[Spisu]

nige1, on 2016-July-14, 10:31, said:

An intelligent player at Reading Bridge Club claimed that the rule of restricted choice was bunkum. We offered to set up a typical matrix, then shuffle the remaining cards and back our respective judgements at £10 per deal. After some homework, our friend declined our challenge

Why would he play differently if RC is "bunkum"? The odds are 2:1 without RC. The big question for some here is why aren't the odds 4:1 if RC works? Louis Watson recommended it strongly in 1934 describing the first finesse as a preparation as I recall.

[Spisu]

shyams, on 2016-July-15, 06:40, said:

Seriously?!?! You don't see the logical dissonance between this post and your opening post?

If the answer to the "2 kids/ boy girl puzzle" is 2:1 for the second child to be the opposite sex, why is the answer to "2 missing honors/ left right puzzle" not 2:1 for the missing honor in the opposite hand?

You're confusing the issue of the odds for a second finesse and the validity of RC. The odds are 2:1 for the second finesse whenever a W does not have both honors and an E wins the first trick. That has nothing to do with pointing out the fact that 50% of all plays of 1 of 2 equals in open play come from combined honors undermines RC.

[Stephen Tu]

It's not 4:1 because apparently what your idea of how one method of RC is calculating isn't actually what RC is saying, you are totally confused about it. Let's take AJT9x opposite xxxx, finesse twice vs. finesse follow by drop, and specify that first finesse lost.

You are saying, it was either Kxx- Q Qxx- K or xx-KQ. So 2-1 for two finesses. (Not exactly 2-1 because 2-2 breaks are slightly more frequent than 3-1 because of the effect of other cards, needing 13 cds in hand. But 2-1 is a close enough for an estimate, and for the rest of this post I will just use this as an approximation). That's certainly reasonable, we aren't disagreeing with you at all about this, it's totally valid way to come to correct conclusion about strategy.

The other method of calculating is saying:
First finesse lost to the Q. Therefore it was not Qxx-K because duh. So it's either Kxx-Q or (xx-KQ AND opp chose the Q, which as a baseline guess we'll take as 50%)
So 1 case to half of 1 case, or 2-1, same as your conclusion
Or:
First finesse lost to the K. Therefore it was not Kxx-Q because duh. So it's either qxx-K or (xx-KQ AND opp chose the K, which as a baseline guess we'll take as 50%)
So 1 case to half of 1 case, or 2-1, same as your conclusion

Or: First finesse lost to the Q. Therefore it was not Qxx-K. Let's not assume opp plays randomly perfectly even from KQ. Let's say they play p% the Q, and 100-p% the K.
So if first finesse lost to the Q, it's 1 case (Kxx - Q) vs. p/100.
But first finesse lost to the K, it's 1 case (qxx - K) vs. (100-p)/100.
So overall, it's going to be 2 cases vs. ((p/100) + (100-p)/100). Well ((p/100) + (100-p)/100) = 1. So again, 2-1, for any value of p, same as your conclusion.

It's 2:1 any way you calculate, so I don't see your beef with the alternate methods of doing the calculation.

The only time one starts to get a different conclusion about strategy is when you start considering that specific 2-2 break is slightly more likely than a specific 3-1 break, combined with an opponent who is not randomizing hardly at all and playing say 95% the higher honor, or 95% the lower honor.

[Spisu]

Zelandakh, on 2016-July-15, 07:28, said:

Spisu, restricted choice is the bridge name for something called Bayes' Theorem. You can read about BT here. Once you have read that and digested it, please feel free to come back and explain to us why BT is mathematically incorrect. I have no doubt that you will be hailed as one of the greats of the field if you succeed.

As you can see from the theorem itself, it is required for there to be 2 events. This is why it is pointless to discuss RC/BT on the first round of the suit. It is fundamental to understand this before you move on to trying to disprove it. Unfortunately, I am also fairly confident that Bill is correct but would love to be proved wrong.

Well, I merely showed that the first honor play from equals (the underlying basis of RC) is a fallacy based on the reality that such plays from combined equals are 50% of all plays involving 2 equals. Knowing that at or before trick one could be useful to some. But restricted choice is certainly NOT Bayes Theorem as you say, though they do claim it is based on a Bayes Postulate. I have never hinted that "BT" is incorrect and I would think that properly used it is completely valid.

[Spisu]

Stephen Tu, on 2016-July-15, 16:41, said:

It's not 4:1 because apparently what your idea of how one method of RC is calculating isn't actually what RC is saying, you are totally confused about it. Let's take AJT9x opposite xxxx, finesse twice vs. finesse follow by drop, and specify that first finesse lost.

