Bidding 2-Loser Hands - Special Cases

This paragraph (and 2 other paragraphs) added on 6-20-2004: The Eddie Kantar column (I didn't know that - part 11) in the most recent issue of The Bridge Bulletin (June 2004) describes a specialized use of a 4NT opening. (and I didn't know that). A 4NT opening is an ask for specific aces. The responses are 5C=no Ace; 5D-5S=ace of that suit; 5NT=2 Aces; 6C=Ace of clubs. This usage is described in the Encyclopedia of Bridge as the ACOL 4NT asking bid. It makes sense, but apparently is more popular on the west coast than in the midwest, as no one that I play with has ever heard of the convention, or, at least does not play it. So, at least part of the following discussion should be revised.

True 2-loser hands with 2 stiffs typically are not a problem. Holding AKQJ1098 AKQJ x x or something similar, you should have few problems; you simply make a forcing opening bid, and then somehow or other get to an Ace-Asking bid (everything works here!), and then place the final contract. If an opening of 4N is Blackwood, then you can start with that. If your LHO bids 6D over your 2C opening, then ...

More problematic are 2-loser hands like the following:

Hand 1: You have{void} KQJ1098 KQJ10987 {void} and are the dealer. If you go slow, opponents may find a save, or even a 6 spade slam! If you bid 6 hearts, you might suffer a diamond ruff on opening lead. Opening 2 clubs (strong) might result in some kind of Notrump tragedy, if partner assumes that you have to have some hcp count such as 22+ (or some other arbitrary number in the range of, say, 19-21, for unbalanced hands with 9+ playing tricks). An opening of 5H presumably asks only about high heart honor(s), and 5D is preemptive. (I know, Precision with both support asking and control asking bids will work, but how many play support asking bids now-a-days? And what do you do if your LHO bids 4S over your 1C opening and partner passes?)

June 20, 2004. A 4NT (ACOL) opening agreement as described in the 1st paragraph can be of value for Hand 1, especially if one of your suits is spades. If you have any other 2-suiter, you likely are not worse off, unless the opponents can figure out that you are going set and can double you. I like it, but do not hold your breath waiting for this hand to come up. [Aside: 8 days after I wrote the above, I held KQJ107xx AKQ108 void x, but was playing with a partner with whom I had not discussed the 4NT specific Ace ask. I got too high, as partner had the Ace of Diamonds instead of a black Ace. Sigh.] You are more likely to hold Hand 2 below.

Hand 2: You have AKQxx {void} K10 KQ109xx and are the dealer. Playing standard methods, you open a club, your new, but experienced, partner responds one heart, you reverse into 2 spades, and now partner shows spade support (bid of 4S, almost surely 4 card support). The 4-card support indicates that both spades and hearts can be treated as no-loser suits, and that it is very reasonable to assume that the minors are one-loser suits. You have not discussed continuations after a reverse (you first met your new partner just a few hours earlier). Is 4S stronger than 3S? Does 4S deny an ace? With a brand-new partner and without prior discussion, what is more likely? (I know, we should have discussed this, but we didn't!)

If you have specific methods for showing specific aces, then the above hands may not be a problem for your partnership. But, most general bidding systems do not have methods for hands like Hand 1 above (of course, specific partnership agreements may handle the above hands). And I suspect that very few partnerships would have an agreement for a hand like Hand 1, and it is very unlikely that a pair that spent 20 minutes filling out a convention card would be sure how to handle Hand 2.

As a practical matter, I think it best to open 6D or 6H on Hand 1. The ODDS are 1.28:1 that partner has a red ace; giving you just over a 56% chance of slam. Simulations that I have run (about 200 hands) also indicate that you might even be saving against their 5-level contract in spades or keeping them out of a cold slam (perhaps as high as 10% of the time). Thus the probability of a successful contract is likely higher than 60%, and might even approach the mid 60's. Even though Hand 1 is very rare, it does not seem practical to try for the best possible double dummy result.

June 20, 2004. The ACOL 4NT ask gains (perhaps) if partner has no aces, but you might be doubled if partner has the wrong black ace (and you have the red suits). Just bidding 6 seems to still have merit, but you are welcome to disagree.

ODDS as defined here is the ratio of the number of partner's hands that have at least one of the 2 required cards to the number of partner's hands that have neither of the 2 required cards. If the ODDS ratio is greater than 1, then the probability of success is greater than 50%, if the ratio = 1, then the probability = 50%, and a ratio less than 1 results in less than a 50% chance of success.

Stated another way: ODDS ratio = [number of hands with exactly ONE crucial card + number of hands with BOTH crucial cards]/[number of hands with ZERO crucial cards]

After the cards have been dealt, but before the bidding so that you know nothing about partner's hand (partner could have been dealt any 13 of the remaining 39 cards)Then, the odds (NOT the same as percentage chance of success) of partner having at least one of the crucial cards, when you know NOTHING about partner's holding in the 2 suits, is computed as:

Formula 1: [C(2,1) x C(37, 12) + C(2,2) x C(37, 11)] / C(37, 13)

This becomes (2 x 13 / 25) + (13 x 12) / (26 x 25) = 1.28 after cancellation of common terms.

In the above formula, C(N, R) represents the number of ways that R items can be selected from a set of N items (the terminology used in the Bridge Encyclopedia is nCr). Thus, C(37, 13) represents the number of hands that partner might have that contain 13 cards out of the 37 cards that you are NOT interested in (in other words, partner's hand is worthless to you).

