Q or 9, is there a difference

[kgr]

AQ982

vs

762

--

You play MP's, other suits are not important.

You let the 7 run for Easts T.

You come in your hand in another suit.

Is there a % difference for number of tricks in playing the 9 or Q now?

Is this a restricted choice situation?

 

Thanks,

Koen

[pclayton]

Darn, I tried to input this into SuitPlay and it wouldn't accept a suit with 2 2's :P

 

I think its correct to double hook the 8. By the way, start with low to the 8, and run the 7 the 2nd round. This allows you to stay in the same hand and finesse against KJxx a 3rd time.

[Guest Jlall]

AQ982

vs

762

--

You play MP's, other suits are not important.

You let the 7 run for Easts T.

You come in your hand in another suit.

Is there a % difference for number of tricks in playing the 9 or Q now?

Is this a restricted choice situation?

 

Thanks,

Koen

This is more complicated than you would think in theory because of the potential of RHO to play the king from KT doubleton on the first round, and the potential of LHO to play the J from Jxx on the second round when RHO does win the ten, etc etc etc.

 

In real life it is probably right to play the ace on the second round if neither of these falsecards are possible in your opinion.

 

If the latter falsecard (H from Hxx initially on the second round) is possible and the former (K from KH) is not which happens often against good players then your best bet is to hook against the 8 next.

[Guest Jlall]

In theory I think it's best to just assume you will lose to KH on your right always. And stiff H is more likely than JT dub so against optimal defenders you probably should also put in the 8 on the second round too. But I still think against 95 % of the people you will play against you do best to play the ace on the second round if LHO has followed low twice and play the queen if LHO puts up an honor on the second round.

[cherdano]

Playing the queen the second time only wins against JT doubleton. Running the 7 wins against T singleton, which is more likely by restricted choice. Playing to the ace wins against KT doubleton, which is even more likely by a small margin, so ace seems best.

 

Oh, of course I missed the false-cards...

[foo]

AQ982

vs

762

--

You play MP's, other suits are not important.

You let the 7 run for Easts T.

You come in your hand in another suit.

Is there a % difference for number of tricks in playing the 9 or Q now?

Is this a restricted choice situation?

 

Thanks,

Koen

AQ98x

+

76x

 

After T1 in this suit, (Which you did =right= if you have the entries. You are trying to get LHO to cover.)

the possiblities left where it matters what you do are:

 

Jxx+KT ~3.4%

Kx+JTx ~6.8%

Kxx+JT ~3.4%

KJx+Tx ~6.8%

KJxx+T ~2.8%

 

The best line is to put the "x" in your hand on the table intending to cover whatever LHO plays as cheaply as possible from Dummy.

 

Assuming the suit is not -+KJTxx or x+KJTx, you have ~55% chance of 4 tricks.

 

No, this is not a Restricted choice situation:

=you are missing 3 important cards, not 2 @ T1 in this suit.

=the 2 honors that remain missing are not neighbors in the suit (AK, KQ, QJ, JT)

[Edmunte1]

Here are all cases with their probabilities:

 

KJT --- 32 3.4%

KJ3 --- T2 3.4%

KT3 --- J2 3.4%

JT3 --- K2 3.4%

KJ2 --- T3 3.4%

KT2 --- J3 3.4%

JT2 --- K3 3.4%

K32 --- JT 3.4%

J32 --- KT 3.4%

T32 --- KJ 3.4%

 

T2 --- KJ3 3.4%

J2 --- KT3 3.4%

K2--- JT3 3.4%

T3 --- KJ2 3.4%

J3 --- KT2 3.4%

K3 --- JT2 3.4%

JT --- K32 3.4%

KT--- J32 3.4%

32 --- KJT 3.4%

 

KJ103 --- 2 2.8%

KJ102 --- 3 2.8%

KJ32 --- T 2.8%

KT32 --- J 2.8%

JT32 --- K 2.8%

 

2 ---KJT3 2.8%

3 --- KJT2 2.8%

T -- -KJ32 2.8%

J --- KT32 2.8%

K --- JT32 2.8%

 

KJ1032 --- void 2%

void --- KJ1032 2%

 

After we have lost to ten, in the first round and play second round (let's suppose also West played 2 on the first round), we are down to these cases:

 

1)KJ2 --- T3 3.4%

2)K32 --- JT 3.4%

3)J32 --- KT 3.4%

 

4)J2 --- KT3 3.4%

5)K2--- JT3 3.4%

6)32 --- KJT 3.4%

 

