kfgauss

2♣ is not forcing, but is very encouraging. While it's not my choice, it is reasonable. Partner's hand is definitely good enough that he really must bid over 2♣. Andy

You're right that by logic it can't really be pure penalty, but it should be passable. I wouldn't try it with 4 clubs in any case (nor with this pickup, or any other perhaps). Maybe a defensive 4-4-2-3 or a good 4-4-3-2 would be an appropriate hand. Andy

Your partner seems not to know how to bid and is basically at fault on all of the hands. You did make a few bids I would not have, though: 1. I'd rebid only 1H... if partner passes, you're quite high enough. No need to GF yet. Partner should rebid some number of notrump instead of 2S (asking), and certainly your 3C (the correct bid at that point, probably) was forcing. 2. Partner seems not to have understood that you showed 4-4 in the majors. Even so, you'd expect a 1S bid over 1H. Your bidding is perfect up until the double. I see no need for that, especially with what you expect to be a 10 card fit and only a singleton in their suit. Pass and 4C are both reasonable. 3. 3C is fairly aggressive, but it had no impact on partner's insane flyer thereafter, unless he took it as strong or some such. You should correct 5S to 6H immediately, though (and in fact possibly try for 7 given your ♣AK), as you definitely prefer partner's (presumably longer) hearts to his spades. It's partner's business that he made you choose at that level, and he must have a really good hand (which he doesn't). 4. 3C is stayman, but given your partner's previous actions, 4C isn't too surprising. Good thing you landed vaguely on your feet. Andy

Probably you mistook the "1" and "0" of diamonds as distinct cards :lol: (it does make the diamond suit look longer at first glance, actually). Anyways, I'm a 2♦ bidder. Partner won't tend to pass when we have something, and this will afford me better description of my hand (more room) and more safety when we have a misfit. Well done to get to the slam. Of course, who's to say if you'd held back a bit you wouldn't have subsequently made a slam try over partner's 3♦ (which partner is presumably accepting). Andy

I'd play: over 1H: double = hearts 1S = good takeout higher = as over 1S (i.e. 1N = distrib takeout, 2S = nat, etc. if you normally play 1N = nat, then still play that) over 1S: double = spades 1N = takeout (even if you play 1N = nat normally) higher = as over 1H (2N might be bid more frequently because there's no good/distrib takeout distinction at a lower level) I think over 1H, you should definitely "make them pay" by being able to make a heart overcall at the 1-level (by doubling), and keep the rest of your bids the same (1S replacing the takeout double). Over 1S, the distinction between two forms of takeout isn't worth losing showing spades at the 1-level. For competitive purposes, I'm happy my opponents are playing this. I get to "overcall" either major (by doubling) and the only thing I lose is a good/distrib takeout distinction over 1S (or 1N natural, if that's what I play). I think gaining the heart overcall of spades is worth more than losing whatever it is I've lost. Andy

The theorem isn't paradoxical, but the "this statement is false" idea is used in the proof. I, perhaps somewhat sloppily, referred to this idea as a paradox. Helene expands a bit on how this comes in to the proof above: you figure out how to write the statement "this statement is unprovable" in your formal system (or, perhaps more precisely, "this statement is unprovable in this formal system"). Then there can't be a proof of this in the formal system, but this means it's true (and, of course, unprovable). Yes, but doesn't that make it merely an example of "proof by contradiction"? (NB I understand now what you meant by "at the heart of it it's paradoxical", so no need to argue further on that ;-). --Sigi Yes, certainly. I intended "paradox" to refer to the "this statement is false" idea, as this is a well-known paradox. Perhaps I sloppily applied the term to the theorem/proof.

I hate this problem, but will perhaps double. Double is pretty dangerous, as they'll be quick to (redouble and) double at these colors. A good agreement about running will, however, probably be helpful: If pard bids 2♣, and the opps double: Running to 2♦ shows equal length (in the reds) or longer diamonds Redouble shows longer hearts (the more expensive strain) This scrambling agreement is discussed in the current Bridge World in an MSC write-up (on the auction P 2S P P; X P 3C X; ?). (I'm not certain why this is the agreement they mention [or how standard it is] instead of xx = equal or longer ♥ and 2♦ = longer diamonds, but in any case, it helps us here.) Andy

3♣, planning on rebidding 4♣ (which is NF) unless something exciting happens. Andy

This isn't quite correct. Sets are allowed to be elements of themselves. The resolution to Russell's Paradox involves making sure that sets never get "too big." The "set" of all sets that don't contain themselves is in fact not a set, but what mathematicians call a class (and the sorts of things you can do with classes are more restricted than those for sets). Andy

The theorem isn't paradoxical, but the "this statement is false" idea is used in the proof. I, perhaps somewhat sloppily, referred to this idea as a paradox. Helene expands a bit on how this comes in to the proof above: you figure out how to write the statement "this statement is unprovable" in your formal system (or, perhaps more precisely, "this statement is unprovable in this formal system"). Then there can't be a proof of this in the formal system, but this means it's true (and, of course, unprovable). Andy

