Restricted Choice

[Spisu]

Except we have math on our side.

 

I would like to see some valid math on RC that doesn't just piggy-back on the a priori odds and claim to be a principle. Just remember that RC math claims that one equal seen/played means you can chop away at the changes the other is in the same hand to make the numbers work. But statistics show unequivocally that one of two equals comes equally as a lone or combined honors in general play and there is no such basis for a general principle.

 

That "principle" just may be as if when your AQ wins a King wins a finesse through West, to assert that the 100% odds some opponent would hold that K had been reduced to 50-50 by your enlightened perception that only one opponent can hold that King.

[1eyedjack]

At the commencement of this thread you claimed that RC was "a fallacy". Now after three pages of arguments your position seems to be diluted to (distilled into my own words) "RC is an expression of established math in an alternative way".

 

Sorry, but to my mind a "fallacy" is a doctrine that generates incorrect (ie "false") results. False and fallacy are derived from the same etymological roots.

 

There is only one "math". It is all internally consistent. It all starts with a priori probabilities, and it all adjusts those a priori probabilities (ie eliminating impossibilities) as information develops. If your only objection to RC is that it expresses in concise terminology the effect of that process, then to my mind that falls short of concluding that it is a "fallacy".

[Spisu]

No, we aren't drawing false conclusions. We are drawing correct conclusions, in fact the same conclusion. You are apparently blind to the fact that both arguments are drawing the *same* conclusion, not different conclusions, and that both are correct.

 

If you lose a finesse to an equal. it is *not* "as likely to be a loner or from combined honors". It's twice as likely that the honor was a loner. That is the whole gist behind restricted choice. A loner will be played 100% of the time. Combined, that particular honor would be chosen to play approximately half the time against an opponent that randomizes. It's only "as likely to be a loner or from combined honors" if an opponent nearly always chooses a particular honor from the combined honors and he plays that honor he favors. Then a second finesse is basically 100% against the disfavored honor, but only 50% against the other. Combined odds will still be 2:1 to succeed.

 

I don't know who you think is saying that an equal is as likely to be a loner as combined. That implies second finesse is only 1:1, not 2:1. RC is not claiming that. It is claiming 2:1, exactly the same as you. The principle is correct.

 

The second finesse is 2:1 to succeed using both arguments. This is a correct conclusion. Neither is false. If RC gives the exact same answer as your 2:1 answer, how is it a false conclusion? If your argument is right, and RC agrees with same answer, then either both are right or both are wrong! Logically RC cannot be wrong while your argument being correct if it is claiming the same result!

 

Wow. You have it all backwards. It is I who believe a lone honor is twice as likely a prioi to win a double finesse. Restricted Choice tries to have it both ways by first asserting they are of equal frequency (lone vs from combined) BUT FOR their claim that plays from combined equals can be expected to occur at only half the frequency of their expectation. That is what restricted choice is all about.

 

So by dividing by two, they come up with the right answer (which is coincidentally my answer).

 

And I have never said that a double finesse is as likely to lose to a lone honor as a combined honor. Again, that is restricted choice's first of contradicting positions. (They say odds should be 50% but then chop it down to 1/3 because, you know..

 

Lone and combined honors first plays in open play come equally in frequency and can be in either hand. A finesse targets ONE hand which is half as likely, or 50%.

That seems to be ignored in RC but dividing the hands by 2 might be a clue.

 

Your apparently not a teacher with your ** "If your argument is right, and RC agrees with same answer, then either both are right or both are wrong!"

 

So you think if I say 2+2=4, and RC says no it's because it's actually 2 cubed divided by the number of integers being added, also giving 4, that both are right or wrong? I see...

[Zelandakh]

That is what restricted choice is all about.

