antonylee

I considered naming the thread "double before Walsh transfer" to go together with https://www.bridgebase.com/forums/topic/82604-after-walsh-transfer-gets-doubled/ :-) We currently play transfers over 1m-(X) (1♦=4+♥ (if m=♣), 1♥=4+♠, 1♠=transfer to NT, keeping a natural redouble) and 1m-(1red) (X=4+{red+1}, etc.). This is in the context of a natural-or-balanced club, unbalanced diamond. We play what I think are standard-ish followups to the transfer, namely transfer completion at the 1-level showing 3-card support. Given that we otherwise play T-walsh where transfer completion shows (usually) a weak notrump, I wonder if (for memory reasons) we could instead keep the out-of-competition scheme here? e.g. after 1♣ (1♦) X* (p) 1♥ = weak-notrump, etc. More specifically, when it is righty who enters into the auction (as in the other thread), I do realize that it is important for opener to immediately clarify the degree of fit, given that there are significant risks that lefty bumps the auction to an uncomfortable level (e.g. 1♣ (p) 1♦ (1♠); some-nondescript-bid (3♠) ???). However the situation I'm discussing here is quite different: righty has already passed, and it is relatively rarer for lefty to bid twice by himself, so we could just consider that the subsequent auction is likely uncontested and treat it as such. Thoughts?

Thanks for the input.

I held xxx AJx J9x A763 over declarer. When partner returned the ♣T covered by the J, I was worried that declarer had the 8 and somewhat half-heartedly tried to cut off declarer's communications by returning a diamond. Certainly not the best play, but I wonder if partner could have returned e.g. the 8 -- given that declarer wasn't able to conceal the 2 at trick 1, perhaps him playing the jack would suggest having neither the T nor the 9.

We play T-walsh where transfer completion shows a min bal. hand (without 4-card support) or an unbal. hand with clubs and 3-card support; 1NT shows 17-19 bal. Any suggestions on how to handle a double of the transfer bid (preferably independently of whether the double shows just the transfer suit, or is T/O of the suit implied)? Options seems to include: a) XX: 3crd support (support redouble); 1M: 11-13 min bal. with a stopper in the transfer suit and only 2crd support; pass: 11-13 min bal. without a stop; 1NT: 17-19 bal and too bad if no stop. b) 1M: as without the double, but promises a stop; pass: as 1M, but without a stop; 1NT: 17-19 bal. with a stop; XX: 17-19 bal. without a stop. c) 1M: min 3crd support; pass: min without 3crd support; 1NT: 17-19 bal. with a stop; XX: 17-19 bal. without a stop. and likely a few other permutations. Thoughts?

[hv=pc=n&s=sk654h984dkckt984&e=st8hkq532daq8642c&d=e&v=n&b=2&a=1hp1sp2dp2nppp]266|200[/hv] IMPs (but feel free to comment if scoring matters). Trick 1 goes ♣3(4th best) ♥2 ♣K ♣2. What do you return and why?

I don't know if mycroft is referring to https://github.com/anntzer/redeal/ (of which I am the author) but if python is your thing: from redeal import * strains = "NSHDC" table = [[0 for strain in range(5)] for tricks in range(14)] def do(deal): for strain_idx, strain in enumerate(strains): tricks = max(deal.dd_tricks(f"1{strain}N"), deal.dd_tricks(f"1{strain}S")) table[tricks][strain_idx] += 1 def final(_): print("\t".join("#" + strains)) for i, row in enumerate(table): print("\t".join(map(str, [i, *row]))) run with $ python -mredeal /path/to/script.py -N "AKQT6 - KT5 65432" -S "854 KJ A8432 QT7" -n 1000 --verbose you get Using default for initial. Using default for accept. # N S H D C 0 0 0 4 0 0 1 1 0 43 0 0 2 1 0 102 0 0 3 23 0 268 0 0 4 67 0 336 0 0 5 90 0 247 0 0 6 90 1 0 3 0 7 29 22 0 32 10 8 617 140 0 178 131 9 82 562 0 533 441 10 0 247 0 254 381 11 0 28 0 0 37 12 0 0 0 0 0 13 0 0 0 0 0 took 71s for 1000 boards; could probably cut that in half by just reusing the work to solve the board in both directions...

[hv=pc=n&s=s9hjt9432dq732ckq&n=sakjt4haq8d654ca4&d=e&v=n&b=2&a=p2hp4hppp]266|200[/hv] Teams. On the ♣J lead you can either go for the simple heart finesse (with a fallback on ♦AK onside available), or try ♠AKJ pitching diamonds -- possibly cashing the ♥A first. While it may be possible to list all winning cases for the second line, the enumeration gets quickly unwieldy -- especially if you have to do it at the table! Any suggestions on how to practically evaluate the winning chances of that approach?

