and J by 1/3 since RHO with all of 432 is supposed to randomly play 2 of the 3, and 2/3 of those holdings he will play the 4 as one of the cards.

 

Now, if RHO can be counted on to always follow up the line with his lowest cards, then I suppose if he plays 2 & 4 then the drop is 100%!!! I hope most of my opponents aren't so accommodating but I've never taken close statistical watch of this. So the lesson for the defenders is to randomize your spot card play just as you do your touching honors play. I know sometimes I don't do this quite as often as I ought to be doing.

 

Now will people please stop posting factually if they aren't *damn sure* they fully understand how this all works? If you don't fully grasp this and think something is wrong with an analysis, why don't you post in the form of a question, ask "why am I getting a different answer from you guys, this is what I think, what's the flaw in my logic or am I right?". If you are getting different answers than the vast majority, more likely you are wrong than everyone else is wrong. If you are right and can produce a convincing argument without holes, or can clearly identify the holes in the majority reasoning, the logical among us will be swayed to your side. But coming here & just putting up junk statements with little justification is annoying.

[ASkolnick]

I understand the problem with the logic with the

 

AQ543

 

JT976

 

but if I have Kx. I do not have a choice at that present moment. And once LHO follows, the possibility of 0-3 offsides has now been eliminated.

 

 

Instead of being insulting, here is the problem with your logic about eliminating 1/3 of JT possibilities.

 

Let's talk about possible holdings which are relevant:

 

Let me be more specific. You have the

J T 4 3 2.

 

J Finesse

JT Drop

JT4 Drop

JT3 Drop

JT2 Drop

 

Will determine whether you play for the drop or the finesse.

 

Give me any order you play the cards, I don't care if you are false carding or not.

 

If RHO plays the 2 and the 3,

 

J Finesse

JT Drop

JT4 Drop

JT3 NOT POSSIBLE ANYMORE

JT2 NOT POSSIBLE ANYMORE

 

If RHO plays the 2 and the 4,

 

J Finesse

JT Drop

JT4 NOT POSSIBLE ANYMORE

JT3 Drop

JT2 NOT POSSIBLE ANYMORE

 

If RHO plays 3 and the 4,

 

J Finesse

JT Drop

JT4 NOT POSSIBLE ANYMORE

JT3 NOT POSSIBLE ANYMORE

JT2 Drop

 

In no cases does playing the two spot cards eliminate a case for JT tight since there is only once case of JT tight which would be JT opposite 234 regardless of which card he plays.

 

Why would I need to eliminate 1/3 of the JT possibilities?

 

If he plays the 42, JT tight is still just as likely

If he plays the 32, JT tight is still just as likely

If he plays the 43, JT tight is still just as likely

 

And besides, there is only 1 holding of JT tight, so how can I eliminate 1/3 of those cases?

 

In all cases there are two cases where its right to play the drop, and 1 case where it is right to take the finesse. So, why is it relevant what order RHO plays in? We can take into consideration the number of cards left to be seen if you want to adjust it slightly. 11 cards remain of RHO versus 12 cards remain of LHO.

 

Now, if you tell me, "the person always gives count" or "the person falsecards x% of the time" then you are talking differently.

 

I am not sure how I can be more explicit than this.

[Finch]

Now, if RHO can be counted on to always follow up the line with his lowest cards, then I suppose if he plays 2 & 4 then the drop is 100%!!! I hope most of my opponents aren't so accommodating but I've never taken close statistical watch of this. So the lesson for the defenders is to randomize your spot card play just as you do your touching honors play. I know sometimes I don't do this quite as often as I ought to be doing.

This is a very important point.

Once they've learnt not to give true count all the time, people _do_ tend to play blindly up the line in this kind of suit. It's well worth watching the pips very carefully indeed - this inference can be more important than all the 'how often would they falsecard' debates.

[Hanoi5]

Are they foreign?

 

Is this bridge xenophobia?

[Finch]

Are they foreign?

 

Is this bridge xenophobia?

I thought this was already explained.

 

Justin's normal algorithm for deciding if his opponents are likely to falsecard or not is based on how good he thinks they are. If he doesn't recognise them, he will usually assume that they aren't any good, as he thinks he recognises most good players. However, if they are foreign visitors then

 

i) he probably won't recognise them however good they are (unless they are really world class names)

 

and

 

ii) they are (on average) likely to be stronger players, because they've taken the trouble to travel a long way to play in the tournament

 

and so are more likely to falsecard then the initial assumption (I don't recognise them therefore they won't falsecase) indicates.

[jdonn]

I am not sure how I can be more explicit than this.

You were very explicit, but still wrong. The fact that the counter-example wasn't 100% analogous doesn't matter. You are making the standard wrong argument against restricted choice. I was about to try to fix it, but I wrote it out then looked and Stephen had already written almost exactly the same thing. So what can I say, good luck to you.