No it isn't. You simply do not understand the concept. The correct answer is not arrived at coincidentally. If it were there would also be situations in which it did not give the correct answer. When you are able to grasp the concept of the underlying maths, Bayes Theorem as already stated, you will see why it is a perfectly valid calculation. Most bridge players find this method of calculation simpler than the alternative one. It does not matter which you use, both are valid. I challenge you to find a single case where RC does not produce the correct answer - you will not succeed. :lol:

[Spisu]

No it isn't. You simply do not understand the concept. The correct answer is not arrived at coincidentally. If it were there would also be situations in which it did not give the correct answer. When you are able to grasp the concept of the underlying maths, Bayes Theorem as already stated, you will see why it is a perfectly valid calculation. Most bridge players find this method of calculation simpler than the alternative one. It does not matter which you use, both are valid. I challenge you to find a single case where RC does not produce the correct answer - you will not succeed. :lol:

 

It does matter as a principle when in general play you see a card having an equal played from a player just following suit. If you try to apply the restricted choice principle to that card, you must accept that it was 2:1 a lone equal. But absent conventional play, no one can or at "has" been able to dispute that frequencies of lone and combined honors exist at ca 50-50 in the opponents hands, contrary to RC. If one isolates one opponent hand into a subset of the deal, things change. That does not create a principle that applies to all hands.

 

"I challenge you to find a single case where RC does not produce the correct answer" you say? Challenge met and defeated. (And yes it is what RC was invented to do, trying to set the odds for the second of a double finesse at 1/3 when they asserted the faulty dilemma that there were only 2 otherwise "equal" possibilities, such as K or KQ.)

[Spisu]

I think you need to do it step by step.

 

1.......void....KQxx....4.783...1.....4.783

2.......x.......KQx.....6.217...2.....12.435

3.......xx......KQ......6.783...1.....6.783

4.......Q.......Kxx.....6.217...1.....6.217

5.......Qx......Kx......6.783...2.....13.565

6.......Qxx.....K.......6.217...1.....6.217

7.......K.......Qxx.....6.217...1.....6.217

8.......Kx......Qx......6.783...2.....13.565

9.......Kxx.....Q.......6.217...1.....6.217

10......KQ......xx......6.783...1.....6.783

11......KQx.....x.......6.217...2.....12.435

12......KQxx....void....4.783...1.....4.783

 

Spisu do you agree that this chart look like the a priori distribution of all the split possible when we are missing KQxx. Note that the 4th column is the permutations.

 

3--KQ5 is slightly less likely than KQ--53

 

However note taht on line 2 x--KQx is multiplied by 2 to represent both 3--KQ5 & 5--KQ3 added together.

 

Odds vary with one's intent and interest, such as if some number of missing cards are not on your radar they will normally each be 50-50 to be in one opponent's hand...But if 2 cards are jointly significant, and the first is 50-50, the second becomes a a 25% joint probability.

 

That's why I started out here citing only the 2 equals (which was a bit confusing to some). But I did this because a principle affecting only those two cards was in question...The expectation was that some uninteresting number of sidekicks would exist just to keep the 2 equals from possibly falling together when you have all those 11 card suits.

 

Exactly how that disinterest which includes ALL other distributions calculated would work out is not exactly clear, BUT the calculations in the ACBL Encyclopedia (under Restricted Choice) use the same expectation frequencies I do here. Those seem to be that an AK combined will be in both hands 50%, and be divided in two ways between the hands 50%, see 2011 ACBL Encyclopedia pg 458...Also, specifically under "Restricted Choice", you can see they show 800 hands divide 200-200-200-200 as AK A/K K/A KA. Also of note is that it sems 200 hands with plays from "AKs" are omitted from the totals verifying RC which makes the number of specific honor plays from combined AK appear to be half the rate of an associated honor. Please check that out if you've got any edition of that ACBL book, because it goes back many decades. See if you view it that way. Feedback welcome.

[eagles123]

Come on mods put a stop to this shite and ban the (not very good) troll

[1eyedjack]

"I challenge you to find a single case where RC does not produce the correct answer" you say? Challenge met and defeated.

Challenge met? Possibly, I have difficulty dissecting your grammar. Maybe English is not your first language, but I just don't understand some of your posts.

Challenge defeated? You do not get to pass that verdict.

