[kenrexford]

FrancesHinden, on Oct 20 2008, 08:05 AM, said:

kenrexford, on Oct 20 2008, 01:33 PM, said:

Cascade, on Oct 19 2008, 04:00 PM, said:

kenrexford, on Oct 20 2008, 08:33 AM, said:

20 HCP.  7 Controls = 23.3-20=3.3=+1=21.  Massoively interesting 6-card suit = +1=22.  Open 2♣...2NT.

You use "=" in a way that I do not understand.

7 controls is average for a 20 count.

Therefore what we have extra is the trick taking potential of the fifth and sixth club.

What is worse (potentially for NTs) is that we have two doubletons.

Actually, 6 controls is average. AA+KK+QQ+JJ = 20 = 6 controls

The expected, modal and median number of controls in any 20-count is 7 (expected number of controls is slightly more than this, 7.0491113...)

It's 7.6 in any 21-22 HCP hand.

Really?

That's interesting, actually. Intuitively, I see how this works (I can reach 20 with many more combinations that feature a disproportionate lack of quacks). This seems to bolster, at least in my way of thinking, a 19-20 range for 2NT openings, as the standard 19-20 count would "usually" be upgraded by my analysis, and the usual 21-count upgraded out of 2NT. This seems to play out in practice, as most of the "20-21" 2NT openings in one huge vugraph set I analyzed a while back were 19-counts.

I suppose a strict 20-21, then, would be truly a 21-22 for me, which suggests a +1 addition to usual thinking when responding to the non-upgrader.

Hmmmmm.

[Cascade]

Average controls are almost linear according to this rule:

Controls = 0.4 * HCP - 1

The rule works pretty well from about four or five HCP up to 30 or so HCP.

[Cascade]

Here are the numbers:

The first column is HCP the second average controls and the third the estimation 0.4*HCP - 1

0  0                   
1  0
2  0
3  0.320197044  0.2
4  0.803729146  0.6
5  1.060914761  1.0
6  1.355393266  1.4
7  1.801655786  1.8
8  2.190594955  2.2
9  2.538883885  2.6
10 2.935475254  3.0
11 3.372256015  3.4
12 3.752359681  3.8 
13 4.151926356  4.2
14 4.580263399  4.6
15 4.989941622  5.0
16 5.390077208  5.4
17 5.811109123  5.8 
18 6.234429522  6.2
19 6.627308396  6.6
20 7.049111347  7.0
21 7.464047395  7.4
22 7.859952752  7.8
23 8.258721634  8.2
24 8.672874887  8.6
25 9.056879765  9.0 
26 9.418196211  9.4
27 9.835929205  9.8
28 10.16498546  10.2
29 10.50830338  10.6
30 10.88095995  11.0
31 11.12181475  11.4
32 11.46308036  11.8
33 11.67227191  12.2
34 11.84656557
35 11.96153846
36 12
37 12

[Cascade]

Here are the expectation of various honours and controls. They all form fairly linear relationships with HCP.

Kings and Jacks are interesting. Expected Kings are almost exactly 1/10 of the number of HCP. Expected Jacks is always one almost independent of the HCP.

Ace = 2 King = 1 Controls

Controls = 0.39714762550297 * HCP - 0.9375869248794

Controls = 0.4 * HCP - 1 (rounded)

Ace = 3 King = 2 Queen = 1 Controls

Controls = 0.68297078407368 * HCP - 0.71629434772955

Controls = 0.7 * HCP - 0.7 (rounded)

Aces

Aces = 0.15080137869129 * HCP - 0.50274312636775

Aces = 0.15 * HCP - 0.5 (rounded)

Kings

Kings = 0.099957885163922 * HCP - 0.000661360065993

Kings = 0.1 * HCP (rounded)

Queens

Queens = 0.046115866115133 * HCP + 0.54936553633723

Queens = 0.05 * HCP + 0.5 (rounded)

Jacks

Jacks = -0.000338630416366 * HCP + 1.01687231334264

Jacks = 1 (rounded)

Any Spot Card e.g. Tens

Tens = -0.035225468 * HCP + 1.364856184

Tens = -0.04 * HCP + 1.4 (rounded)

[the hog]

P_Marlowe, on Oct 19 2008, 07:23 PM, said:

After a 2NT opening, responder may or may
not check for a minor suit fit.

After the 1C opening, I dont think you will be
able to come up with a realistic auction.

With kind regards
Marlowe

I certainly think you will Uwe.
1C (1H) X {or 1S}, where the x or the 1S bid DENIES 4+S and shows both minors.
From here on I think it is relatively easy and there are a number of pathways leading to 6. Personally I would bid 4C now.