7)KJ32 --- T 2.8%

8)2 --- KJT3 2.8%

 

Cases 4,6,8 are losers, and on case 5) the King just shows up on second round, so we're down to:

 

1)KJ2 --- T3 3.4%

2)K32 --- JT 3.4%

3)J32 --- KT 3.4%

4)KJ32 --- T 2.8%

 

So the finesse is obvious, being 9.6% to 2.8% chance (77.4%)

[Edmunte1]

The problem is what finesse:

 

In case 1(3.4%), we have obviously to put the Queen

In case 2(3.4%) we have to finesse the Queen

In case 3(3.4%) we'll lose no matter what finesse we're playing, unless we're playing the Ace

In case 4(2.8%) we have to finesse 9. So the right play should be:

 

a)If the Jack appears finesse

b)if the second small card appears you have 3 possible lines

- playing the Queen for 35.4% chance

- playing the Ace for 35.4% chance

- playing the 9 for 29.2% chance

[Guest Jlall]

The problem is what finesse:

 

In case 1(3.4%), we have obviously to put the Queen

In case 2(3.4%) we have to finesse the Queen

In case 3(3.4%) we'll lose no matter what finesse we're playing, unless we're playing the Ace

In case 4(2.8%) we have to finesse 9. So the right play should be:

 

a)If the Jack appears finesse

b)if the second small card appears you have 3 possible lines

- playing the Queen for 35.4% chance

- playing the Ace for 35.4% chance

- playing the 9 for 29.2% chance

no.

 

Oh, of course I missed the false-cards...

[Edmunte1]

This was just maths, ignoring the human factor. But:

 

-In case 2)K32 --- JT 3.4% ---->East sometimes can put the Jack, sometimes the Ten. But one issue that is usually misunderstood is that apriori case 3)K32--JT 3.4% is more probable than 4)KJ32---T 2.8%. But after the first round played the probability for 3)K32--JT has diminished to 1.7% (west decided to play T), and the probability of 4)KJ32---T remained the same, and that's all restricted choice is about.

 

- In case 3)J32 --- KT 3.4% ----> this is right if west never falsecards with the Jack. Now depends of the level of the opponents, but i guess this falsecard is way below average, being something like 25% chance, so the probability of this case should be 3.4% *75% = 2.55%

 

So some final calculation should look like:

a) If jack apears finesse( you'll win case 1) and lose 25% of case 3))

:) If the second small card appears:

-play 9 for 40% (2.8%/(2.8+2.55+1.7))

-play A for 36%

-play Q for 24%

[Stephen Tu]

No, this is not a Restricted choice situation:

=you are missing 3 important cards, not 2 @ T1 in this suit.

=the 2 honors that remain missing are not neighbors in the suit (AK, KQ, QJ, JT)

 

Actually, it absolutely is a restricted choice situation. The criteria you listed are not particularly relevant to whether restricted choice applies. What is relevant is that a defender won with one of equals, if he had more than one he had a choice of what to play, e.g. if he had JT, while if he has only one of the two he has to play it all the time. One can adjust for this in the calculations either by halving the probability of the relevant holdings where defener has both JT compared to stay T stiff, or by just solving for all holdings, treating the missing honors as K, H, H.

 

As Justin says, it mainly depends if 2nd hand is good enough to falsecard from Jxx/Txx the second round to protect his partner's possible KT/KJ, making it look like he started with KJx/KTx. If not, play the ace, if so, hook low, since then KJxx/KTxx are more likely than Jxx/Txx.

 

Since defender played low the second round, really the only relevant holdings are:

(a)J32/T32

(b)KJ32/KT32

( c )K32

 

P(B) > P( c ), so low is better than the Q.

(a) has to be weighted as to frequency of the falsecard. If % falsecard > ~17%,

P(B) > P(a), low better than ace. otherwise ace is better.

 

Note that if you don't know your opps, it's "safer" to give them credit for knowing to falsecard. If you are wrong, and they never falsecard, you are only giving up 1.1% on your line vs. the weak player exploitative line. If you insult them by playing the ace, if you are wrong & they are actually always falsecarding, you give up 5.65% vs. the game theoretically optimal line (hook low).

 

I never know how much to insult my opponents. Are there enough bad players out there to go for the 1.1% edge as a default play?

[foo]

Edmunte1,

you've made some serious calculation errors.

AQ98x+76x means Our "x's" are the 2 and 3 (doesn't matter which is in each hand)

 

That leaves the other "x's", the 4 and the 5, out.

