Romanized Blackwood follow-ups

Well, I seem to have been laboring under a misunderstanding for the last ten years or so. The error in my ways was finally brought home to me at a club game with my partner Len a few weeks ago. It was actually the second time in maybe a year that this had happened (different partners) and I began to sense a pattern.

The issue arises because the Romans have an unusual counting system – and I don't mean all of those Xs, Ms, Ls, Is, etc. No, they count like this, using what is technically a modulo-three system with an inversion of the second and third digits. 0 (or 3 or 6, etc.), 2 (5, 8...), 1 (4, 7...). No wonder they have so many car accidents – they have no idea how fast they're going.

I'm kidding of course. This counting scheme, which so far as I know is only used in bridge and even then only when checking controls for a slam, is actually very well-suited to the problem at hand, especially when counting key cards [as in Roman key-card Blackwood], as opposed to just aces, of which there may be any number between zero and five. The two (and five) key card responses are imbued with an extra piece of information: possession of the trump queen (after the other responses, asker must inquire if he needs to know).

After a (Roman) 4NT Blackwood enquiry elicits one of the ambiguous responses: 5♣ or 5♦, there is a potential problem. Whose responsibility is it to continue bidding when responder actually has the higher number of "aces"? The books are generally quiet on this point, although some of them imply that it is asker's responsibility because they make comments like "Asker should know from the auction what the response shows." Only Fred Gitelman (as far as I have seen) casts any doubt on the efficacy of this idea by observing that there have been some spectacular grand slams doubled down 3 after this occurrence.

For my part, I learned somehow – I know not whence or from whom – that asker should assume the lower number and sign off. And therefore that responder would continue bidding with the higher number. This wouldn't work however when clubs or diamonds is the agreed suit and responder bids our suit at the five-level. Now, opener must step up and make the decision because a pass will end the auction unilaterally.

But when a major suit is agreed, the notion that responder will continue with the higher number is still plausible (though apparently not guaranteed). As an example, here is our auction from the most recent debacle: 1♦ – 1♥ – 2♥* – 4♠* – 5♣* – 5♥ – all pass. 2♥ showed 15-17 "dummy points" since we play 12-14 no-trump; 4♠ was kickback for hearts; 5♣ showed zero or three key cards. Is it conceivable that I would be asking for key cards with none of my own? What kind of key-card-less hand could I possibly have that wants to be in slam opposite a likely balanced hand 15-17? On the other hand, could partner possibly have a hand with no key cards? Not in this universe.

I was adamant that it was responder's responsibility to get us to slam with the higher number and that it was not required for asker to have to try to work out responder's possible hands. But unfortunately I have found no support for my position in the literature.

In this particular case, it was obvious to each of us that we had all or most of the key cards. My hand: ♠QT ♥AQ84 ♦KT6 ♣AK54 (two key cards + Q). Partner's hand: ♠AK53 ♥KT75 ♦AJ5 ♣J2 (three key cards). Assuming that partner did indeed have three, I would have played in 6NT rather than hearts. In fact, it makes 7NT with careful play. But it makes only 6♥ because the trumps split 5-0. In practice, I didn't play the hand well and only just made my contract for a poor match-point score (although not a zero). I would have likely have made 12 tricks in no-trump.

So, let's take an objective look at the situation. There's no possible hand for responder that has zero key cards given this auction. So let's look to see if there's any bad hand that asker might have had consistent with the bidding. For 5♥ to be the proper contract, asker would have to have no key cards at all and yet be invoking the key-card ask. Asker might have this hand: ♠QJT2 ♥QJ84 ♦KQ6 ♣KQ. Is it conceivable that this hand would be trying for slam opposite a hand with15-17 dummy points? It is not. But what if asker had a more distributional hand such as ♠Q2 ♥QJ98432 ♦– ♣KQ54. Note that splintering or asking with "exclusion key-card Blackwood" would not be appropriate here since opener began with 1♦. Could this hand be looking for slam by jumping straight to 4NT? Yes, it's just possible but extremely unlikely.

I think there's another, more useful, principle that can be applied here, and which I wrote about a while back in Aceless Wonders. This principle can be generalized to the following rule: do not indicate a strong hand without an ace. Or, to put it another way, all aceless hands are (relatively speaking) weak. So, strength-showing bids (incidentally, these are almost the same set of actions which Eddie Kantar considers to be triggers of the 1430 asks versus 3014 asks in his system of Roman Key-card Blackwood) all promise at least one ace:

• opening with 2NT, (strong) 2♣ (or 1♣ if playing a big club system);
• (opener) reverses;
• strength-showing jump (in the case of responder, this would have to be a strong jump shift);
• (opener) makes a game-forcing cuebid in competition;
• asking for key-cards (if none of the previous triggers have already occurred).

Using this principle, I don't think it is possible to get into a bad slam (the type where you are missing three key cards). In other words, key-card asker can always bid on – assuming the higher number of key cards – unless key-card responder has never implied an ace by making a strength-showing bid. In such a case, it now becomes responder's responsibility to continue bidding when holding the higher number of key cards. There is however still an outside chance that each partner has just one key card and that the partnership bids up to the slam level – but this would require both partners to have overbid their hand in the same auction.

The more significant problem with this rule is that both partners have to be on the same page (this page, in fact). I suppose then that a practical solution is the following: each partner has the responsibility of bidding on unless it is probable that the partnership is missing three key cards.