[Stephen Tu]

shyams, on 2016-July-16, 10:01, said:

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]

[quote name='1eyedjack' timestamp='1468683797' post='891556']
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]

eagles123, on 2016-July-16, 09:26, said:

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]

1eyedjack, on 2016-July-16, 11:50, said:

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]

1eyedjack, on 2016-July-16, 05:50, said:

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]

Phil, on 2016-July-16, 12:20, said:

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]

1eyedjack, on 2016-July-16, 09:43, said:

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]

Stephen Tu, on 2016-July-16, 09:57, said:

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]

Spisu, on 2016-July-16, 08:18, said:

"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]

Zelandakh, on 2016-July-16, 17:53, said:

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]

Stephen Tu, on 2016-July-16, 09:57, said:

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?

[Zelandakh]

Spisu, on 2016-July-16, 21:22, said:

Maybe it would help you read what the "Rule of Restricted Choice" definition actually is and then read what I have said.

Oh I have read both. As a mathematician I understand BT and RC - now I want to understand your position in defining it as "bunkum".

Spisu, on 2016-July-16, 21:22, said:

I have not said that RC gave "false odds" on a finesse, but that the basis for the "rule" is a fallacy.

So are you saying that RC gives the correct result in all cases? Is your position akin to saying that gravity does not exist because it is really just a bending of space-time and that we should therefore not use Newton's laws?

[Spisu]

Zelandakh, on 2016-July-17, 04:22, said:

Oh I have read both. As a mathematician I understand BT and RC - now I want to understand your position in defining it as "bunkum".


So are you saying that RC gives the correct result in all cases? Is your position akin to saying that gravity does not exist because it is really just a bending of space-time and that we should therefore not use Newton's laws?

As you probably know, "Bunkum" was not my word. It was one of your cohorts who used the word (that's why it was in quotes if you didn't notice), and I cited him to disagree with his point.

And, as to Newton vs Einstein's general theory, there is no comparison. My point is as I've said before here...If you say 2+2=4 because 2 cubed divided by the # of integers added also equals 4, then an honest mathematician has a duty to speak up. I did not say you can't get a right answer by the wrong means.

But if you understand Bayes and RC, please share how actual Bayesian statistics impacted RC other than being used incorrectly to presume a player from double honors does so randomly but with no updating or testing to verify that presumption. Actual facts would be appreciated.

[shyams]

eagles123, on 2016-July-16, 09:26, said:

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

Mods, pretty please?

[Spisu]

1eyedjack, on 2016-July-16, 13:58, said:

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.

Please come back if you have something to add. BTW, did you see the challenge on the ACBL Encyclopedia Example 1 under Restricted Choice? Explaining why the numbers don't seem to add up would be a big "something".

[Zelandakh]

Spisu, on 2016-July-17, 05:33, said:

As you probably know,
<snip>

What I know is that you quoted me but failed to answer my question. So here it is again: are you saying that RC gives the correct result in all cases?