starfruit

[hv=d=s&v=b&n=saxxxhxxdkjtxxcxx&w=sjtxhxxdxxxxcqjxx&e=sxxxhxxdaqxxcaktx&s=skqxhakqjtxxdcxxx]399|300|Scoring: IMP[/hv] N E S W -- -- 1♥ P 1♠ P 2♣[1] P 2♦[2] P 2NT[3] P 3♣[4] X P[5] P 3NT[6] P P P Playing with screens: [1]: 16+ any or 4+♣ [2]: 8+ any [3]: N to W: 5♥ 4clubs exactly, S to E: ♥ 5+ 1 suiter FG [4]: Relay [5]: S to E: weakness in ♣, N to W: perhaps giving him a chance to XX with ♣ [6]: N to W: Unsure, N to W: ♣ stop Play: W led ♣x which E took with K. W continued with a ♣A. E returned ♣ to W's QJ. He did not cash ♦A, hoping to get 1 extra trick Meanwhile, dummy pitched 2♠ and declarer pitched a ♥ after showing up with only 3♣ W switched to a ♠. Has E/W been damaged by this misinformation? W is suggesting that the misinformation has led to him being unable to figure out that declarer actually had. He thought that likely declarer will not have sufficient tricks (only 5♥) and everything might hinge on finding the ♦Q. Had he knew that declarer has a ♥ 1 suiter, he would not have think that way and played partner for the ♦A. N/S feels that once declarer has shown up with only 3♣, it should be clear to W that there has been misinformation (Is he able to call for director at that moment?). In addition, N suggested that he did not receive any unauthorized information (screens) and had been bidding according to system (hence 3NT without ♣ stop). Hence, this would be more akin to a situation where S misbid his hand. In addition, E had been given the right explanation and he is in fact in a position to take advantage of the situation (taking 3NT-1 when 4♥ would have been iron clad). Can we treat this disadvantageous position as the "penalty" for this series of misinformation. Hopefully not too confusing up till this point... Thanks for any views given! :(

The guy who taught me this called it a "tracer bid", but I'm not sure where he got that from. A quick search on google reveals that there is an article from The Bridge World magazine named "The Tracer Bid", written by Krishna Vahalia. It's the Februrary 1987 V 58 N 5 issue btw. Unfortunately I don't have that issue. . . (I'm only 4 that year lol) so maybe someone else can check this out.

Thanks for all the help given :) I managed to contact Anna for help. She helped me to convert the card to PDF. . . with all the suit symbols intact! Wonder how she does that? I'd better switch to the Word version next time. . .

Hi, I'm currently using the WBF convention card editor version 2.18 to make a new convention card in preparation for an up coming event... The problem which I encountered comes when I tried to create a PDF version of the supplementry notes by using the "print to file" option. Turns out that the output file did not have any of the "♠♥♦♣" symbols on them, and instead they were replaced with dots. Anybody has any idea how to make the symbols appear on the generated PDF files? I heard that the problem concerns OS of the computer used, and that there'll not be any problems if I'm using winME/98 (I'm using win XP now) but unfortunately I can't find any available computer with that setting. P.S : I've seen alot of the "experts" convention cards (example meckwell) having a heading of "WBF convention card 2.19"... but I can't seem to find any version of the editor higher than 2.18. Anyone knows how that version of card is made? Wonder if that version can solve my current problem... Thanks for any attempt to help :blink:

Now for the other side of the problem : [hv=d=e&v=n&s=sxhqxdjt9xxckjtxx]133|100|Scoring: IMP[/hv] I was the one who made the 5♣ call, thinking that our opponents are cold for 5♠ and so hoping that they'll make a wrong guess to lvl 5/6. So unfortunately as the cards lie we're cold(?) for -650 for 5SX= or -500 for 6CX-2 Though in reality declarer finesse the wrong side for the ♥Q (obviously a major blunder, but he was just a beginner) So the other question is : what would you do over 3♠ with this hand?

Agree. I'm actually more interested in whether if 2NT is ok as a strong 2 suiter though. I think at this vulnerability + the fact that we're holding minors makes 2NT less desirable.

P.S. View page 2 for the other side of the problem [hv=d=e&v=n&s=sxhxxdakqxxcaqt9x]133|100|Scoring: IMP[/hv] The bidding went : N E S W -- 1♠ 2NT 3♠ 5♣ 5♠ ? The agreement was to play 2NT as 2 lowest unbid suits, either very weak or very strong. So now here we are, stucked with a guess. Seems to me that either Pass, X or 6♣ are possible calls. . . But since partner's 5♣ seems weak, looks like 6♣ would be rather unlikely to make. Thus I think the question is whether we should make some move to show that we in fact have the stronger version of 2NT overcall... But a X might fetch us a 5♠X = or 6♣X -3 B) (Switch the vul and perhaps it's easier to X since we don't mind partner bidding 6? Or did he have that option at all?) What do you think we should do? By the way do you agree with the initial 2NT overcall?

