Spisu

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.

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.

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.

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.

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.

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

"Had to play it" suggests random discovery in this case, unlike other most other so called RC situations. So in 2 of 3 cases you've found a singleton. Notably the originators changed their tune in this case and abandoned their focus on specific cards played and came up with "Quack". The odds are 2:1 for the randomly discovered play of a singleton.

Random choice of play is an oxymoron. And mathematical validity of the sort of weather prediction or a batting average predicting a specific at-bat hit can exist. These can be calculated/updated with the valid use of Bayes, but the RC chose to assume randomness with no basis. But assumptions require verification...You might assume as a starting point that one player chose 50-50 between equal honors, then keep records of his plays over many hands to see, and update that 50% to whatever (THAT is a Bayesian update)...And then you would know one person's proclivities..Then you could do the same on all the bridge players you ever play with...or, 1. you could do what RC did and presume, then claim it is mathematical but isn't, or, 2. just accept the fact that equals are divided 50% and the targeted hand has both honors only 25% when the targeted hand is known to have any honor (all a priori 2:1).

I am familiar with Bayes and the restricted choice folks'claim it is based on the Bayes postulate known also by some as the "Equidistribution of Ignorance". The only thing missing I've noted is any confirming data to support a valid basis for that other than the claim. Note that assuming an opponent plays randomly is not mathematically valid. So if you have evidence of a Bayesian basis other than presumptions or guesses, please share. I'd love to see it. BTW, the freq dist for lone vs double honors shows KQ, K/Q, Q/K, QK each 25%...so the frequency of a specified honor such as the K in East is 1/2 of 25% or 12.5%, while a KQ is 25%, twice that of a specified lone honor. The math of RC depends on your not noticing the frequencies are different. For 400 hands, East holds 400 total honors 200 are divided( 100 Ks and 100 Qs) and 200 KQ combined honors. Simple math for first finesse shows 100 lone Ks played will match with 100 Ks from KQ (1:1), 100 Qs match with 100 Qs from KQ, leaving 100 KQs to nail you in the 33% of losing second finesses. There is no "half" relationship of KQ.

This is a good example but probably not in the way you intended. It illustrates the fallacy regarding the RC claim equal cards are distinguishable (well, to declarer but not defender, a notable contradiction). The equals quality of statistical significance for the boxes is that 2 boxes are the same (each Au or AG) and 1 is different (AU/AG), and nothing to do with Au/Ag itself. It's 2:1 when you accept the obvious "sameness" that the other coin will be the same as you see. The equivalent in RC (had they been more astute) would have been to seize on the fact that a K and Q are equals like the "sameness" of 2 Au and 2 Ag over 1Au/Ag, then not going astray at seeing a gold (He played a KING!), and flying off at a tangent over what the specific significance of that gold (KING!) might be...and getting "creative" in the process.

Sorry if I was impatient. But if you read a sentence or two into restricted choice in most places, it professes that plays of one of two equals reduces odds the other equal is present by half. But the fact is and the point I presented is that 50% (+-) of all 2 honors divided between hands (normally with side cards of course) are combined honors, and nothing can reduce the frequency of an honor played from a combined honors' hand below its frequency/odds of 50% (they are the only two cards, combined, and in half of all such hands)....nor can a single (lone) honor increase the frequency of divided hands above 50%. There are of course consequences of this which should be apparent. Nothing I have said or written suggests impacts the odds for the second finesse which are easily seen a priori, independent of either equal played. But RC does not set those odds.

I am surprised at the comments here. This is not that hard. The odds are obviously 2:1 for the second of a double finesse after losing the first finesse, and I have never said a word to the contrary. Why not give a thought that your example was perhaps an actual reason for those odds, contrary to the claims of RC which asserts the odds are based on play from equals.

A good starting point for you might be to work on understanding that the "First Play" of that suit that I specified clearly does not come "..After a round of the suit has been played" as you suggest.