Indeed, memory is particularly an issue here because these are rather rare auctions. I'm not sure things are that easily adaptable for us to 1N openings, though, given that we play a structure based on 1N-2r; 2M-2N* = forcing 5431s which we quite like as well (and is quite common in France).

Ah, swapping 3s and 3n over 3d is a nice and simple idea, thanks :)

We play 4red = transfers ("standard american"), 4♣ = 5♠5m (4♦ = what's your minor, 4♠ = fit) -- we "always" distinguish 5M5m from 5M4m by directly bidding the minor over the transfer with 54, using 2N-4♣ to show 5♠5m, and 2N-3♦; 3♥-3♠*; 3N(misfit)-4m to show 5♥5m (if opener has no ♥ fit).

4♠ is played from the wrong hand only with 5♠5♥ where there's no ♥ fit, and with slam-forcing 4♠5♥, so overall not so often. (53)Ms either way are handled by transfering to the long major; opener relay-breaks at 3(M+1) with 2M5oM and responder's next bid is a transfer. So e.g. 2N-3♥; 3N! now P=to play, 4♣=♦, 4♦=♥, 4♥=♠, 4♠=♣ (probably it would be best to play e.g. 4♣=♣ and 4♠+ = various ♦ hands, e.g. "♦, responding to a hypothetical 4♥ kickback" but these are exceptionally rare and not really worth the memory load).

3♣: 1 or both 4cM, or 5♠4♥ ... 3♦: <4♥, <5♠ ... ... 3♥: 4♠ ... ... 3♠: puppet to 3N -- to play, or slammish 5♠4♥, or others. ... ... 3N: 5♠4♥, NF ... 3♥: 4+♥ ... ... 3♠: "do you have 4 or 5♥?" (actually this has problems similar to the originally given auction, do you bid 4♥ with 5 or something else?) ... ... 3N: to play ... 3♠: 5♠ 3♦: transfer, possibly 4♠ as well ... 3♥: mostly forced ... ... 3♠: "do you have 2 or 3♥?" -- then over 3N (=2♥): 4m=55, 4♥=5♠5♥, 4♠=4♠5♥ but too strong for 3N. ... ... 3N: 5♥4♠, NF The general structure can be remembered as "After 3♦ by either player, invert the meanings of 3♠ and 3N" (a trick taught to me by my former partner).

Mostly because that's "standard" and we never discussed other superaccepts... We actually play 2N-3♦; 3♠ = 5233 exactly, which helps when responder is 3♠5♥ (transfer followups). Do you think the wrongsiding of games after a superaccept at 4(M-1) is worth the few (well, I don't have a good intuition of how many) additional slams you get to?

Over 2N-3♦; 3♥ (nearly forced), we currently play 3♠* = asking whether opener has 2♥ (3N) or 3+♥ (4♥); 3N = 4♠5♥ NF (if slammish, a 4♠5♥ responder starts with 3♠ instead), together with (some kind of) muppet Stayman. The downside is offering the chance of a lead-director double (or lack thereof) over 3♠; in exchange, we get to right-side 4♠ when that's the right contract, and can also distinguish between 5♥4m (directly bids the minor over 3♥) and 5♥5m (goes through 3♠* first) (if opener has no fit) -- so overall we quite like this. However, here in France, it seems extremely common, including among good players, to instead play fitted transfer acceptances, i.e. 2N-3♦; 3♥ = fit, 3N = no fit. You lose the possibility to signoff in 3♥ with weak hands, but instead get a full cuebidding level between 3♥ and 4♥ on slam hands. In our system, over 3♠*, opener could also choose to cue just to cater for the possibility that responder is slammish (so 4m = minor cue, the spade cue is not available but such is life), but that seems just like gratuitously leaking info to the opps most of the time. Always bidding 4♣ with a fit only leaves one intermediate non-signoff (4♦) to responder so that doesn't seem to buy much either. Any thoughts on the best way to use the half-level (4♣, 4♦, 4♥) that we have available with fitted hands after 2N-3♦; 3♥-3♠*?

After 1m-1M; 1N-2D! (GF checkback), there is usually plenty of room available, e.g. 1♣-1♠ 1N-2♦! 2♥-2♠ (4 hearts, maybe 3 spades (let's say you respond up the line); 5 spades, no heart fit)... now you have everything from 2N to 3N available (and even all the way to 4♠ when holding 3♠). Recently we had a hand that led us to add the agreement that in this auction, a jump to 3N over 2♠ shows a fit but with a NT-oriented hand, but this is pretty ad-hoc. Any thoughts on how to best exploit the available space? (Things applicable in an unbalanced diamond context for 1♣(2+)-1M; 1N-2♦ would be appreciated too.)