Thus a holding like KJx+Tx= KJ4+T5 or KJ5+T4.

As you correctly calculated, each specific 3:2 holding is 3.4%.

Therefore a 3:2 holding with an unspecified spot card in either hand is 6.8%.

Similarly, each specific 4:1 holding is 2.8% and one that has an unspecified spot card in either hand is 5.6%.

 

After LHO plays small (the 4 or 5) and RHO plays the T, all holdings where this is not true are impossible, leaving

KJ54+T 2.8%

KJx+Tx 6.8%

K54+JT 3.4%

J54+KT 3.4%

Kx+JTx 6.8%

54+KJT 3.4%

x+KJTx 5.6%

Our plans do not matter in the x+KJTx and 54+KJT cases, so ignore them.

 

Small from our hand, =intending to cover LHO's card as cheaply as possible=, !not! finesse,

 

a= Wins when we play 8 and LHO plays 2nd Hand Low: KJ54+T (2.8%)

Loses to K54+JT and J54+KT (6.8%)

In the rest of the holdings, it is impossible for LHO to play a spot on T2.

If we play the Q here, we effectively give up another trick in both the KJ54+T and the J54+KT cases. Slow loser or fast loser, it's still as 2nd loser.

Same Story if we bang down the A and it does not crash the K.

 

b= Wins when we play Q if LHO plays J: KJ54+T, KJx+Tx (9.6%)

Loses to J54+KT (3.4%)

 

c= Wins if we play A if LHO plays K: K54+JT, Kx+JTx (10.2%)

 

 

Stephen,

Go look in something like _The Encylopedia_.

If RHO had both the K and the T, they would =always= win with the T. Therefore you have no additional information about the location of the K. Therefore this is !not! a Restricted Choice situation as to which hand has which remaining honor or even if both missing honors are in the same hand on T2.

The win of the T on T1 does nothing to help us discern who has what honor on T2.

(Contrast this to what we would know if =LHO= had played the T as 2nd Hand.)

Only if it did would this be a Restricted Choice situation.

[kenberg]

Whether you call it restricted choice or not the language of restricted choice seems useful. The outstanding cards are the K, the J,T, and two spots. RHO holding the J and T, with or without the K, could have played either one. It makes sense to collapse the J and T to two tacks (so the play of one tack is evidence that lho holds the other).

 

Now what. Let me ignore the possibility of falsecards at least for the moment.

 

 

At crunch time, we have seen all the spots and rho played a tack on the first round. If rho holds both remaining cards it doesn't matter what we play. So we assume rho started with either 1 or 2 cards.

 

If he started with a stiff tack (two ways) we pick up the suit by playing low (and only by playing low). If he started with K-tack (two ways) we pick up the suit by playing the ace. If he started with both tacks (one way) we pick up the suit by playing the Q.

 

My conclusion: Barring false cards, what we should not do is to play the Q. Maybe low, maybe the ace, but not the Q. A slight preference (by empty spaces) for the ace.

 

Falsecards? Too tough for me until I have some more coffee.

[foo]

The reason you do !not! play the Q on T2 unless you absolutely have to is that if the Q only captures "air" or gets killed by the K, you can no longer win 4 tricks in this suit.

 

Same logic about not capturing a big enough card holds for the A here.

 

If Restricted Choice was in play, you'd be able to have a decent sense as to whether RHO started with T, JT, or KJT. The win of the T on T1 gives you =no= additional information to help you make that decision.

After all, you always "knew" that RHO having KJT was more unlikely that the other possibilities.

[kenberg]

The reason you do !not! play the Q on T2 unless you absolutely have to is that if the Q only captures "air" or gets killed by the K, you can no longer win 4 tricks in this suit.

 

Same logic about not capturing a big enough card holds for the A here.

 

If Restricted Choice was in play, you'd be able to have a decent sense as to whether RHO started with T, JT, or KJT. The win of the T on T1 gives you =no= additional information to help you make that decision.

After all, you always "knew" that RHO having KJT was more unlikely that the other possibilities.

Forget KJT, there is nothing you can do about that, You lose exactly two tricks in the suit however you play. The relevant holdings are stiff T, JT, and KT. But the three holdings are not equally likely since with KT or stiff T the play of the T is forced (at least if we rule out the K falsecard from KT) while with JT it is not.

 

Restricted Choice seems to me to be a reasonable shorthand for describing this situation but it's possible to go through the analysis without ever mention this contentious phrase.