I may be more familiar with the 2 kids/ boy girl puzzle issue than you. Many might say that actually seeing the boy would be random discovery and make odds 50-50...So I try to use more easily perceived sources of non-random data such as meeting your new neighbor you know has two kids and you see a pink girl's bike. Then you would have justification that you very likely had learned indirectly (non-randomly) that one was a girl, which makes the odds 2:1 for a boy and girl to be the 2 kids.

Your is a non-sequitur because I have not even mentioned the wisdom of plays. Restricted choice has probably endured because the "finesse twice" is valid but arises elsewhere. It is the rationale of RC that is error, and the fact is that it piggy-backs on the actual valid basis for that strategy which is an a priori stategy.

Seriously? You cannot recognize 4 a priori possible holdings showing how 2 cards divide between 2 hands? There is no other way two cards divide. Did you think the King could divide by mitosis and have 2 in one hand and a Queen or 2 in the other? Get this please: Two cards divided between 2 hands can only be 2 in either or 1 in each, in 2 ways. That is 4 possibilities Funny that you show up insulting but utterly clueless.

The "rule of restricted choice" claims that play of one equal by an opponent decreases the chances the other equal is in that same hand. But 2 equals (say K-Q) divide between 2 opponents hands in only 4 ways, West K-Q, K, Q, zip, and East the same in matching hands. Each of the 4 is a 25% possibility. So 50% of cases will see both honors (or any equals) in the same hand and being led or otherwise played from half of all hands. So half of all first plays of an equal come from combined honors. So restricted choice is a fallacy. Any questions? Edit to add: I'm surprised to see comments below talking about singletons and cards already played. Perhaps I overestimated the concept of "expert". This post very simply shows that 2 equals (regardless of any accompanying cards) must eventually be played, and when one is played, it will come 50% from combined honors and 50% from divided honors. And that disputes restricted choice claims. If that's not clear, at least I tried. Edit to add: I thought it would also be obvious that a "first play" of a suit does not "follow" or otherwise come subsequent to that first play of a suit but, for the bewildered, that is indisputable fact.

Nicely stated...It's rare among the restricted choice fallacies that anyone knows the key to it all is a relevant honor by an East rules out both honors in West. This eliminates that 25% from the equation and, since divided honors are not mutually exclusive but inclusive, their 50% likelihood continues against the 25% a priori for both honors in East. So you've got an automatic 2:1 for a second finesse when it's an AJ10 sort of a double finesse. Here though I would have tried to guess who might be short, then played the top honor first that allowed a finesse through the other hand.

I've been forced to avoid doubles in favor of a simple overcall (whatever the hand strength) if I have a good suit of my own, just to keep the robot from going bonkers and rebidding some ragged 4 or 5 suit card suit [in competition with your solid suit(s)] all the way to the seven level....Yes, bidding clubs (K10xxx), unsupported, to the way to seven with a weak hand as I bid my solid 6 card major! And just recently I held 9 hearts (AKQxxxxxx), and the robot bid an unsupported 6 card club suit, bid for bid, to seven...Yes, I was mad as heck, and bid 7 hearts, going down of course. But unfortunately the robot may do insane things when you overcall also...Like a couple weeks ago RHO opened 1D, I overcalled 1S, 1NT, P, 2D...then P,P,P...Yes dear hearts, the robot never bid, and passed out to the opponents 2 diamonds in the balancing seat holding 7 hearts to the AKQ! Yes "he" let the opponents play 2 diamonds with that holding even "knowing" his partner had values also. On one occasion I opened 1D, the robot bid 1S, 1NT by me...then P,P,P. The robot then unveils 7 spades AKQ10xxx....too "weak" to rebid apparently. And there are many more examples... I hope the latest improvements work for the robots because the argument that it all averages out is a lame one, especially when you see bidding mutilated in a way that not even the worst partner you ever had would do. A well bid and played hand is a work of art and symmetry, and seeing one butchered is a jolt that upsets you, and can affect your subsequent play.