You are saying, it was either Kxx- Q Qxx- K or xx-KQ. So 2-1 for two finesses. (Not exactly 2-1 because 2-2 breaks are slightly more frequent than 3-1 because of the effect of other cards, needing 13 cds in hand. But 2-1 is a close enough for an estimate, and for the rest of this post I will just use this as an approximation). That's certainly reasonable, we aren't disagreeing with you at all about this, it's totally valid way to come to correct conclusion about strategy.

The other method of calculating is saying:
First finesse lost to the Q. Therefore it was not Qxx-K because duh. So it's either Kxx-Q or (xx-KQ AND opp chose the Q, which as a baseline guess we'll take as 50%)
So 1 case to half of 1 case, or 2-1, same as your conclusion
Or:
First finesse lost to the K. Therefore it was not Kxx-Q because duh. So it's either qxx-K or (xx-KQ AND opp chose the K, which as a baseline guess we'll take as 50%)
So 1 case to half of 1 case, or 2-1, same as your conclusion

Or: First finesse lost to the Q. Therefore it was not Qxx-K. Let's not assume opp plays randomly perfectly even from KQ. Let's say they play p% the Q, and 100-p% the K.
So if first finesse lost to the Q, it's 1 case (Kxx - Q) vs. p/100.
But first finesse lost to the K, it's 1 case (qxx - K) vs. (100-p)/100.
So overall, it's going to be 2 cases vs. ((p/100) + (100-p)/100). Well ((p/100) + (100-p)/100) = 1. So again, 2-1, for any value of p, same as your conclusion.

It's 2:1 any way you calculate, so I don't see your beef with the alternate methods of doing the calculation.

The only time one starts to get a different conclusion about strategy is when you start considering that specific 2-2 break is slightly more likely than a specific 3-1 break, combined with an opponent who is not randomizing hardly at all and playing say 95% the higher honor, or 95% the lower honor.

Personally I am interested in the basic principles without complicating distributional factors. We need to get the principles right, and the outlying hands with added complexities can come next. So the hand you show is not on my radar at the moment. But it's not just "a different way to get the numbers". People are drawing false conclusions in general play when they see an equal that is as likely to be a loner(divided honors) or from combined honors. Double finesses target a hand and force a play from one hand which is perforce half as likely to hold both honors as the 2 opponent hands' combined. Principles need to be correct.

[Phil]

Spisu, on 2016-July-15, 15:55, said:

It is patently absurd to come up with a stipulation and then claim it must be true unless someone can prove it's false.

Except we have math on our side.

[Stephen Tu]

Spisu, on 2016-July-15, 20:02, said:

People are drawing false conclusions in general play when they see an equal that is as likely to be a loner(divided honors) or from combined honors. Double finesses target a hand and force a play from one hand which is perforce half as likely to hold both honors as the 2 opponent hands' combined. Principles need to be correct.

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!

[1eyedjack]

Spisu, by now I expect that you have had a chance to digest my post in another thread, here

http://tinyurl.com/hjps5mv

That post contrasts two situations which have in common that you hold king to 5 opposite ace to 4,

They differ in that in case 1 the only significant missing card is the Queen, but in case 2 you are missing both Queen and Jack.

In both cases you cash the ace and then lead low toward the king.

In both cases 2nd hand fallows low to both tricks. 4th hand follows to the first round with a low card in case 1, and with Jack in case 2.

I attempt to demonstrate in that thread, that in case 1 the drop is about as good as the finesse on the second round (slightly better in fact but I gloss over that), while the finesse is heavy favourite in case 2. The only difference in the underling conditions is that in case 2, the fourth hand's play of Q on first round is potentially from equal cards (QJ) where in case 1 the first round play of x is not. I cannot escape the conclusion that the difference in initial underlying conditions is the cause of the difference in the resulting probabilities, which is all that PRC states. DO YOU DISAGREE?

Incidentally, was there any point to your post #38 in this thread? If it was simply to preserve my comment for posterity lest I have some inclination to remove it, I can assure you that I have no such intention.

[benlessard]

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.

[Zelandakh]

Spisu, on 2016-July-15, 16:56, said:

I have never hinted that "BT" is incorrect and I would think that properly used it is completely valid.

Could you provide an example of a case where the correct application of RC provides the wrong answer. That you do not think RC is BT tells me that you simply do not understand it mathematically so I think we will find your error fairly easily from such an example.

[Spisu]

Phil, on 2016-July-15, 21:11, said:

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]

Stephen Tu, on 2016-July-15, 22:37, said:

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]

Spisu, on 2016-July-16, 05:59, said:

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.

[Spisu]

Zelandakh, on 2016-July-16, 07:30, said:

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.

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]

benlessard, on 2016-July-16, 01:12, said:

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]

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.

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]

Spisu, on 2016-July-16, 05:59, said:

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.

Quote

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]

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

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!