# In bridge, how would you bid the “worst” 14 point hand?

You are playing 15-17 point no trumps, five card majors, four card diamond bids, and three card club bids listed on many convention cards. And you have an unusual 14 high card point 4-4-3-2 hand that basically violates all the rules, like this:

(s) QJTx (h) QJxx (d) KQx (c) Kx.

Give me another T (or even a 9) in one of the major suits, and I might consider this a "weak" 15 for 1NT. Upgrade my spade Q to a K (and compensate by downgrading K of clubs to Q of clubs), and I would consider bidding a "strong" four card major. But as the hand stands, I wouldn't do either of these things.

What would most people do? Perhaps a three-card one diamond is the least of evils? Or would "pass" be a viable option, even though most people wouldn't pass 14 point hands? (Downgrade one of kings to a Q, and I'd pass the resulting 13- point hand.)

• Depending on what would surprise your partner least, bid 1D or 1C. If you've established that a 1D bid means 4 of them, then I go with 1C here--partner knows you would bid that with 4432 or 4423. Hopefully partner bids one of a major. The hand is difficult to bid, but far too good to consider passing. – sitnaltax Dec 29 '13 at 19:33
• @sitnaltax: One club! I had entirely overlooked this option. – Tom Au Dec 29 '13 at 20:33
• @TomAu: As others have said, this is not a matter of judgement, but of agreement. Just agree upon something with partner. If you are asking which agreement is better, there is no objective answer, IMO. – Aryabhata Dec 31 '13 at 20:11

I regard that as a comfortable, though minimum, 1NT opener playing 15-17 !NT, and open it accordingly.

If you wish to improve your Notrump hand evaluation, use a 54321 point system (for Ace through 10) and perform this transformation on the result: multiply by 2; add 1; and divide by 3 dropping any remainder. When applied to this hand provided we get an initial count of 22, which becomes successively 44, 45, and then 15 under the described transformation.

• Actually, I put in the T because I thought it would take the hand just "under," not to, 15 (a second T or well-placed 9 would have put it over). But what you do without the T? – Tom Au Dec 29 '13 at 23:33
• Without the ten, the count described above yields just 14 instead of 15, so I open 1D planning on a 1NT rebid. However your choice to have the ten in the spade suit (or not) really swings more like 1.5 points than 1.0. The QJTx holding in spades is really closer to 4 points than 3, raising the original hand as given slightly above a rock bottom 1NT opener. – Forget I was ever here Dec 29 '13 at 23:37
• OK, a "random" T (e.g. KT of clubs) doesn't do so much but T behind QJ in spades is worth quite a lot, right? I also consider QJTx worth about four points. – Tom Au Dec 29 '13 at 23:40
• @TomAu: Actually I consider QJT almost 0.5 point weaker than QJTx; the small card is very valuable both for its length, and for its ability to improve communications. Often this will reflect in the 321 distribution count, but may have to be applied manually in some cases as an adjustment. – Forget I was ever here Dec 29 '13 at 23:42

You bid it like you’d bid any other minimum balanced hand with no five-card major.

Open one of a minor. Rebid NT, raise partner’s major, or pass partner’s 1NT as appropriate.

Which minor you open is a matter of partnership agreement. The usual agreement in the US is that 1D can be on three with this shape, but there are some who prefer to have 1D always be four, and so open 1C here.

Passing is a poor plan, even if you made it slightly weaker, as is upgrading it if it had slightly better spots; it’s a perfectly respectable minimum balanced hand, and pretending it isn’t is just going to lead to reaching the wrong contract.

• @ji8e: (Mostly) good points. IMHO, the spots could be worth something. I play that T9 is worth a point behind a QJ (essentially you have three "jacks,") but not in say, KT9. Likewise, TWO QJT sequences are worth a little extra. Again, the Ts really upgrade the Js. – Tom Au Dec 29 '13 at 20:39

This is a curious question. You note that the context is five card majors, four card diamond bids, and three card club bids. When you agreed to play this system you must have realized that you would occasionally pick up a 4=4=3=2 hand. What was your plan? Never mind the ones with 14 HCP, how about 12 or 13 HCP?

I play 4-card diamond suits with one partner. We have agreed to open 1C when 4=4=3=2 with 12-14 or 18-19 HCP. What you should open is a matter for agreement between you and your partner.

• With 12-13 HCP I would probably "punt" by passing. The meaning of the question was, what do you do when you have a rare hand that falls into the "cracks" between your various conventions? My "convention card" would read, "15-17 point NTs, five card majors, four card diamonds, three card clubs." But I reserve the right to deviate with a hand like the above (for the question, I deliberately constructed one that is on the edge in every way.) But thanks for a good answer. – Tom Au Jan 2 '14 at 21:34