 

If you are confronted with this holding repeatedly and you play small to the 8 losing to either the ten or the jack, and next lead toward the board and see another spot, you have to decide what to do. Presumably you do not vary your response based on whether you see a J or a T on the first round. The Q will be right when and only when rho holds the JT tight. The A will be right when rho started with KJ or KT. It's true that with the current hand we can rule out KJ tight, but since the T and J are effectively indistinguishable this doesn't matter.

 

 

Thus: The strategy of play small first and if it loses to J or T then later play the ace will win twice as often as the strategy of play small first and if t loses to the J or T then play the Q (the first strategy works every time rho is dealt KT or KJ, the second when he is dealt JT). Similarly, playing small first and low the second time will win twice as often as low and then Q.

 

So: Should rho, holding KT tight, play the king the first round? Seems as if he should. although if he plays the T there is a possibility declarer will run the 9 next time. But if rho would play, or would sometimes play, the K from KT then this screws up the analysis when the T is played since it was based on the play of the T from KT being forced.

 

Anyway, whether you call it Restricted Choice or not, the fact that the J and T can be played interchangeably when both are held does affect the analysis. Restricted Choice is a version of Bayes' theorem from probability, looking at the probability that rho holds the J, given that he played the T. As long as he can play either J or T when holding both, but only the T when holding the T w/o the J, this probability is not 1/2. Falsecards make it tough though.

[foo]

Yes, I understand that what Bridge players call Restricted Choice has its roots in Bayes Theorem.

However, Bayes Theorem is based on you getting new information that allows you to adjust the probabilitites.

That is not true here. When RHO wins T1 with the T, you have no new information to help you in any meaningful way on T2 after LHO plays the 2nd x.

 

Remember, the goal here is to take 4 tricks in this suit. Not simply to avoid losing a trick on T2.

KJ54+T 2.8% You must play 8 or 9

K54+JT 3.4% You must play Q

J54+KT 3.4% You must play A

54+KJT 3.4% You are scrod.

 

No matter which choice you make, it works in 1 of 3 cases that matter and fails in the 2 others. ...and you have no idea based solely on the play in this suit which it is.

 

 

Now, there =is= a way to use Bayes Theorem here. Don't play T2 of this suit immediately and instead get a count on the hand by playing elsewhere.

The theory of Vacant Spaces and simply counting who has show up with what =can= help you decide if LHO is more likely to have 4 cards or less in this suit as well as telling you whether it's LHO or RHO who rates to have the K.

[Guest Jlall]

I never know how much to insult my opponents. Are there enough bad players out there to go for the 1.1% edge as a default play?

hi, so much garbage in this thread it was nice to read your post.

 

To answer your question, an emphatic yes! Most people will never play the jack or ten on the second round. Even your run of the mill expert actually. If I was playing vs a random person who I'd never seen/heard of even at the national level (if it was the first day of the event) I would definitely assume they could not make this falsecard even with me needing to be right more than 4 out of 5 times to break even.

[Stephen Tu]

Go look in something like _The Encylopedia_.

If RHO had both the K and the T, they would =always= win with the T. Therefore you have no additional information about the location of the K. Therefore this is !not! a Restricted Choice situation as to which hand has which remaining honor or even if both missing honors are in the same hand on T2.

The win of the T on T1 does nothing to help us discern who has what honor on T2.

 

Foo, again you have no idea what you are talking about. Yes, it is true that since the K & T are not equals, winning the T does not affect the probability of the location of the K. But it absolutely DOES have effect on who has the J, since the J & T are equals. If you do not take restricted choice into account, one might do something silly in the calculations like conclude, "the probability of JT doubleton is greater than that of just stiff T, therefore hooking the Q is better than hooking low 2nd round", which is absolutely the wrong conclusion.

 

As I said, to do calculations properly you either halve the probability of the holdings where RHO has both JT (in this case, with LHO following low 2nd round, the only relevant one is JT tight), or count BOTH possibilities T stiff / Jstiff, KT doubleton / KJ doubleton, treating the equal honors as the same card that you can't identify which is which. That is what restricted choice is all about. If you don't understand this, you don't really understand restricted choice.

 

 

It's not supposed to matter whether RHO plays K or T the first round from KT tight, if his partner is competent. Playing the K is just a Grosvenor, in practice you will play him for K alone. The important ? here is whether LHO is good enough to play J from Jxx 2nd round, which is mandatory if you know this combo.

[Guest Jlall]

It's not supposed to matter whether RHO plays K or T the first round from KT tight, if his partner is competent.  Playing the K is just a Grosvenor, in practice you will play him for K alone.