Something related : http://forums.bridgebase.com/index.php?sho...ic=17680&st=15# Though the difference is that my situation is vul against not B) But I suspect most would advocate only an overcall on a good 6 card suit unless you have alot of HCPs. (From the responses of my post) KQT86 just doesn't seem to cut it with only 12HCP

Hmm. . . are those percentages meant to be for ♥ length? as in. . . 0♥ = 9% 1♥ = 73% 2♥ = 11% 3♥ = 7%

Might be a good idea to ask for their cardings : Some play that the ♦J lead denies a higher honor. In that case you should rise with the Ace and play a ♠ down. . . then guessing what to do. I think I'll try the ♠A and if it fails, probably to try a ♥ finesse to get rid of a♦, later trying a double finesse in ♣. You risk going 2 down for this line, but you might want to note that if you do lose a ♦ chances are pretty dim. However, if the lead does not deny ♦K, probably best to finesse as if it works your chances are much better. If the ♦Q does take a trick, I'd play along elimination + throw in line as mentioned by pclayton.

This looks awfully similar. . . I had ♠Qxxxx ♥QTxxx ♦xx ♣x or something similar last time, and I tried a X. . . but partner had ♠Ax ♥xx ♦Kxx ♣AKQxxx And rebid 3NT :P

Alright I've given the original probability thing a small thought : Doesn't the fact that LHO will always play T if he has it (as given from the original post) makes it even chance whether, after playing the T, that he has either Tx or QT or QTx? At that point of time there's only 1 x outstanding so the probability of an original holding of Tx vs QT should be equal. It can't be 2 against 1 for Tx against QT since one of the Tx case is eliminated after RHO has played it. Similarly, if we compare Tx against QTx or QT against QTx for LHO, at first it seems that it's less likely for LHO to have QTx since that requires 2 of the remaining cards to be placed with him, but i think the fact that LHO still has 1 more card than RHO counters it. I dunno, but I still feel that the way I calculate the odds for the the remaining trump position is correct (no offense) since it takes into concern that some original holdings are already out. . . However, for your other point I can't really catch it for the moment (busy with work). Will try to look it up when I'm free. :P

Hmm. . . I think I got what you meant for the first part. But for the second part I'll think about it later when i'm free :huh:

They're based on before declarer decides whether to play K/J, where West has 11 unknown cards while East has 10 unknown cards. There are 2 remaining ♥s outside, so total possible combinations for their position = 21 x 20 which gives 420. For holding (1), East holds the remaining 2 cards and there are 10 x 9 possible holdings where the 2 ♥s will fall into those 10 "holes". This gives 90 current /420 total... Hmm. . . perhaps there's something wrong with this reasoning? :huh:

Here's my 2 cents : 1)We start by considering only the position where you have to make an immediate decision whether to finesse or drop 2)West has 11 cards while East has 10 cards 3)We then consider all the possible holdings left which satisfies the original assumptions (T from QTx, QT, Tx) Here are the possible holdings (Bold denote cards played) : West East T Qxxxx Tx Qxx QT xxx QTx xx 4)Next calculate the probability of these holdings (now showing only cards left) West East % -- Qx 90/420 = 21.43% ----(1) x Q 110/420 = 26.2% ----(2) Q x 110/420 = 26.2% ----(3) Qx -- 110/420 = 26.2% ----(4) *Note that they add up to 100%* 5)Now consider the 2 lines on the trump suit alone Line 1) --> Finesse the J followed by K, and doing your ♦ stuff if still possible Line 2) --> Play K for the drop, followed by ♦ discards if the Q doesn't appear Line 1) picks up (2) immediately, with extra chances for (1). However, you have no chances against (3) + (4) assuming they will cash their ♠A (which shd be obvious). Line 2) picks up (3) immediately, with extra chances for (2) + (4). However, you have no chances against (1) provided defender is alert to ruff your 5th ♦ with a small trump 6)Actual calculation of the odds : Line 1) --> 26.2% + chances that RHO has holding(1) + ♦xxxx (x%) Line 2) --> 26.2% + chances that RHO has holding (2) +♦xxxx(z%) + LHO holding (4) + ♦xxxx (y%) 7)Calculating x% and y% and z% x has a priori of holding (1) which is 21.43% So now LHO has 10 Unknown cards (1trump 1♣ 1discard) compared to RHO's 8 Unknown cards (he has 4trumps and at least 1♣) Actual calculation = 8.17% x 21.43% = 1.75% Compare with : y has a priori of holding (2) which is already 26.2% So now LHO has 9 unknown cards (3trumps 1♣) compared to RHO's 10 unknown cards(2trumps 1♣) Actual calculation = 10.84% X 26.2% = 2.84% similarly, for z: West has 10 unknown cards (2 trumps 1♣) while East has 9 unknown cards (3 trumps 1♣) Actual calculation = 10.84% x 26.2% = 2.84% Clearly, y + z has higher% (more cases!) Summary : Since the trump T play CAN be from QTx + Tx (which "normal" defender isn't capable of doing), odds goes towards playing for good breaks (3-2 in trumps, otherwise 3trumps with 4♦s). This is essentially due to the probability of QTx offside and Qxx onside being very close. Final : Line 1) = 26.2% + 1.75% = 27.95% Line 2) = 26.2% + 2.84% x 2 = 31.88% P.S. perhaps you're better off relying on your natural instincts (or what others call "table presence") LOLLLLLLL ;) I think the ability to be able to falsecard from QTx matters alot here, just as what others mentioned. If you think it's a hard play, the odds actually tilts towards Line 1) since the extra chances for Line 2) will be decreased, making the 2 lines even closer. Hope I haven't made any calculation errors, will check when I'm back home :P