TimG

I once had opponents give each other a high five across the table. I was sort of amused, but my partner was rather upset.

I don't think he's quite good enough, partner would have still had an opening bid without the ♠Q and ♥A, for instance.

Wouldn't 1♣-(1♥)-P-(1♠); P-(2♥)-P-(P); DBL be for takeout? It may be reasonable to worry about partner being stuck and not being able to handle the auction if you don't bid 1NT over 1♠, but a 2♥ rebid by intervenor doesn't seem to be one of the problems. While I'm making sure my understanding of the standard treatment of doubles is OK, this double: 1♣-(1♥)-P-(1♠); P-(1N)-P-(2♠); DBL would be penalty, wouldn't it?

You need a new partner. Rude opponents are one thing, a rude partner is another entirely.

4♥. If partner cues 4♠, 4N would now be RKC for clubs, right? If partner can't cue 4♠, I think I'd just shoot out 6♣.

I don't mean to be snide, really, but isn't that two tries that are available?

Yes, this sort of study has been done; the information is useful to the defenders. It probably does not fully make up for the advantage of playing in the 44 fit. But, I think it is closer than most people think.

I think lots of players make fewer honest signals when they think their opponents are paying attention. As long as it does not become an agreement that they switch to upside-down once asked, I think it is OK. In the case that you speak of, it is unlikely that the intermediate player is paying any attention except in the most obvious of situations (like suit preference when giving a ruff) and I expect your better opponent would have given a true signal in a situation such as that.

Gee, these look very familiar...

South to start things off, no enemy bidding: Axxxx xxx Kx AQx x AKQJx AQxx Jxx

I think not using Stayman will often be a winning decision, but it carries some downside, too. One hand is insufficient to draw any firm conclusions.

Isn't the modern tendency to double when it comes back around when one has made one of these stronger than usual preempts? I think too many people think that responder has the last chance. Maybe he should after a preempt, but that doesn't seem to be the modern approach.

I'm not sure this is true. Suppose that we stick with your assumption that the raindrops are falling vertically, but change your assumption about the body being a rectangular prism. I think I can describe it best by considering a two dimensional coordinate system. Suppose that our "person" is represented by a line from (-1,0) to (0,2), that the rain is falling at 2 units per second and that the "person" moves to the right at one unit per second. As the person moves from (-1,0)(0,2) to (0,0)(1,2) and raindrops that started to the left of (or above) the "person" will not be able to catch up with the "person" as it moves out of the way, and the "person" won't be able to catch up with any of the raindrops that started to the right of (or below) the "person". Converting this to 3-D, the only raindrops that hit this person will be those that hit the upper side of the prism; there will be no raindrops that hit the front (or back) side. But, change the rate at which the person moves or the angle of the lean will mean that there will be some raindrops hitting the front (or back) side. Isn't this the same as leaving the person vertical and adding a horizontal component to the falling raindrops? So that if the horizontal component of the rate of fall of the raindrops were identical to the rate of movement of our vertical person, there would be no front (or back) side raindrop hits? I have no idea at what rate raindrops fall, how fast I can run, or at what "lean" I would run at. But, I do know that when I am trying to stay relatively dry, I not only go quicker than a walk, but I also duck or lean my upper body forward. The lean/speed combination ought to have an affect on the real world experiments.

I was mistakenly trying to solve the problem for 4^4444. Probably still wrong...

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 was confused by the wording, too, but interpreted it the same way as you, so I guess it wasn't as unclear as we thought.

I'm going with ......

Average high school student from what country?

Standard over here is that a direct X of 1♠ is penalties. OK, so how do 'normal' people show a minimum 1354 hand in this auction, if pass-then-pull shows a stronger hand? If you can't sit for the double (and with a singleton spade it seems reasonable not to want to defend 1S), then you have to bid after 1D-DBL-RDBL-P.

I have this agreement with at least one partner. Pass = weak is easy to remember. I used to think it also could help in tempo sensitive auctions. But read a pretty convincing argument that this is not the case (maybe from jlall).

But, he reportedly did make this same comment approximately one year ago regarding Hillary's health care plan

I think 2NT should show a good hand with diamonds well stopped (had enough points to bid over 1♦ but was stuck for a call). Add the, I dunno, QJ of diamonds to the 2NT bidder and it makes more sense. How can you not pass with that hand?

Pass, and we don't even have a tough lead (since partner will be on lead).

I would have thought that if they were good enough to be trumps at the six-level opposite a void, responder would have bid 4C at his second turn. Surely there are hands where responder would want to bid his clubs three times on KJ-8th. I agree with Helene, opener should bid 3D at his second turn. But, I expect some would consider that to show a solid suit.