No, then I did not summarize it clearly enough. It goes like this: Consider the countable set of all proofs, natural number, statements, predicates and propositions that can be written in a particular language. Let's enumerate that set. Consider the predicate: q(n): the system contains no proof for the statement p_n(n), i.e. the n'th predicate with n as an argument. Since we just described the predicate q without using any tricky stuff, we can assume that q belongs to our enumerated set. Let's call it's index k, i.e. p_k = q. Now consider the statement q(k). True or false? The answer is that it is true but that the system can contain no proof for it. This is contrary to how Sigi phrased it. Sigi says that there are truths that are hidden for us. But this is not hidden for us, it's just hidden for the system. Penrose claims that a Turing machine cannot convince itself that q(k) is true since it's system contains no proof for it. Yet with human mathematical insight (something external to the system), it can be proved. This surprises me since Goedel's theorem, as I understand it, is what Sigi explains. Besides, Penrose's book could easily be assumed to belong to the enumerated set. This would turn q(k) into a paradox so it only proves that there is some logical fault in what I just wrote. Maybe there is a logical fault in Penrose's book or maybe I just misunderstood it. Anyway, I do not think that it's particulary relevant to the issue of consciusness. (See posts from kfgauss ans joshs, I agree with both of them). I'm not really an expert on this, but here goes: You ask two separate questions. I'll try to paraphrase (tell me if I don't state them correctly). 1) How is it that we can prove q(k) with our human mathematical insight, yet Sigi claims certain truths will be hidden from us due to the formal system we choose to work with? I believe there's another point of confusion running around here, which is that mathematicians sometimes work with axioms and sometimes work within models. When working with axioms, there is no truth beyond provability, but there are undecidable statements. In this case, we may have some intuition that we'd like something undecidable to be true. Then one might add an appropriate axiom (that one intuitively likes) so that one can prove it (e.g. the Axiom of Choice and, somewhat less frequently, the Continuum Hypothesis are occasionally used). When working within a model, however, we've essentially chosen a mathematical universe and any statement we can make will be true or false (in this model). If this model is big enough (contains a copy of the natural numbers etc) then the incompleteness theorem will apply. Going to a bigger model, one will be able to prove some things one previously couldn't, but now there will be more statements, etc. A nice example from the Wolfram page on Gödel's Incompleteness Theorem is that apparently the consistency of arithmetic can be proved if one uses transfinite induction, but if one wants to prove the consistency of arithmetic + transfinite induction, one would need to expand further. (I'm not a logician, so some terminology and/or claim may be slightly off. Don't ask me what the difference between "model" and "formal system" is, or if there indeed is one.) 2) Couldn't you just add (the ideas in) Penrose's book to a formal system, so that it would be able to prove its own q(k) because it would have the idea that such q(k) are true? I'm not really sure what you mean by add (the ideas in) Penrose's book to the formal system. Probably the answer is that this just doesn't make sense and/or can't be done. Andy

Presumably mr1303 responded 3D (the correct response to stayman with this hand) and played there, partner still believing a diamond preempt opposite, instead of bidding & making 3NT. The point appears to be that the captain of the other team didn't "get" UI and was surprised that this hand wouldn't bid 3NT once it found out about a misunderstanding. Andy

The first hand I definitely pass. The second hand is quite a dangerous hand. It's clearly dangerous to bid, but it's also dangerous to pass -- you can easily give up a double part-score swing (I guess these aren't so huge at total points, perhaps) or miss a game. I'm somewhat timid and will pass because I'm red and I can't get out in clubs at the 2-level, but I have no idea what's right. A simulation could be very interesting. Andy

Another reason to play puppet stayman! :D Andy

You mean Douglas Hofstadter. Richard Hofstadter appears to have been a historian. I also quite liked that book. It paints a pretty picture of things, though one doesn't really have any way of knowing whether it's an accurate picture. Lots of other good stuff in there too. Andy

This convention (called reverse flannery) is a somewhat common bidding gadget, used somewhat independent of system (i.e. it's not precision-specific). The point of the convention is that one can't really show 5♠-4♥ hands easily with standard bidding, especially if partner now bids 2 of a minor. However, 4♠-5♥ hands are quite easy, as you can find a spade fit over your 1♥ response, so if partner bypasses 1♠ to bid 2m you know you can ignore spades. In either case, there isn't much of an issue over a 1N rebid, as checkback stayman can be used on invitational hands, and with the weak 5♠-4♥ hands you can pass, rebid 2H, or rebid 2S as you see fit. (I guess there's a slight gain from having 2H on this auction promise 5-5 if you adopt this, but that's pretty minor.) As with any convention, there are all sorts of follow-ups to deal with, and the gain is pretty infrequent. Unless you're really fine-tuning things, I wouldn't recommend playing this (and other meanings for 2H and 2S may arguably be better anyways). Andy

