EricK

4NT is always to play after minorwood if playing MP pairs.

I see no reason why we should be at all lenient to somebody who makes a claim without stating a line of play. Even if dummy's good, it really saves no time to say "dummy's good" rather than "3 spades, 2 hearts then a club" (or "trumps from the top then everything else is high" or any other simple statement of what tricks you are claiming)

What happens if the defenders' hands are swapped so that if the hand were to be played out, she would see that her trump wasn't good before making the discard from dummy?

4♥ for me. It's likely enough where I want to play that I might as well bid it straight away and make the opps guess.

It seems highly unlikely to me that the scoring table we now have is in any way "the best possible" (almost no matter how that is defined). Unfortunately we are probably stuck with it in the same way we are with the QWERTY keyboard.

ok, fair enough. No more such threads from me in the future. FWIW, I wasn't trying to justify 3NT, but to subject 1NT to a little more scrutiny than it had been getting. If South chooses to pass (which would be many people's first choice, or second choice after 1NT) then the bidding could easily go 1NT 2♣ 2♦ 3NT. If South opens 1♦ which is the "standard" opening on this hand, it might easily go 1♦ 3NT. Avoiding 3NT on these isn't easy at MPs playing a standardish system (whether or not you play a weak NT!)

[hv=d=s&v=n&n=sxxxhaqxxdatckjxx&s=sahkjxxdqxxxcqtxx]133|200|Scoring: MP 1NT* 3NT *12-14[/hv] Very new partnership, not much discussion about methods. South, who judged that things hadn't been going well so far, decided to open a weak NT on a 1-4-4-4 hand with singleton A. North, who judged that things hadn't been going well so far, decided to jump straight to 3NT rather than look for a ♥ fit. A spade was led and only 8 tricks materialised. Opinions please!

I would upgrade this hand if playing 15-17.

1. I don't know what I "should" bid. What I will bid is 2♦ and then support ♠ on the next round. This hand looks to be worth about 20HCP to me. 2. If I rebid 2♣, 3♣ would show about 10-11 points and ♣ support, it would not be forcing. In response I suppose I would rebid 3♦. 4♣ is an odd bid. In response ! would cue 4♦ and see what happens. 3. Over 2♠, 3♣ shows, I suppose, 10-11 and probably 5♣ and maybe only 4♠. In response I would, again, cuebid 3♦. 4♣ is another odd bid. It sounds like a cue bid with ♠ agreed. I might venture 4♦ 4. I don't like making a non-forcing bid on this hand, and so I bid 2♦. When partner bids 1♠ there is no way I am stopping short of game here. He either has 5♠ or a fit for one of my minor suits.

So, have I got this right, doubler might be about 12 or 13hcp with 4 spades and advancer might be about 5 or 6 with 4 spades and you don't propose that either should compete with 2♠? This may be workable at IMP, but at MP you're just handing the opps a likely top in my mind. Everybody and their mother where I play, including most of the LOLs, are going to be bidding 2♠ and they are either making it or only one off, while 2D, plus one like as not, will be making the other way. The only bad thing that can happen is if you really are one off and the opps find the MP double - but in practice this is more of a concern in expert circles than at every day bridge tables. Nick You haven't got this right. I am suggesting that doubler advances with 4 spades. I am not sure anybody is suggesting that responder to the doubler should rebid his spades on just a 4 card suit and 5 points - although maybe some are.

That can't be true, can it? Doesn't it imply that the sum of the digits of any multiple of 9 is 0? Even if I am mistaken and 9 mod 9 = 9, what of 19? 1 + 9 = 10; 19 mod 9 = 1. Maybe the sum of the digits of the sum of the digits eventually equals the number mod 9 when carried out enough times so that the sum is less than 10. i.e the sum of the digits for 19 is 10, the sum of those digits is 1 which equals 19 mod 9. But, the problem stipulated only two steps. I meant that a number and the sum of its digits are congruent to each other modulo 9 (i.e. leave the same remainder upon dividing by 9) or in more general terms: Let N be a positive integer and let D be the sum of the digits of the representation of N in base b. Then N≡D(mod b-1) This means, when applied to the problem in the OP, that 4444^4444, the sum of the digits of 4444^4444, the sum of the digits of that number, the sum of the digits of that number and so on, all leave the same remainder upon dividing by 9. Which, as can be easily shown, is 7. The second half of the proof showed that 7 itself was the only option for the sum of the digits of M because the next option, 16 was already too high.

Yes, you are right. It should be 45.

I didn't realise people played the minimum response as so wide ranging. Certianly 8 points and 5 card spades is too strong for 1♠ "in my bidding book". But isn't the issue here that in direct seat, we can afford to pass with minimum-ish hands and only 3♠ because partner will bid again with 5♠ and a reasonable hand in context; whereas if we have 4♠ can we afford to pass since partner is unlikely to bid again if he only has 4♠ as well?

I agree with the answers given already. Hidden working:

To what extent is partner "still in the race"? Can he not justly argue that he has shown his suit (as you requested), and also shown his limited strength (by not jumping). Is he not justified in passing if 2♦ is passed round to him almost no matter what he holds?

I'm wary of saying that anything with regards to bidding is mandatory, but this comes close. Partner's 3NT could be based on a number of hand types some of which have support for your suit, some of which don't (eg solid minor and two outside aces). If you don't want to trust partner in this situation then you shouldn't have opened 3M in the first place.

I think the solution is:

I assume we must use all four numbers:) In that case:

If you want a hard problem, use the same numbers (100,25,10,8,7,5) to make 923

You don't have to use all the numbers though, do you? In that sense (100-25-8)*(7+5) is the simplest solution. How easily gettable that is within the 30 seconds allowed is debatable. Probably the easiest solution to find quickly is 100*8+7-(25+5)/10 or 100*8+7-(25-10)/5, but under time pressure it's easy to panic and think you "obviously" can't make 3 with 25,10 and 5.

For a solvable numbers game, that is pretty hard.

In fact, isn't 2NT irrational as to play? So shouldn't it be artificial? Maybe a 2-suited GF. Can there be a 2-suited GF hand which passes over 1♦?

It's not wasting bidding space if it shows a hand, or some important feature of the hand, which couldn't be accurately shown at all by following the cheaper route, or would actually end up being shown at a higher level if the cheaper route were followed. One possible use for this sequence, for instance, would be to promise a solid heart suit.

Why play "artificial" Drury at all? Why not play 2-level responses (including 1♠-2♥) as fit non-jumps? Wouldn't that give opener useful information along with the knowledge that you have a good hand with a trump fit? I suspect this is more useful than immediately knowing the degree of trump fit (as with 2 way Drury) or just knowing of the trump fit (as with 1 way Drury).

I cared about masterpoints when I was young and improving. Now I know how good I am (better than the average club player, not in the same league as any "good" player) I don't really see the point. Looking back, I don't really see the point of caring when I did, but I can't change that!