 

I direct you once again to this post, which you have so far ignored perhaps because it cannot be refuted

 

http://tinyurl.com/hjps5mv

 

I make three statements below, and you are invited to state with which you disagree as accurate. You may select more than one, but you must select at least one in order to establish the fallacy of the principle.

 

1) In the first of the two examples examined, the probability of the drop and finesse are broadly equal

 

2) In the second example examined, the finesse is about twice as likely to succeed as the drop.

 

3) The only difference of significance between the two cases (that might affect the above two observations) is that East's first card in the second example is equal to the missing card, but not in the first example.

[Stephen Tu]

Restricted Choice tries to have it both ways by first asserting they are of equal frequency (lone vs from combined) BUT FOR their claim that plays from combined equals can be expected to occur at only half the frequency of their expectation. That is what restricted choice is all about.

 

RC is calculating the frequencies *after the first finesse has lost*, and *after you have led the second card and 2nd hand has followed low*, and you are at the decision point of whether to try to drop the remaining honor, or finesse 2nd hand for it. At this point, a very large number of the a priori probabilities have been eliminated, dropped to 0%. You have eliminated all hands where 2nd hand has both honors (since RHO won the first trick). You have also eliminated all hands where 2nd hand had a void, a stiff, or a doubleton honor . The remaining possibilities will then be the only possible layouts, expanding to cover 100% of the problem space, retaining their relative proportions given by the a priori probabilities.

 

At this point, there are two ways to make the calculation, and both are valid. In one, you do not identify the honor that won, and count that lone honors are dealt roughly twice as often as the combined situation, for a 2:1 result. In the other, you DO identify the honor that won, which eliminates one of the lone honor possibilities. So in actuality there are only two possibilities, the specific lone honor you saw and combined honors and your opponent chose to win with that specific honor. They are dealt at relatively equal frequency. But your opponent now basically controls how often your 2nd finesse wins depending on which honor won, depending on their tendencies. They can't affect your overall success rate if you stick to your strategy and always finesse, but they can absolutely affect your success rate for a particular honor showing up. Let's say they always play K from KQ. Then if you lost to the K, you will absolutely find that the second finesse only succeeds about half the time, not 2/3 of the time. But if you lost to the Q, now the second finesse wins all the time, since they are never playing the Q from KQ so it has to be Q alone. RC just usually describes the averaged situation where an opponent randomizes and gives the combined odds of 2:1, rather than specific odds of the second finesse succeeding depending on the exact percentage of an opponent's choice from doubleton honor and which of these honors won.

 

Lone and combined honors first plays in open play come equally in frequency and can be in either hand.

Not after the first finesse lost to 4th hand they don't.

[shyams]

Come on mods put a stop to this shite and ban the (not very good) troll

I can upvote this only once. So I am +1 'ing it here for additional oomph. Hope others join in as well!

[Stephen Tu]

I can upvote this only once. So I am +1 'ing it here for additional oomph. Hope others join in as well!

 

I don't think being confused and stubborn is the definition of a troll, or is ban worthy. A troll is traditionally defined as someone who is posting things he knows to be false for the purpose of starting an argument. There doesn't seem to be any indication that Spisu doesn't actually believe he is correct.

[Spisu]

Challenge met? Possibly, I have difficulty dissecting your grammar. Maybe English is not your first language, but I just don't understand some of your posts.

Challenge defeated? You do not get to pass that verdict.

 

I direct you once again to this post, which you have so far ignored perhaps because it cannot be refuted

 

http://tinyurl.com/hjps5mv

 

I make three statements below, and you are invited to state with which you disagree as accurate. You may select more than one, but you must select at least one in order to establish the fallacy of the principle.

 

1) In the first of the two examples examined, the probability of the drop and finesse are broadly equal

 

2) In the second example examined, the finesse is about twice as likely to succeed as the drop.

 

3) The only difference of significance between the two cases (that might affect the above two observations) is that East's first card in the second example is equal to the missing card, but not in the first example.

[1eyedjack]

That is the second time you have simply reposted my comment in its entirety without in any way responding to it. The earlier occasion was post #38. What is the point, pray tell?

