Jump to content

How strong should partner be


pooltuna

Recommended Posts

Just strong enough for me to make 5, as that's what I'm going to bid.

I see, so holding

[hv=s=saj8h863dakj4caj9]133|100|[/hv]

 

and having his telepathy turned off (translation the ICQ server had a denial of service attack) he is supposed to guess to bid 6?

Link to comment
Share on other sites

Just strong enough for me to make 5, as that's what I'm going to bid.

I see, so holding

[hv=s=saj8h863dakj4caj9]133|100|[/hv]

 

and having his telepathy turned off (translation the ICQ server had a denial of service attack) he is supposed to guess to bid 6?

This is not a double of a 4H opening.

Link to comment
Share on other sites

Just strong enough for me to make 5, as that's what I'm going to bid.

I see, so holding

[hv=s=saj8h863dakj4caj9]133|100|[/hv]

 

and having his telepathy turned off (translation the ICQ server had a denial of service attack) he is supposed to guess to bid 6?

This is not a double of a 4H opening.

You've been playing in the buffalo chips so long you are starting to throw them around everywhere if you don't think this is a X of a 4 bid. I have to ask, no matter how implausible it is, what call do you suggest South make?

Link to comment
Share on other sites

This slightly dubious calculation may help to explain why it's unwise to bid more than 5:

 

There are 22 missing HCP outside hearts, comprising three aces, one king, two queens and three jacks. Suppose that partner has 15 HCP, and has no honours in hearts. Partner's average number of aces will be 15/22 x 3 ~= 2.

 

For slam to be good, you need at least an average of 2.5 aces. That requires his average HCP to be 2.5 x 22/3 ~= 18.

 

Even taking into account that partner will place more weight on aces than on lower honours, it seems well against the odds to hope for three aces opposite.

Link to comment
Share on other sites

Easy 5. Not so much because we have the strength, but because we have the shape.

 

You won't reach perfecto slams most of the time when your opponents open at the four level. That's life.

 

Ron, the hand pooltuna posted is a routine double of 4.

Link to comment
Share on other sites

Easy 5. Not so much because we have the strength, but because we have the shape.

 

You won't reach perfecto slams most of the time when your opponents open at the four level. That's life.

 

Ron, the hand pooltuna posted is a routine double of 4.

Is Ron the blind one with the occassional acorn diet?

Link to comment
Share on other sites

Over 4Hx you pass with all hands where you think you dont make anything and YOU DONT CARE if they can make 4Hx or not. Therefore a double of 4M preempt should show some hope of beating 4H facing very little.

 

Like a better than average strong NT or better. Or a shapely hand with at least 3 defensive trick.

 

5C making 6 is what i would expect at most expert tables.

 

 

IMO partner is favorite to have 3 aces here but he might easily have just 2 clubs. or you might have no wya to pitch the D. Give me the K of D instead of the K of H and ill surely bid 6.

Link to comment
Share on other sites

This slightly dubious calculation may help to explain why it's unwise to bid more than 5:

 

There are 22 missing HCP outside hearts, comprising three aces, one king, two queens and three jacks. Suppose that partner has 15 HCP, and has no honours in hearts. Partner's average number of aces will be 15/22 x 3 ~= 2.

 

For slam to be good, you need at least an average of 2.5 aces. That requires his average HCP to be 2.5 x 22/3 ~= 18.

 

Even taking into account that partner will place more weight on aces than on lower honours, it seems well against the odds to hope for three aces opposite.

This is worse than slightly dubious.

 

Hands with a few honour cards and lots of spot cards are more likely than hands with lots of honour cards and few spot cards.

 

Effectively this means that hands with a higher number of high cards are much more likely to be made up of aces and kings than queens and jacks. In fact it turns out that the average number of aces is very close to:

 

ACES = 0.15 * HCP - 0.5

 

ACES(5 HCP) = 0.15 * 5 - 0.5 = 0.25

 

ACES(10 HCP) = 0.15 * 10 - 0.5 = 1 (Most people won't find this surprising)

 

ACES(15 HCP) = 0.15 * 15 - 0.5 = 1.75

 

ACES (20 HCP) = 0.15 * 20 - 0.5 = 2.5 (People find this more surprising)

 

etc

 

This affect will be the same if we already know where some of the honour cards are.

 

So that when there are relatively more aces missing they are much more likely to be with the hand with a greater number of high cards. That is I think your numbers are lower estimates for the number of aces given the assumptions about how many high cards partner has.

 

I don't have a spreadsheet set up for the precise assumptions you made. But given our hand - 3 kings and a queen - and not discounting the heart honours I calculated that a 15 HCP hand opposite has an average of a little over 2.4 Aces and an 18 HCP hand opposite has an average of almost 3 Aces (2.98...).

 

While unlikely there is also the possibility that partner has a heart void.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...