Well, yeah when you say if his partner is competent, but imagine you are playing with a client who's never capable of falsecarding and you're playing against jeff meckstroth. Then if you knew Jeff had 3 in his hand and you were looking at AQ98x in dummy you would win the king on the first round to save yourself (unless jeff gave you so much credit that he realized that this was your best shot with KT and...). Of course this is a completely artificial situation.

[Stephen Tu]

To answer your question, an emphatic yes! Most people will never play the jack or ten on the second round

 

Yeah, I suppose you're right. I have too much of a tendency to think that since I study combos like this & remember the right falsecards in most of these situations, that my random supposedly decent opponent has also. Need to take the more "assume opponent is terrible until proven otherwise" approach.

[foo]

Stephen, I believe _The Encylopedia_ and a few other references explain why your reasoning is flawed and that this is not a Restricted Choice situation.

More like Monty Hall or "The Lady and the Tiger". It's an illusion that you get any new information because you don't get any information that =matters=.

"The difference that makes no difference is no difference."

 

 

...and contrary to Stephen's occasionally rude comments, I=do= know what I'm talking about here so I'm going to clear up something that has not been made explicit until now.

 

The true falsecard situation in this combination is when an opponent holds Jxx or Txx in front of AQ98x.

 

If you know your opponent will never falsecard with this holding the proper play on T2 is to play the A.

 

The play of the 8 or 9 is effectively insurance against a possible falsecard by LHO through dropping the J or T as if forced to here.

 

Winning the K when you don't need to holding KJ or KT is a con that is far more easily seen through than the J or T from hxx falsecard.

 

The bottom line is not to commit yourself until you have to in these situations. Get a more detailed count on the hand as much as you have to or can afford to before touching holdings like these a 2nd time.

[Stephen Tu]

Foo, if you truly understood restricted choice, and were actually right about this, you could make a logical argument why I am wrong, or deconstruct my statements and tell me where I am wrong. Instead you just say "encyclopedia explains why", and aren't pointing out where my logic is faulty. I call BS, there is nowhere in the encyclopedia that explains why I am wrong, because I am not wrong. If you really believe this is so, please pick out which specific statement(s) I make below is/are incorrect, that you disagree with, and explain why. If all you can come up with is "I think this book says you're wrong", then you don't truly understand what those books are talking about.

 

In your posts you keep on listing

the JT tight possibility as 3.4%,

and the T stiff possibility as 2.8%.

, but not mentioning the possibility of J stiff.

 

Where restricted choice comes in is in what probability you assign to JT tight. You cannot assign the full a priori probability of 3.4% to JT tight, because opponent played the T, and he will not play the T all the time from that combination! He is supposed to play it only about half the time. If he did play the T all the time, one could employ the counter strategy of hooking the Q when the T appears, but hooking the low when the J appeared, and make him worse off than someone who randomized their play from JT. But real opps randomize, and the best you can do is consistently pick up both stiff T and stiff J, and give up to JT (if comparing the hook Q & hook low lines). Now for hook low vs. play ace, as discussed above that depends on the falsecarding frequencies. The KT/KJ possibilities are not relevant when comparing the finesse lines since both lose 2 tricks in that case.

 

The right way to think about these restricted choice problems is to either compare probability of opponent holding JT AND he chose to play T (50%) = 1.7%

vs. the possibility of T stiff,

OR to take the full 3.4% probability, but count both the T stiff & J stiff cases, treating both honors as identical since the goal is to optimize your strategy for both cases, T appearing or J appearing first round from RHO, and logically the strategy should be the same since those cards are effectively the same here, assuming the defender is not pursing some strange never or always falsecard strategy from JT tight.

 

How am I wrong here? Where in the encyclopedia does it say this reasoning is flawed?

 

If restricted choice did not apply, and the JT tight probability was not reduced properly, the conclusion that would be reached that it is superior to play the Q than to hook low again, because 3.4 > 2.8. But as stated above it does apply, and this isn't true.

 

You DO get information that matters here when RHO wins the T. The info gained is that T stiff is more likely than JT doubleton, because part of the time from JT doubleton he would have played the J, whereas from T stiff he has no choice.

[foo]

Foo, if you truly understood restricted choice, and were actually right about this, you could make a logical argument why I am wrong, or deconstruct my statements and tell me where I am wrong.

 

In your posts you keep on listing the JT tight possibility as 3.4%, and the T stiff possibility as 2.8%, but not mentioning the possibility of J stiff.

Logic point 1: The play of T1 of this suit makes it =impossible= for the suit to have split KT54+J or J+KT54 :D There can be no J stiff.