 

Post #8 might be a third example but I took #9 as the response intended to be tacked on to #8

[gordontd]

Come on mods put a stop to this shite and ban the (not very good) troll

Does anyone remember Reef Fish, I think from rgb?

[Spisu]

That is the second time you have simply reposted my comment in its entirety without in any way responding to it. The earlier occasion was post #38. What is the point, pray tell?

 

Post #8 might be a third example but I took #9 as the response intended to be tacked on to #8

We have had raging thunderstorms and power interruptions, but I have an assignment for you I'll post when the storms are over.

[Spisu]

At the commencement of this thread you claimed that RC was "a fallacy". Now after three pages of arguments your position seems to be diluted to (distilled into my own words) "RC is an expression of established math in an alternative way".

 

Sorry, but to my mind a "fallacy" is a doctrine that generates incorrect (ie "false") results. False and fallacy are derived from the same etymological roots.

 

There is only one "math". It is all internally consistent. It all starts with a priori probabilities, and it all adjusts those a priori probabilities (ie eliminating impossibilities) as information develops. If your only objection to RC is that it expresses in concise terminology the effect of that process, then to my mind that falls short of concluding that it is a "fallacy".

Do you have a problem with language? I have never hinted other than that RC is a fallacy. Maybe you misunderstood when I said dividing by 2 reduces a quantity by half. And RC does that quite accurately if not appropriately.

[Phil]

Congratulations, you made it to my signature.

 

That's a pretty rare feat.

[1eyedjack]

I think we just have to agree to differ. It is vanishingly unlikely that anything could be added that has not been said before in this thread. I certainly have nothing more to add so shall leave you all to it.

[Spisu]

Congratulations, you made it to my signature.

 

That's a pretty rare feat.

 

The fact is indisputable that if 2 specific cards are involved in your strategy the odds they are both in one opponent's hand are 1/2x1/2, or 25%, and the odds for the second only change when one of the two is identified by random discovery (and only then can the second card become 50%). For any unspecified run of the mill cards, odds are 50% each.

 

Science is apparently not your strong suit. Congratulations on your banner of shame.

[Spisu]

Challenge met? Possibly, I have difficulty dissecting your grammar. Maybe English is not your first language, but I just don't understand some of your posts.

Challenge defeated? You do not get to pass that verdict.

 

I direct you once again to this post, which you have so far ignored perhaps because it cannot be refuted

 

http://tinyurl.com/hjps5mv

 

I make three statements below, and you are invited to state with which you disagree as accurate. You may select more than one, but you must select at least one in order to establish the fallacy of the principle.

 

1) In the first of the two examples examined, the probability of the drop and finesse are broadly equal

 

2) In the second example examined, the finesse is about twice as likely to succeed as the drop.

 

3) The only difference of significance between the two cases (that might affect the above two observations) is that East's first card in the second example is equal to the missing card, but not in the first example.

 

As I have said at least a couple times the point here is the basic assumption of RC that play of one equal gives mathematical basis to a certain percentage exclusion of the other. For that reason, I stipulated and later explained "I" was addressing the very very basics. Your 9 card suit incorporating RC, random discovery and distributional elements in your 24,000 hands has different issues that would be interesting if we ever got beyond the most simple fundamentals. (Like, if you can't deal with basic science, then stay away from quantum theory or probability maybe.)

 

Here's a very simple for YOU (or anyone here) at the simplest level: The ACBL Encyclopedia has had an explanation/proof for "Restricted Choice" in editions going back at least to the 3rd (1973), and in the newest, 11th, 2011. Find some edition, first check out Example 1 (it's in all earlier editions I have seen) and explain how the data supporting the basic principle of RC would have fared had the author not excluded 200 plays from AK holdings on a basis akin to "you lose those anyway". Second, explain why the author, in what should be a simple example of leading to QJx from xxx to set up one trick, added a faulty dilemma of making it QJ9 to shift the focus over to whether to play the "9". I would really like your opinion (no joking) on how adding those excluded plays (half of them quite interestingly) from double honors would have affected the final numbers.

