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What do you bid?


rduran1216

  

25 members have voted

  1. 1. pass or 2c?

    • pass
      15
    • 2c
      10


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16-18 is not a common range. If partner has the maximum we should be in game, especially if he shows up with 4 spades. If he shows 4 hearts I can invite (2NT), if he shows 4 spades I'd be tempted to drive to game right away and if he bids 2 I'd be tempted to pass. 2 for me.
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This type of problem is why my wife and I play that...

 

1NT-2

2-2 and

 

1NT-2

2-2

 

...are sequences that show unbalanced hands with about this range and 4-5 spades. It seems to work well.

 

With the balanced hands with four spades, after the 2 rebid by Opener, bidding 2NT only costs when Opener has both majors and a minimum, and only if a spade contract is actually superior to the notrump contract, and only with those hands where partner with a minimum does not apply Crane's idea.

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I don't think it is so obvious to bid over 1NT.

 

You have a maximum of 25 HCP (with your quota of 10s and a few extra 9s) and, if you do find a 4-4 fit, it might be a very weak 4-4 fit.

 

Maybe vul at IMPs it might make more sense to move over 1NT, but at matchpoints 1NT is probably as good a contract as any.

 

I would pass 1NT.

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You have a maximum of 25 HCP

 

 

They're even teaching beginners these days that game is good on 25 points: I watched a youtube clip of Andrew Robson's "Play Modern Bridge," a series introducing rank beginners to the game it seems, and he was teaching that 25 points meant game (in NT or a major). Link to his channel (see "Clip1").

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They're even teaching beginners these days that game is good on 25 points: I watched a youtube clip of Andrew Robson's "Play Modern Bridge," a series introducing rank beginners to the game it seems, and he was teaching that 25 points meant game (in NT or a major). Link to his channel (see "Clip1").

Yes, but 12 opposite 13 is better than 7 opposite 18. And some of the time that you invite with 7, you'll be in 3NT with 7 opposite 17.

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I agree with ArtK78. Partner is far more likely to have 16 than 18. Even if partner does have the maximum to get us to the Magic 25 threshold, 3NT might not play so well with a likely natural heart lead from either side, so we might beat all the tables where the auction starts 1m-1S-2NT.

 

As an aside, I always thought the "25 for game" number was based on the likelihood that a combined 25 will provide enough tricks for game about half the time. Is it based on something less abstract than this? Or is it just the "Distilled Wisdom of the Ages", and the lowering of the number from 26 (a la Goren) to today's more common 25 simply represents increased aggression in modern bidding and an acceptance that bidding a few more bad games is worth it in the long run?

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Partner is far more likely to have 16 than 18.

 

Care to elaborate?

 

Or is it just the "Distilled Wisdom of the Ages", and the lowering of the number from 26 (a la Goren) to today's more common 25 simply represents increased aggression in modern bidding and an acceptance that bidding a few more bad games is worth it in the long run?

 

I don't remember the exact numbers but I think a vulnerable game should be bid at a 45% rate of success. A non-vulnerable game is closer to 50%. So yes, bidding is more aggressive than during Goren's days but they're not 'bad games', they're just more edgy to get closer to the success rate.

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Yes, but 12 opposite 13 is better than 7 opposite 18. And some of the time that you invite with 7, you'll be in 3NT with 7 opposite 17.

Nevertheless you probably have to try with this hand because with a fit the value of the singleton increases

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Yes, but 12 opposite 13 is better than 7 opposite 18. And some of the time that you invite with 7, you'll be in 3NT with 7 opposite 17.

 

 

I agree with ArtK78. Partner is far more likely to have 16 than 18. Even if partner does have the maximum to get us to the Magic 25 threshold, 3NT might not play so well with a likely natural heart lead from either side, so we might beat all the tables where the auction starts 1m-1S-2NT.

 

As an aside, I always thought the "25 for game" number was based on the likelihood that a combined 25 will provide enough tricks for game about half the time. Is it based on something less abstract than this? Or is it just the "Distilled Wisdom of the Ages", and the lowering of the number from 26 (a la Goren) to today's more common 25 simply represents increased aggression in modern bidding and an acceptance that bidding a few more bad games is worth it in the long run?

 

I'm sure someone's done simulations, but I don't know the results. Certainly your last phrase is how I'd think about the declining number. Note that double-dummy results apparently favor the defense slightly (i.e. declarer typically does a bit better), if taken before the opening lead. See that thread by bluecalm that awm cited in the long bidding vs play thread.

 

In any case, this hand is worth more than 7 given that 10 and all those 9's in 4 card suits with other honors. In fact, K&R gives it 8.4.

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http://hyperphysics.phy-astr.gsu.edu/hbase/math/immath/gauds.gif

 

Gaussians distributions ain't discrete

They also extended from -infinity to +infinity

And they're symmetric

 

Oh yeah, in this case the output is constrained...

However, your basic intutition is dead on

 

My hand generators are back at home.

However, I was bored enough to write one in MATLAB

 

Points = zeros(39,1);
Points(1:3) = 4; % 3 Aces
Points(4:7) = 3; % 4 Kings
Points(8:10) = 2; % 3 Queens
Points(11:12) = 1; % 3 Jacks

%% shuffle

simlength = 100000

MC_Result = zeros(simlength, 1);

for i = 1:simlength
   
   index = randperm(39);
   Points = Points(index);
   MC_Result(i) = sum(Points(1:13));
   
end
   
mean(MC_Result)   

Output = 0:40;
Output = Output';

for i = 1:41

   Output(i,2) = length(MC_Result(MC_Result == i-1));
   
end

Output(17:19,:)


ans =

         16        4141
         17        2832
         18        2050

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Admittedly it's not gaussian, but gwnn's post gets the essential point across. Kudos for the quick code, though; I'm all for actual data over heuristics.

 

Alternatively, you can just go to wikipedia:

 

p(16 hcp) = .0331

p(17 hcp) = .0236

p(18 hcp) = .0161

 

So 16 hcp is more than twice as likely as 18.

 

Of course, these numbers will change a bit for balanced hands instead of all hands, but presumably not too much.

 

Added: Impressively, Pavlicek gets it without simulating.

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Admittedly it's not gaussian, but gwnn's post gets the essential point across. Kudos for the quick code, though; I'm all for actual data over heuristics.

 

Alternatively, you can just go to wikipedia:

 

p(16 hcp) = .0331

p(17 hcp) = .0236

p(18 hcp) = .0161

 

So 16 hcp is more than twice as likely as 18.

 

Of course, these numbers will change a bit for balanced hands instead of all hands, but presumably not too much.

 

Added: Impressively, Pavlicek gets it without simulating.

 

You need to factor in that one person (us) is already know to have a 7 point hand. This is what Hrothgar does above.

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Phil you're really fast! anyways, I'll note in passing that I'm pretty sure hrothgar's original code didn't take it into account and he edited it later, that's why we both just said that it's .33:.16 after hrothgar and the websites too!
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Phil you're really fast! anyways, I'll note in passing that I'm pretty sure hrothgar's original code didn't take it into account and he edited it later, that's why we both just said that it's .33:.16 after hrothgar and the websites too!

 

I was referring to his edit.

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