[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]

Hannie, on Nov 18 2007, 01:57 PM, said:

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]

ASkolnick, on Nov 27 2007, 03:22 PM, said:

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]

ASkolnick, on Nov 27 2007, 03:22 PM, said:

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.

[FrancesHinden]

Stephen Tu, on Nov 28 2007, 11:51 AM, said:

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]

Quote

Are they foreign?

Is this bridge xenophobia?

[FrancesHinden]

Hanoi5, on Nov 28 2007, 04:39 PM, said:

Quote

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]

ASkolnick, on Nov 28 2007, 09:54 AM, said:

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]

Quote

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!!

dburn, on Nov 28 2007, 01:11 AM, said:

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?

Quote

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]

kgr, on Nov 28 2007, 03:24 PM, said:

Quote

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...

[han]

jdonn, on Nov 28 2007, 03:41 PM, said:

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

Good thinking Josh, then he really got us. Instead of scoring only two tricks the dirty scoundrel may get a full two tricks if we get it wrong!

[jdonn]

Hannie, on Nov 28 2007, 05:10 PM, said:

jdonn, on Nov 28 2007, 03:41 PM, said:

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

Good thinking Josh, then he really got us. Instead of scoring only two tricks the dirty scoundrel may get a full two tricks if we get it wrong!

If anyone should get a little humor without requiring small yellow faces, it should be you mister.

[awm]

So here's a random question...

Your opponents reach the contract of 7♥. From the bidding they are almost certainly on a 4-4 fit. You are east; you can see:

---

After winning the opening lead in dummy, declarer calls for the ♥A. Which spot do you play?

My feeling was that it made absolutely no difference and I should play a random spot, but declarer (Barry Rigal) commented that he "wrote the odds on this combination" and that one should "never play [[the spot I played]]" from this position.

[gerry]

My favourite restricted choice analogy is the following:

Jim is on a TV game show and to claim his prize has to choose between one of three doors. One door has a million bucks and the other two conceal a lolly. There is an added help though, after Jim has picked a door the host opens one of the other doors with the proviso that he does not reveal where the money is, that is he opens a door to reveal a lolly.

Now Jim is allowed to choose again, does he stick with his original choice or change to the remaining unopened door?



Clearly he should change as in the likely event (2/3) that he originally choose a lolly, the host has now essentially shown him where the money is by opening the other door with a lolly. Only in the case where Jim chose correctly the first time will changing his choice cost.

[han]

How expensive are the lollies?