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I can help to translate :) Title: 4 Aces in a trick Have you ever met a deal which all 4 Aces appear in the same trick? Here N declaring a 3NT contract and have already taken 6 tricks, while opponents have taken 3 tricks, and thus come to this 4-cards ending, with N on lead: [hv=pc=n&s=shadck32&n=sahkjdqc]133|200[/hv] Opponents are baring ♦AK and ♣A only in the minor, and the remaining cards are ♥ (with the ♥Q) without any ♠ left. At this point declarer can only see two Major Aces as the obvious tricks (due to the blockage problem in ♥) and is pretty desperate. Declarer try his luck by leading his ♠A, and E unexpectedly discarding the ♦A! With this good news, now declarer carefully unblocking his ♥A as planned, and after a long and hard thought, W discarding his ♣A out of the blue!! Three Aces are discarded following the ♠A lead. Declarer now can read the situation clearly, exiting with the ♦Q, with W winning with the ♦K and returning ♥2 immediately. Declarer take the finesse and successfully take the 9th trick and made the contract. The actual end position is like this: [hv=pc=n&s=shadck32&n=sahkjdqc&e=sht43dac&w=shq2dkca]399|300[/hv]

When holding Axxx trump with 4-1 break facing a 5-3 fit, it is usually adopting a forcing game (as the opening lead suggest) and thus by ducking twice one can effectively pose a dual threat: - eliminate the trump policeman if declarer insist to draw three round of trumps, and continue the forcing game - ruff declarer's winners with the small trump if declarer stop to draw trump For the actual hand it is quite interesting if NS is defending 3♠ instead: South need to take the ♠A on the 3rd trick to gain a vital tempo and force declarer in ♥ but not ♦. Of course it will be quite double dummy as declarer is hiding a huge ♣ suit. For the declarer side, one should immediately turns to develop his side trick once finding that trump break 4-1. Drawing the 3rd round trumps will be fatal. For the second deal EW should aim for 7♦ if ♠ behaves. And one should guard the possible 4-1 breaks in ♠ when playing 6NT, by playing small to J first - i.e. crossing to dummy in the 2nd trick.

If the sequence 2NT - 3♦ - 3♥ - 3♠ shows 55+Ms, then opener rebid scheme afterward should be something like this: 3NT = To play, no interest in Ms 4♣/♦ = Max, fit ♥/♠ 4M = To play When in a two suiter sequence without prior agreement of trump, the first priority should be setting the trump. When there is slam interest, one will plan to bid the following below game level: 1. Forcing Raise (as 3x/4x) 2. Advanced Cue (above 3x) 3. Fit bid (above 3x) When there is not sufficient space below the game level, option one will not be available and take the next one. Sometimes, like in this case, even option 2 is not available, and therefore the 4m should be used to set trump only. For the actual hand, depends on your method whether you have something like 6KCBW which allow you to find out all the missing M honors and the m controls - given that 4♣ is really showing a maximum ♥ fit.

A more accurate translation should be: Some people said that Minor Suit Stayman is employed by old-fashioned, natural players, and listed many experts like, e.g. Zhao Jie (also known as Jack Zhao) etc. are employing four suit transfer (with improvement of course) to show that it is the modern trend. What are your thoughts on this?

There should be a bridge club in City U which you can join. Also there are regular events hold by HKCBA and BridgeHouse etc. Have fun in HK! P.S. I am not from City U.

The first lead does not matter here. In the actual play line actually quite accurate to let W get in at trick 4 and return ♠ at trick 5. All E need to deduce is that N hand has no more developed tricks to discard S hand loser (from the play that ♥ is discarded as losers and ♣ is well guarded). Therefore all E need to do is to draw all declarer's remaining trump and left dummy with 2 inevitable ♦ losers. Just as simple as that.

