obscurans

I've been trying to cook up a strong diamond system of my own, based on Zar's idea about a strong opening, and a catch-all medium opening. I found out quickly moving the strong one above the medium makes it simpler: 1♣: any medium (say 14-16) non-NT hand or minimum (11-13) 63/54 minors, 31 majors 1♦: any strong (17+) or weak NT, F1, all others nf 1♥: min 4-5 no 6c suit 1♠: min 4-5 deny 4♥, no 6c suit 1N: "14-16" no 4cM, 5422, 6322 allowed 2♣: min 6♥ or 7+♥, void 2♦: min 6♠ or 7+♠, void 2♥: min 6♣ 4c suit or 7+♣, void 2♠: min 6♦ 4c suit or 7+♦, void 2N: min 55 minors 3X: min 7+, no void There's some symmetry when responder relays 1♦ (waiting) over 1♣ and opener simply bids normally, having gotten extra strength off his chest. As much as I'd like to have nat 2-level openings (read: GCC), I need the extra space to get a weak misfit/inv+ relay for second suit. I've stuck the flat minor 1/2-suiters into 1♣ (rebid NT 'offshape' nf) so we should be able to scramble some fit from the 2M openings. If nat (and probably flat) responder is under much more pressure to shut up and leave in the M misfit. "Minimum" can be stretched out with shape, so the 2+lv openings is basically constructive preempt/intermediate. System emphasis is heavily on bidding exact shape. This is completely untested and unfinished, I have no idea how and if it works.

Actually 700k is a pretty good sample for global statistics on hands. I can even pin down the occurrence of insanely rare hands (6610 or worse, 16+HCP) with 100M or so hands generated. It's the same thing with pollsters grabbing about 1k people and generalizing to however many million there are in your country. The assumption that your random deals are similarly dealt to those at the table is sufficient to get 1/n variance behavior.

Yes, ninjaed, I missed that. I plead 3am your honor. Every one in the given library, all 717102 of them.

OK, I've regained partial sanity. MP measures only differences, so in fact you are asking: IF we have EXACTLY X HCP, THEN we have a 50+% chance of making 3NT: find the minimum X that still fits (back to 25, yay). Since we suppose the 'normal' people will bid 25HCP games anyway, your anti-field behavior is limited to 24HCP and 23HCP bids... which are less than 50% shots. Solved. Now, at IMPs vul, doing the same thing and reading off the columns, 24HCP gives a 37.7% chance of making 3NT, so it's possible that you can bid 24pt 3NT red. This number is so close to the mark (you stand to gain 10 for 3N= vs 2N+1, lose 6 for 3N-1 vs 2N=, so you need a chance of 6/(6+10) or 37.5%) that it's well within experimental error. I'm running a couple million more hands first. Back to the sensitivity problem, which measures how many games you can safely bid (knowing only combined HCP), at 25 it's 70.9%, so you will miss 30% of games. This is the true yardstick of how good your hand evaluation is - I'll be optimizing that and be back with results (much) later.

I have a heretic assertion to make (and I disbelieve it as strongly as everyone else): 23HCP makes for a 3NT bid. For simplicity, let's assume matchpoints and ignore extra undertricks for extreme overbidding, so what you want is: IF we have X total HCP between us, THEN we have a 50% chance of making 3NT. Find X. I'm using the double dummy library provided by Prof. Matt Ginsberg (of GIB fame) found at http://www.cirl.uoregon.edu/ginsberg/gibresearch.html. I took every deal, computed N/S and E/W total HCP, then recorded their double-dummy tricks possible for all 4 declarers. So an entry is in the form (total HCP of pair, tricks possible at NT). Whether the deal should be played in NT, whether they have slam, or if they should be even declaring (read: we have 3HCP together) is ignored. If the tricks differ depending on right-siding, the result for the stronger hand declaring is weighted 2/3, the weaker one 1/3 (if equal, 1/2 each). All entries are collected and summed up for statistics. Fundamentally, this is a binary classification problem. You have a hidden variable (whether we have 9+ tricks at NT) and an observed variable (total HCP), and you want to predict the hidden using the observed. Unless you have a better idea, we predict all hands with X+ total HCP as making 3NT and any less as not making 3NT. This gives rise to a confusion matrix: True negative (number of hands where we have LESS than X HCP and LESS than 9 tricks at NT) False negative (number of hands where we have LESS than X HCP and AT LEAST 9 tricks at NT) False positive (number of hands where we have AT LEAST X HCP and LESS than 9 tricks at NT) True positive (number of hands where we have AT LEAST X HCP and AT LEAST 9 tricks at NT) Then we can calculate a boatload of different statistics from the matrix. The most important one, corresponding to the question asked above, is the positive predictive value, TP/(FP+TP) - verify it matches the description. We want the minimum X such that PPV is at least 50%. And here's where the wheels come off - the magic number is not 25, it's 23. If you restrict it to hands where double-dummy shows 7-11 NT tricks... it drops to 22. After a couple hours of not believing and hence fruitlessly debugging the program, I gave up. Simply put, when people state that 25HCP is required for NT game, they seem to be answering the question: IF we have 3NT, THEN we have X+ HCP 50% of the time, where X is indeed 25. But that is backwards, and in fact is calibrating the sensitivity, TP/(FN+TP), which measures how many games have 25+HCP. Now, if you haven't written me off as a complete lunatic yet, is that why bridge is getting more and more aggressive? The people who bid closer to 23HCP games do better in the long run? Is the benefit of double dummy (remember, defenders also get to play perfectly) really a full trick on average? PS: Data. HCP #no 3NT #hv 3NT 0 0 0 1 42 0 2 204 0 3 498 0 4 1416 0 5 3726 0 6 8292 0 7 16254 0 8 30240 0 9 51648 0 10 84996 0 11 131166 0 12 185604 0 13 258696 0 14 341816 10 15 430580 46 16 517026 126 17 599910 576 18 661065 2007 19 700184 6088 20 709570 16598 21 668130 38142 22 584836 78236 23 464634 135852 24 322037 195115 25 196346 234280 26 105431 236395 27 50292 208404 28 22352 163252 29 9610 121556 30 3762 81234 31 1310 50338 32 409 29831 33 74 16180 34 2 8290 35 0 3726 36 0 1416 37 0 498 38 0 204 39 0 42 40 0 0

If you're going to X slam for penalties, there's something to be said about 5♦ X. 80%+ to West.