But the above formula would only apply if you were the dealer, and no one else had bid. If partner dealt and opened with 2 spades or 3 clubs, you would "know" that partner could not have both red aces (typically by agreement), and that the chances of even one red ace must be much less, because partner has a "known" minimum number of black (irrelevant) cards, and therefore fewer "slots" or places for a red ace.

Thus, for Hand 2, the odds are not as great, given the bidding. I knew about the odds for a hand like Hand 1, and knew that the odds for Hand 2 would be lower, but I estimated (at the table) that the odds were better than one to one. A later calculation showed that 1:1 odds are a little high. The odds that partner has at least one useful ace are at BEST about 0.96 to 1, given the above bidding for Hand 2.

So, how would you compute the odds when the bidding has revealed a certain number of cards in irrelevant suits in partner's hand? Following is an analysis that works for both Hands 1 and 2 as described above.

The calculations below assume the following.
1. You have 2 locked up suits (solid from the top or void)
2. You have 2 one-loser suits (missing one critical card, anything from a stiff to KQJ109876 or AQJ10987)
3. The possibility of partner having a void in one of the one-loser suits is ignored.

Thus, you are only interested in the likelihood ratio of partner's having the critical card in one or both of your one-loser suits (and you assume no adverse ruff on the opening lead)

Suppose that Suits A and B are the suits where you know or assume that you have no losers, and Suits C and D are the suits where you need a particular card (one in each suit) from partner to insure no losers. For the above 2 hands, the particular cards required are aces, but it is not required that the cards be aces.

Assume:
X = known or assumed total cards in Suits A and B in partner's hand, as revealed by the bidding, if any (partner's holding in these 2 suits is either known or not relevant)
Y = total cards in Suits A and B in your hand
Z = 26 - Y - X = known or assumed total cards in Suits A and B in opponents' hands
S = 13 - X = available places in partner's hand for one or both of the crucial cards in suits C and D
If you are the opening bidder, then X is zero.

A useful formula for computing the odds for the above assumptions is:

odds:1 = 2 ( 13 - X) / (25 - Y - X) + [(13 - X) (12 - X)] / [(26 - Y - X) (25 - Y - X)]

(The above formula is the one that results after cancellation of common terms.)

The above can be somewhat simplified using the definitions of Z and S.

Odds:1 = 2S / (Z - 1) + [(S) (S - 1)] / [(Z) (Z - 1)]

For Hand 1 above, X is zero (assuming that you are the dealer), and Y is zero (you have 2 voids); thus Z=26 and S=13.

The Odds are given by (2 x 13) / 25 + (12 x 13) / (26 x 25) = 1.28 exactly. You can also obtain this answer for odds by computing the probability that partner has neither critical card (two/thirds x 25/38), and then converting the probability of not having either card to an odds ratio.

For Hand 2, we must make some assumptions about how many hearts and spades are in partner's hand given the bidding. Most would agree that this sequence of bidding shows at least 4 hearts and most likely exactly 4 spades. Thus partner likely has 8 or 9 cards in the majors, and 4 or 5 cards in the minors.

Assuming partner has 8 cards in the majors, then opponents have 13 cards in the majors (26 - 8 - 5).
Then Z = 13 and S = 5, and the odds:1 are equal to: (2 x 5) / 12 + (5 x 4) / (13 x 12) = approx. 0.961:1, or about a 49% chance of success.

If partner has 9 cards in the majors, then Z = 12 and S = 4. Then the odds are:
(2 x 4) / 11 + (4 x 3) / (12 x 11) = 9/11 or 9:11 for (success) or 11:9 against (failure). The probability of success is 45%.

It is not inconceivable that partner has 4 spades and 6 hearts. Then the odds are even worse, at:
(2 x 3) / 10 + (3 x 2) / (11 x 10) = approx. .655:1 or a little less than a 40% chance of success.

The above calculations assume that partner's bidding is not influenced by the presence or absence of one of the 2 critical cards (to be sure, partner cannot know that these 2 cards are critical). And, the odds might be affected by partner's style (obviously cannot compute this). If the auction suggests that it is more likely that all of partners high cards are in spades and hearts, then the odds are worse. If partner would make a 4S bid with any scattered 8 count and 4 spades, then the odds as computed likely are reasonable. What you would love here is to have some type of control-asking mechanism; perhaps a grand is still possible.

I have never had Hand 1, but have had Hand 2. I bid 6S, down 1 when opponents cashed their 2 minor suit aces. Partner had J10xx AJxxx Qxx J. Certainly enough for game, and we had no agreements as to what continuations meant, so partner made certain we got to game. No problem there. If I had done the above analysis prior to this hand, I might have passed or bid 4N, for a push board (but we were behind in this match...) Yes, I could have used Blackwood, but would be ahead only if partner gave a response of NO aces or TWO aces. A response showing ONE ace leaves me with a guess. (Reflection after a couple of days has finally convinced me that I should have used Blackwood, and given up unless partner has 2 aces, but only because I can clearly sign-off in 5S. Or am I resulting?)

The above analysis is not complete, and has not been subject to extensive review. If you have any comments or corrections or have had a hand like Hand 1, please e-mail the web-master, so that this article can be updated, with credits given. Perhaps someone would like to analyze your chances if you have hand 1, and partner opens 2 spades? You would have to repeat the analysis for each plausible number of clubs that partner might have (say, zero through 4), and you would have to consider the possibility that partner has a GOOD 6 card suit (say, AKJxxx), so that partner then cannot have an outside Ace., AND you would have to remove the Ace and King of clubs from the "irrelevant" cards, and perhaps some more fine-tuning.