[Spisu]

RC is calculating the frequencies *after the first finesse has lost*, and *after you have led the second card and 2nd hand has followed low*, and you are at the decision point of whether to try to drop the remaining honor, or finesse 2nd hand for it. At this point, a very large number of the a priori probabilities have been eliminated, dropped to 0%. You have eliminated all hands where 2nd hand has both honors (since RHO won the first trick). You have also eliminated all hands where 2nd hand had a void, a stiff, or a doubleton honor . The remaining possibilities will then be the only possible layouts, expanding to cover 100% of the problem space, retaining their relative proportions given by the a priori probabilities.

 

At this point, there are two ways to make the calculation, and both are valid. In one, you do not identify the honor that won, and count that lone honors are dealt roughly twice as often as the combined situation, for a 2:1 result. In the other, you DO identify the honor that won, which eliminates one of the lone honor possibilities. So in actuality there are only two possibilities, the specific lone honor you saw and combined honors and your opponent chose to win with that specific honor. They are dealt at relatively equal frequency. But your opponent now basically controls how often your 2nd finesse wins depending on which honor won, depending on their tendencies. They can't affect your overall success rate if you stick to your strategy and always finesse, but they can absolutely affect your success rate for a particular honor showing up. Let's say they always play K from KQ. Then if you lost to the K, you will absolutely find that the second finesse only succeeds about half the time, not 2/3 of the time. But if you lost to the Q, now the second finesse wins all the time, since they are never playing the Q from KQ so it has to be Q alone. RC just usually describes the averaged situation where an opponent randomizes and gives the combined odds of 2:1, rather than specific odds of the second finesse succeeding depending on the exact percentage of an opponent's choice from doubleton honor and which of these honors won.

 

 

Not after the first finesse lost to 4th hand they don't.

 

Sure, if an East always played the K from KQ, when he wins with the K the odds are 50-50 on the other card's location IF he NEVER varies. Not that it matters much unless KQ happened to be doubleton....You'd be nuts to gamble on a 50% low end of the scale that E never varies play, when normal play odds for finessing (in this specific situation) range from 50% and up, better by that "and up" than a finesse for a K when you hold A-Q. Interestingly, if an East accidentally dropped a Q from his hand just before you made a double finesse for K-Q, that would be random discovery that increased the odds for the other honor to be in East's hand to 50%.

 

And you are wrong that at the decision point of the second finesse **"A very large number of the a priori probabilities have been eliminated"**. Very little has been eliminated. The 50% likelihood of divided honors is in full play because an East winning honor is fully consistent with that, as is whichever honor E won with. The 25% chance West had both honors is gone, but that also is consistent and actually essential for divided honors. The belief that very much has changed is bizarre.

 

That is the beauty of a double finesse. The a priori odds favor you at every stage.

[Zelandakh]

"I challenge you to find a single case where RC does not produce the correct answer" you say? Challenge met and defeated.

I am sorry, you will have to bear with me, clearly my English has deteriorated since leaving the country as I cannot work out your example. So please, for my benefit, provide the precise card layout along with the approximate (false) odds given by RC and the correct percentages. Thank you in advance.

[Spisu]

I am sorry, you will have to bear with me, clearly my English has deteriorated since leaving the country as I cannot work out your example. So please, for my benefit, provide the precise card layout along with the approximate (false) odds given by RC and the correct percentages. Thank you in advance.

 

Maybe it would help you read what the "Rule of Restricted Choice" definition actually is and then read what I have said. I have not said that RC gave "false odds" on a finesse, but that the basis for the "rule" is a fallacy. That does not exclude that fallacies cannot be piled together, offset, or eventually match what should have been obvious at the beginning. And some parts of multiple math calculations certainly could have partial validity. Bayes is fine if actually done properly.

 

The words identifying RC by the ACBL Encyclopedia as general "rule" are "the play of a card that might have been selected as a choice of equal plays increases the chance that the player started with a holding in which his choice was restricted".