I don't mention impossibilities because they don't matter...

 

Logic point2: I will not quote _The Encylopedia_'s section on Restricted Choice and the logical fallacies that are sometimes associated with it. I have given the reference. Even the hint as to what the problem is.

 

I have pointed out =repeatedly= that you get no information beyond the original apriori probabilities from the winning of the T on T1 that tells you anything about how to play the suit on T2 in the suit. If there was, it would be possible to construct a logically convincing argument as to how to play the suit on T2 based on that fact. Instead, the best way to play the suit is based =solely= on whether one thinks it is more likely that LHO will falsecard or not on T2.

That is not a Restricted Choice based argument or strategy.

 

The arguments have been made. I have pointed out the flaws in your statements. I given references to back up my analysis of those flaws.

You simply don't want to accept them.

[Stephen Tu]

Logic point 1: The play of T1 of this suit makes it =impossible= for the suit to have split KT54+J or J+KT54 smile.gif There can be no J stiff.

I don't mention impossibilities because they don't matter...

 

The first matters in the context of an overall strategy of what to do when RHO wins either with the T or the J. Logically your strategy should be the same either way.

 

It's OK to say you will ignore the first case, but then in your calculations you also must state the probability of JT doubleton as 1.7%, not 3.4% to be consistent. If you ignore cases where RHO has the stiff J, you also must ignore cases where RHO had JT & chose to play the J instead. RHO will NOT play T from JT 100% of the original 3.4%, he will only play it half of that.

 

Logic point2: I will not quote _The Encylopedia_'s section on Restricted Choice and the logical fallacies that are sometimes associated with it. I have given the reference. Even the hint as to what the problem is.

 

You are dodging. You will not quote, because there is no quote there that supports your thesis and refutes mine.

 

I have pointed out =repeatedly= that you get no information beyond the original apriori probabilities from the winning of the T on T1 that tells you anything about how to play the suit on T2 in the suit.

 

Yes, you repeatedly make statements without any valid logic to back them up. Saying the same thing without any cogent argument to support it over & over again doesn't make it true. Restricted choice absolutely does apply, when comparing the two hook lines, comparing whether low to the Q catering to JT doubleton is better or worse than hooking low catering to T stiff. Do you agree that hooking low is better than hooking Q? If so, how does that follow when you state that the probability of JT is 3.4% while the probability of T stiff is only 2.8%?

 

(Answer: because probability of JT, RHO playing the T is really only 1.7%, which is less than 2.8%, this is an application of restricted choice).

 

Restricted choice does not apply exactly in comparing hooking low vs. playing the ace second round, as the K & T aren't equals & you have to assign some arbitrary probability to the various falsecards based on your experience with players of various skill levels. For this part I will agree with you. But you are wrong that there is no application at all of restricted choice in this problem.

 

I have pointed out the flaws in your statements. I given references to back up my analysis of those flaws. You simply don't want to accept them

 

I made quite a few statements. You picked out exactly one above to pick at here, and I have now explained why you are incorrect. You refer people to the Encyclopedia, but refuse to quote the passage, because there really aren't any that back your claim. I think most people here will agree that you've failed miserably to make your case.

[foo]

bottom p382 "a mistake to avoid" to top p383 Ex 4 and 5

(ex 4 is KJ9+xxx; ex 5 is AQTxxx+xxx)

 

I am !not! wasting my time copying it verbatum for you. I simply will not be goaded into such stupidity.

The rest of the write-up is equally instructive.

 

The bottom line here is that you do not receive enough information from the play on T1 for it to influence the odds enough, and hence your play enough, on T2.

"The difference that makes no difference =is= no difference."

The odds do not change enough based on the play of T1 for it to affect your plans. Therefore this is !not! a Restricted Choice situation.

 

Had the cards been 54+KJT, or x+KJTx would RHO =ever= win T1 with anything but the T?

No. He has no reason to randomize his choice or to falsecard.

 

If Restricted Choice held, you would be able to pick which card to play on T2 based on the play of T1 to maximize your chances. You can not. Therefore it is not an issue.

 

 

Once you can handle it, I strongly suggest you (re)read _Master Play_ by Terence Reese on this topic.

 

..and using impossiblities as "props" for your arguments is such a logical fallacy as to make logical discussion of any topic that you bring them to impossible.

 

I haven't dodged even a little bit. The students who actually want to learn have all the logic and references they need on this topic. Yourself included if you so desire.

I'm done here. Go argue with someone else.