[Stephen Tu]

Why would I need to eliminate 1/3 of the JT possibilities?

 

Askolnick, let me specify your opponents. They are experts, and holding both the J & T, will play each card 50% of the time. Holding the 234, he will choose a random one of the 3 to conceal, choosing evenly among (23), (24), (34) as the set of two cards to reveal.

 

Initially, the opponent will be dealt specifically

1. JT4 3.39%

2. JT 3.39%

3. J 2.83%

 

By your logic, we are done, we compare these numbers.

 

But this is not what happens, because the opponents have choice of plays, and won't play the J all the time on the left (except for #3 where he has to), and RHO won't always play the 2 & 3 on the right. Since you specified that RHO followed with the 2&3, you can't count the times when he will play 2&4 or 3&4. You only get to count the times the J will show on the left & the 4 does not appear on the right.

 

So actually,

1. has to be reduced by factor 1/2, because half the time the opponent is playing the T, not the J. Meanwhile RHO has 23 and is forced to play those two 100% of the time so no reduction for that factor. 3.39/2 = 1.695

 

2. Has to be reduced by factor 1/2, for same reason that LHO might play the T. Additionally, RHO will also play randomly, thus only 1/3 of the time will he not play the 4, so a total reduction of 1/2 * 1/3 = 1/6. 3.39*1/6 = .565

 

3. LHO has no choice of J/T, but RHO has the choice of spots, so reduction by 1/3. 2.83/3 = 0.943

 

 

If you understand the basic restricted choice argument of AKT98 vs. xxxx, Q falling behind the AK, you should get this. One doesn't count all the QJs (which would be > than the stiff Qs a priori), you only get to count half, the amount your opponent will actually play the Q. Similarly, if you treat equivalent low spot cards as specific ones, you can only count the amount the opponent will actually play the ones you specify.

 

Now if your opponent is silly enough to be totally predictable & always plays Q from QJ tight, or always follow with 2 & 3 holding 234, then the analysis changes (restricted choice no longer valid, it is modeled on a rational opponent who will play randomly when he has a choice) depending on which spot cards you see. You'd have a 100% play for the drop if the 2 or 3 were missing, and one could do your style of counting when the 4 was missing. But most suit combo problems specify "best defense". The analysis of what to do vs. potentially very bad opponents can be more complicated, you have to model their behavior to calculate the percentages they will play each card.

 

Let me break this down carefully for you:

1. JT4 3.39% = 1.695% (LHO plays J, RHO plays 2&3) + 1.695%(LHO plays T, RHO plays 2&3)

2. JT 3.39% = .565%(LHO plays J, RHO plays 2&3) + .565% (LHO plays J, RHO plays 2&4) + .565%(LHO plays J, RHO plays 3&4) + .565% (LHO plays T, RHO plays 2 & 3) + .565%(LHO plays T, RHO plays 2&4) + .565%(LHO plays T, RHO plays 3&4)

3. J 2.83% = .943% (LHO plays J, RHO plays 2&3) + .943% (LHO plays J, RHO plays 2&4) + .943% (LHO plays J, RHO plays 3&4)

 

If we are treating spot cards as specific, and looking where the 2&3 are played, we only get to compare the first numbers in each breakdown.

[kgr]

Thanks all for this interesting post!!

The same sort of reasoning applies to this case:

 

AK103

 

Q872

 

The ace is cashed and the nine appears to the left of the ace. If this is a singleton nine one must cash the king next; if from J9xx one must play to the queen. But the nine "must" be played from J9xx in order to create the possibility of taking a trick in the suit, and because the a priori probability of J9xx is three times the probability of singleton nine, one plays to the queen.

Interesting. ..but what about probabilty for RHO to have sigleton x compared to Jxxx?

Q432

 

AJ85

 

When a small card is led to the jack, the nine appears from LHO. What should one do on the second round to avoid losing a trick in the suit?

So T9x is more likely then T9 and you have to play low to the A next?

...but then we also have to compare probably for RHO Kx vs Kxx. Argh..I don't know.

 

The most interesting part for me is the falsecarding, but I don't think I will be able to do it at the table without thinking about it. So you know what you have to do when you play against me. But I will start watching JTx and T9x and prepare my play in advance..so you never know :) (at least my chanches will increase if you think I'm able to falsecard this :) )

[jdonn]

Q432

 

AJ85

 

When a small card is led to the jack, the nine appears from LHO. What should one do on the second round to avoid losing a trick in the suit?

So T9x is more likely then T9 and you have to play low to the A next?

...but then we also have to compare probably for RHO Kx vs Kxx. Argh..I don't know.

And just to make it more confusing, what if the dirty scoundrel has KT9x...