Not sure what the defence should be doing: leading into my suit is probably bad, and trumps might work; once I see the 5-0 leading through W lets me cover their card exactly, so taking the Q might not be so bad (I would draw 3 and ruff once in dummy). Once I'm running the spades, underruffing is the best chance at locking me in the dummy. An X would make me run to spades unsupported, I hold 6 here and even if there is a stack, it's now onside.

At a sectional today:[hv=d=n&v=n&n=sk3hjt82da973cq92&w=s7hq9765dqt65cajt&e=sq954hdk82ck87543&s=sajt862hak43dj4c6]399|300|Scoring: MP Pass - Pass - 1♠ - Pass 1NT - Pass - 2♥ - Pass 3♥ - Pass - 4♥ - AP[/hv]Sitting South, got a "friendly" ♠ lead; when the 50 struck I knew I was down. Took the K, led another heart to dummy (W took the Q), got a spade ruffed. Next, rammed the rest of the spades through, overruffing twice, ruffing back to hand. Escaped for -1, low (not bottom) score. Just wondering if I had my 4♥ bid, and was there any better way at this hand. Double dummy says 2♥ or 3♠ limit.

I'm no serious bidding theorist, but I'd say without a known fit and over interference 5NT should be "pick a slam". There's still 5♠ free for a trump honor ask, although I think its usual meaning is bid slam with 2 and grand with 3. Maybe include the 5-level cuebid as 1-or-2 ask?

[hv=d=w&v=n&n=s-ha876542dt976c72&w=st653hkjt3da5caq8&e=sakqj98742h-dqcj96&s=s-hq9dkj8432ckt543]399|300|Scoring: IMPs Brief history: today I was on BBO, and three of us were just about to leave, when I picked up the East hand. And it was too late to hold partner back for just one more.[/hv] Yes Virginia, it's a 13-fit. Obviously 6 is cold, but how would you get there? I would have definitely made a "strong" jump-shift regardless; probably would have heard a raise, 4 cards substituting an honor, then serious 3NT. The end I'm not sure... 1♣-2♠- 3♠-3NT- 4♣-4♦- 4♥-4NT- 5♥-5NT- 6♥-6♠?

[hv=d=s&v=b&n=sak62h6dak532c854&w=sqj4ht4d9864c9763&e=s8hj98752dqjtckj2&s=st9753hakq3d7caqt]399|300|Scoring: IMP 1♠ -pass-2♦ -pass- 2♥*-pass-3♣*-pass- 3N -pass-4♠ -pass- 4N*-pass-5♣ -pass- 5♦ -pass-6♠ -all pass *: downgraded for the ♦ misfit *: 4sf *: if p hid the ♠ fit, then 4sf shows strength; 0314[/hv] First up, kudos to partner for figuring out what I needed there. Nothing to the play, I even got a friendly ♣ lead. Just wondering how it would go under 2/1.

Spade-club double hook? That's pretty close to 70%. If the spade loses, you draw the rest of trump, run the diamonds and hope the clubs behave. If the first hook wins you cross to the diamond A and finesse the club Q. If that wins you're home, else pull the club A and ruff to hand for a second hook. You'll need 4-3 clubs (60%). If you try to finesse trumps once, there's nothing else to do if it loses but try the clubs; if you pull the ace after it wins, you run the diamonds and hope the last trump's on East who hands you the club suit on a platter. 4-1 W you're never going to endplay anyone, 4-1 E onside means the king dropped first round so you simply hook the club queen for the overtrick. So that's half of 3-2, half of 1/5 of 4-1 for close to 40% (nobody survives a 5-0).

I was considering the strong jump-shift available (mostly SAYC), and hence a major rebid could be a "strong" 5-suit (that AKQxx would probably qualify). Without 2/1 I don't want to make any nonforcing bid before the RKC sequence. Also RKC would do its main job of discouraging pointless slams if ever I missed any of the top three spades. 1♠-3♣-4♠-up could probably happen, although an unhelpful 3♠ is likely.

[hv=d=w&v=n&n=sakqjxxhtxdxxxcjx&s=stxhakqdakqjcaq8x]133|200|Scoring: IMP North-South 1♠-4NT 5♠-5NT 6♠-7NT All pass[/hv] I had the luck to pick up the monster South hand and was wondering how to interrogate partner's spades... when he opened them himself second. The only question is 6N or 7N, so RKC, and he did have he ♠AKQ tight. Either the ♠J or ♣K will freeze 7NT, but I could only ask (and fail to get) the latter. So, in the tank, I was constructing some possible hands. The worst possible (assuming rule of 20) he could put down was AKQxx, the ♣J and a 5-suiter somewhere: ♥ gives me a 3-3 break, a 3-2 and the finesse, minors just a 3-3 or 50% (squeeze on ♦?). If instead of the J I get a sixth spade, that's a 3-2 or 50%. And the decision was to ignore the ♣K completely. Close enough to the business 2/3s or no? And if I chickened to 6N, could partner point to his own ♠J and call that a 13th trick?