Give me a hand where you can see that partner can't be strong with 5-5? Hum.. 17-18 hcp with 1534 and strong diamonds? Say x AKxxx AKQ Qxxx I suspect you mean AKxxx x AKQ Qxxx (partner opened 1♥, you bid 1♠ & 3♦). But seriously, if you have this hand, you would/should bid 2♦ as 4th suit forcing, which kinda removes it from any discussion. Got another example? No, this was an example of a hand where you would supposedly "know" that partner doesn't have a GF 5-5 ♠+♦ hand. I'm not sure why partner can't have something along the lines of AKJxx x J109xx Ax though. Andy

I read parts of this book a long time ago and recall thinking that the philosophy part of it was entirely rubbish. (I'm no philosopher, of course, though.) If I recall correctly, his big philosophical claim is that because of Gödel's incompleteness theorem (yes, at the heart of it is a "this statement is false"-type paradox), there will always be true statements that a given robot can not realize are true, but that humans can. Two (rather different) responses to this: 1) A robot, though deterministic, needn't be a machine for rigorously proving theorems. Humans believe things that they cannot prove (often contradictory things), and it doesn't seem hard to create a (deterministic) robot with this capability. (It does seem hard [as an engineering/computing problem] to create a robot with this ability that will "believe" the correct sorts of things most of the time, but this isn't my problem, as long as I've gotten around his philosophical point.) 2) Ignoring the idea in point (1) that humans believe things they cannot prove, note that he starts with one given robot and notes that humans will be able to realize the truth of some statement that it cannot. He does not, however, complete the argument by arguing/showing that there is no such statement for a given human. (And indeed, though I'm now being somewhat facetious, I'm certain there are [very complicated] statements that are impossible for me to be able to realize are true that robots will.) [There is a slight sticky point here, which is that if he is indeed claiming that all humans will be able to realize the truth of the statement the robot cannot, then surely the same can't be true for any human. However, I don't believe the statement with the proviso "all" for various reasons.] Andy

I really think this shows at least ♠Hx. I do raise on 3 often, but with 3 card support and a good 6 card minor I'll often choose to rebid the minor (if this is where we differ, then I understand somewhat more, but still disagree with 4♠ on xx). This is for various reasons, including choice of games with 5 reasonable but not great spades (say KQ9xx), as well as being able to better judge slam prospects when partner's hand is better overall. Also, as others have noted, one can/should usually take it slower with 6 decent spades. Andy

I also pass, but am less certain about the meaning of 4NT. I wouldn't know what "standard" would be (to play?) and it seems there are several different "obviously correct" scientific interpretations. Anyone have meta-agreements that cover this in a sensible way? Andy

Not only does this lose to Hxx-Hxx when they're aware enough to duck (and hey, we're already assuming they're aware enough to duck from Hx-Hxxx), it also loses to Hxxx-Hx, which none of the other lines lose to either (which they also have to duck from, but this one's easier). This makes it substantially worse. Sure, but absent much info, low to 10, then back & finesse is superior in number of cases and in addition, in the cases it loses, the opps have to find a duck from Hx. Andy

Perhaps you could clarify the alternative line that you think is superior, in order to check against the various other holdings that might gain or lose. My prefered line (low to T, cross back and run J) ONLY loses against Hx with LHO (who ducks the T). But it does not ONLY gain against xx with LHO, so comparing the frequencies of those holdings in isolation is inconclusive. You refer to LHO having singleton H. Although much less frequent than the other combinations, this only seves to strengthen the argument for low to T on first round, followed by crossing and running the J on the second round. LHO holding singleton small is an irrelevancy, as either low to T first or running the J first both succeed. My comment was meant to support your line, which I agree is the correct one (i.e. low to 10, then back & finesse again). The variant line of cashing the Ace after leading low to the 10 (presumably what Josh was suggesting) only does better on Hx-Hxxx holdings and only does worse on xx-KQxx and x-KQxxx holdings. Andy

You're forgetting about x-KQxxx.

My calculations show that Line III is actually best by a small margin. I think you've forgotten some cases that Line I fails: ♦K and 4 or more clubs including the 9 with RHO (5 cases of club layouts). Thus, counting layouts of ♦K and clubs in which Lines I and III differ, I get these 5 where III beats I, and 3 where I beats III: the two you mention with ♣H9xx and ♦K with LHO as well as an extra one with ♦K with LHO and all the clubs with RHO. (As you point out, Line II is substantially worse than either I or III. You're losing in lots of cases where the clubs are offside but the ♦K is favorable and gaining nothing as LHO will return a club if he wins the diamond, and RHO will return a red card and you'll have to take the club hook again anyways.) In conclusion: tough luck, Phil. ;) Andy

Yes, but you've essentially prepaid for the final it seems, so it's twice that if you don't qualify. Andy