Try with the recently practice relay system CHAOS: 1♦ = (10)11+, 5+♥ except 4♥ UNBAL, including 5♥5+♠ ......1♥ = 8+, relay 1♠ = (10)11-14, MIN ......1NT = FG, relay 2♦ = 5 or 7 card ♥ w/ 1♦ OR 4♥ UNBAL ......2♥ = relay 3♦ = 4513 (start to probe 6♠) ......3♠ = RKCB in ♠ 3NT = 1 or 4 KC (good news) ......4♦ = Traditional Spiral Scan 4♠ = w/ ♠Q, w/o ♥K (6♠ very likely) ......4NT = Parity Spiral Scan for ♣K, ♦K, ♥Q 5♦ = ♣ K or ♦K + ♥Q (i.e. ♣ K) ......5NT = Traditional Spiral Scan (confirm ♣ Q not missing, not ♦Q instead) 6♦ = w/ ♣Q, w/o ♦Q ......7♠ = S/O

Unless you badly need a swing, or knowing that N holds ♠Q + ♥J + ♦A, you may adopt an anti-percentage line of finessing ♥J, small to ♦Q to execute the strip squeeze. Otherwise I do not see this is a special playing problem - just finessing ♠Q as pointed out.

Not sure if this counted as Winkle squeeze (wikipedia said it involve 3 suits).

The MP can be simply calculated in the following way: - Assume there are n tables in a tournament, in which each board has n NS scores to compare. - Sort the score in order, assign the MP 0, 2, 4, ..., 2(n-1) to the pairs from bottom to top. EW will score similarly. - For tied scores, score the arithmetic average of the MPs of the related rank. E.g. if 2 pairs are sharing the bottom score, then both of them score 1 MP in that board. - add up the MP for all boards played will be the MP obtained in the tournament. Divide the total MP by the maximum possible MP (which is 2(n-1) multiplied by the number of boards) will give the usual percentage we seen in a MP tournament.

As confirmed by the other post, this is a typical problem about dummy reversal + unblocking. One ♦ honor can be discarded by ♠A, and the other one need be discard when drawing trump, which means that the trump holding in N need to be longer than S by 1 card, i.e. you need to ruff ♠ twice to achieve that. Due to 4-0 break you cannot ruff the opening lead, and you cannot discard the ♠ as mentioned above, therefore you need to discard ♦ on ♣ A and proceed.

Do you mean this? [hv=pc=n&s=sht9876dakcat9876&w=skjt987h5432dckqj&n=saq2hakqjdqjt987c&e=s6543hd65432c5432]399|300[/hv]

翻譯上的小建議： 感覺上的意思是在問您喜歡「付費，沒廣告」還是「免費，但有廣告」的遊戲

Assuming the NT range is 15-17. 1) If RHO does not raise to 3♦, I believe the normal action will be pass. (Not sure if some really aggressive players will invite under this vulnerability). When RHO raise, partner's ♦ length and HCP is expected to be smaller and the probability that you can bring in the ♣ suit runs is higher. So you can certainly construct hands which 3NT makes (♠J10xx ♥AQJx ♦Ax ♣Axx) I think the hand might worth an invite if RHO raise to two level only which hope to stop in 3♣ when partner is minimum. Certainly we cannot do this now. The hand constructed will always require partner to have 5 quick tricks (including ♣A). I vote for a pass. But I am not surprised if some choose to bid 3NT at this color. Another concern is that partner may not have real ♦ stopper. 2) This is a classical 3♥ hand. If you choose to bid 3NT instead, you can give up this convention and adopt other for 3♥. 3) Another difficult decision. Depends if you have some gadgets/agreements here to differentiate a competitive 3♦ or 3♦ with constructive values. I think your hand is not much stronger than the typical reopening double over 1M to take the auction. But game is on, say partner have ♠ xx ♥ Axxx ♦ Axxxxx ♣ K Again I think the first issue is the partner expectation over this reopening double.