 

And I showed you previously that was incorrect as a rule or principle because of the obvious fact that any two cards (equals or not) can only exist in 4 possibilities in two bridge hands...The ACBL Encyclopedia presents these 4 as equally likely...AK, K/A, A/K, KA...So 50% of all hands have combined equals..(for the hostiles, it may actually be 48-52, but please contact the ACBL to vent your rage at their Encyclopedia's restricted choice data based obviously on symmetry).

 

Therefore, half of all first plays from equals randomly divided between 2 hands MUST come from combined honors, with no possible way for such plays to change the frequency of their own existence. Nor could a lone honor in the other 50% of deals increase its own frequency to over 50% of deals at the expense of combined honors. So the "rule" falls on its face. It is statistically impossible.

[Spisu]

RC is calculating the frequencies *after the first finesse has lost*, and *after you have led the second card and 2nd hand has followed low*, and you are at the decision point of whether to try to drop the remaining honor, or finesse 2nd hand for it. At this point, a very large number of the a priori probabilities have been eliminated, dropped to 0%. You have eliminated all hands where 2nd hand has both honors (since RHO won the first trick). You have also eliminated all hands where 2nd hand had a void, a stiff, or a doubleton honor . The remaining possibilities will then be the only possible layouts, expanding to cover 100% of the problem space, retaining their relative proportions given by the a priori probabilities.

 

At this point, there are two ways to make the calculation, and both are valid. In one, you do not identify the honor that won, and count that lone honors are dealt roughly twice as often as the combined situation, for a 2:1 result. In the other, you DO identify the honor that won, which eliminates one of the lone honor possibilities. So in actuality there are only two possibilities, the specific lone honor you saw and combined honors and your opponent chose to win with that specific honor. They are dealt at relatively equal frequency. But your opponent now basically controls how often your 2nd finesse wins depending on which honor won, depending on their tendencies. They can't affect your overall success rate if you stick to your strategy and always finesse, but they can absolutely affect your success rate for a particular honor showing up. Let's say they always play K from KQ. Then if you lost to the K, you will absolutely find that the second finesse only succeeds about half the time, not 2/3 of the time. But if you lost to the Q, now the second finesse wins all the time, since they are never playing the Q from KQ so it has to be Q alone. RC just usually describes the averaged situation where an opponent randomizes and gives the combined odds of 2:1, rather than specific odds of the second finesse succeeding depending on the exact percentage of an opponent's choice from doubleton honor and which of these honors won.

 

 

Not after the first finesse lost to 4th hand they don't.

 

Your belief that the "first play" of two equals comes "after the first finesse lost" is noted, but the Nobel committee would be interested in your disproving time reversal invariance."

[Stephen Tu]

Spisu, let me break this down step by step for you. Please identify the # of the statement(s) you disagree with and why, and what you think the actual answer is.

 

The suit combination in question is AJt9x opposite xxxx in hand, missing 4 cds including the KQ. You finesse, and it loses to the K. You get back to hand and lead a low card, and LHO produces the remaining low card.

 

Out of the following numbered statements, which do you disagree with and why?:

 

1. Originally, RHO will be dealt singleton K, so layout is Qxx - K, approximately the same amount of time as he will be dealt KQ doubleton. 6.22% for the stiff K specifically, 6.78% for the doubleton with both honors. Another 6.22% for the stiff Q, but we know now this is not the case, because the first finesse was won with the K.

 

2. After the first round finesse has lost to the K, and LHO follows low to the second round of the suit, all possibilities have been eliminated, other than xx - KQ and Qxx - K. RHO could not have started with the Q alone since he played the K, the layout can not be Kxx-Q. LHO could not have started with both honors since you already lost one to RHO. LHO did not start with a stiff. LHO did not start with the doubleton qx because he followed low both times. LHO did not start with a void. RHO did not start with a void. Only 2 possibilities left, originally xx-KQ and Qxx-K.

 

3. Against an opponent who plays randomly from KQ, the chance of the second finesse winning is close to 2:1.

 

4. If you agree with 2&3, why is this so, if there are only 2 possibilities remaining, xx-KQ and Qxx-K, which were close to equally likely to begin with, if it is not due to RHO having freedom of choice to win either honor from the doubleton KQ holding?