Moscito Variants: 1♣ = any (14)15+ ......1♦ = FG 1♠ = Two Suiters w/o ♠ or m 1-suiter. Usually 19- ......1NT = Shape Relay 2♥ = ♦ 1-suiter ......2♠ = Shape Relay 3NT = 3370 (+6 = (7330) w/ Low-Shortage) ......4♣ = QP ask. Responder knew slam possible, and 5♦ not bad over 3NT. 4NT = 12 QP (base = 9) ......5♣ = DCB w/ King parity. Responder knew 2 QP, i.e. 1K or 2Q is missing in the 3 suits. Opener next +1 = even K (Missing 1K = Small), otherwise zoom (Missing 2Q = Grand.) 6♥ = +7, odd K with 2♦2♠1♥H Honors. DCB sequence = ♦♦♠♥♠♥. ......7♦ = S/O

♦A, ♦ ruff, ♠A, ♦ ruff?

I believe that GIB use some sort of Monte Carlo technique - typically simulating some hand based on the current information and then do a double-dummy analysis. So inherently there are some randomness inside it. Even you give the identical situation to the same robot repeatedly, the decision may varies from time to time.

Bump B-) You may find me if you have time.

lycier 過譽了 我不是甚麼專家，也只是一位 Moscito 的初學者而已。 純粹為了測試體系和分享。

Moscito: 1♣......... 15+ any ......1♦... FG any 1♥......... 18+ any ......2♣... 6+♠ 1-suiter / (54)+ ♠+♣ 2♦......... Relay for shape ......3♦... 5224 3♥......... Relay for QP. A = 3, K = 2, Q = 1 ......4♣... 8QP (+1 = Base = 6QP, +3 = Base+2 = 8QP) W is missing ♠AKQ ♥AQ ♦Q a total of 11 QP. E shows 8 QP. That means at this point W know that they are missing 3 QP only, which could be 1A, 1K1Q or 3Qs. 4♦......... Relay, initiate DCB with K parity ......4NT... Even K, 1/2 Hs, 0 ♣ H. Now know that E is missing ♠K and 1Q. If that is ♠Q there is grand. 5♣......... Relay, DCB ......5♦... 1 ♠ H E confirm 1 ♠ Honor which has to be ♠A. So W know that E holds ♠Axxxx ♥AQ ♦Qx ♣xxxx But there is no room to investigate the presence of ♦J so can only signoff in 7♣ 7♣......... Signoff

dvd's attempt is a very natural one. One can easily see that the pattern will be 2 members forward -> 1 member backward -> 2 members forward -> 1 member backward -> 2 members forward and to minimize the backward time, we will assign Bill as the man holding the torch and accompany all his family members forward. This will result in dvd's answer. However, we shall note that there is clear wastage when Bill bring his parents forward in the above design. There will be less wastage when both parents go forward together. Namely, Two Brothers forward (3) -> Bill backward (2) -> Parents forward (8) -> Brother backward (3) -> Two Brothers forward (3) and the result will be only 19 mins. Note the order of backward can be exchanged in this design.

Vice squeeze?

我認為這副牌主要所體現的並不是描述 5-6 雙套時描述的問題。純粹個人意見： 1. 東家在第三家開叫 1♠ 比 1♦ 更具阻擊力，南家叫 1♥ 的權利被剝奪 2. 西家 Versace 選擇 Dbl 而不是加叫 4♠ 主要原因並不是因為找不到配合，而是因為有 ♥ 的強度與長度 3. 東家 Lauria 選擇開叫 1♦ 後已準備在 2/3/4♥ 後在同一線位叫 ♠； 我相信如果東家持的是 ♥ 而非 ♠ ，他的選擇也許不同 4. 事實上這類先叫低花 1♦ 再叫 3/4♠ 強烈暗示 5-6 雙套；假如只是 4-5 雙套強牌大可 Dbl 5. 我不清楚 Fantunes 體系，也許他們有辦法叫清 5-6 牌型。此外開叫 1♠ 所傳遞的訊息應與普通自然體系有所不同 6. 假如南北找到 ♦A - ♦將吃 - ♥A - ♦將吃 - ♠A 的防禦，我們會對這副牌有不同的看法嗎？

When the black suits are eliminated, W has no choice but to return ♦ for you. So the double ruffing finesse in ♦ is a 100% play - either onside, or W is end-played.

Very pleased to hear one of the author of Ambra! Hope to learn Ambra in depth :)