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From: Travis Crump <pretz...@techhouse.org>
Newsgroups: rec.games.bridge
Subject: Re: A somewhat odd situation
Date: Wed, 14 Nov 2012 16:12:35 -0500
Organization: A noiseless patient Spider
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On 11/14/2012 11:19 AM, Adam Beneschan wrote:
> On Wednesday, November 14, 2012 8:16:59 AM UTC-8, Adam Beneschan wrote:
>> On Wednesday, November 14, 2012 6:15:14 AM UTC-8, Dave Flower wrote:
>>
>>> On Tuesday, 13 November 2012 09:51:34 UTC, bhmwe...@gmail.com wrote:
>>
>>>
>>
>>>> It's not unusual to use Blackwood in order to check that the opponents don't hold too many aces. But yesterday I used Blackwood to check that partner did not hold too many aces! The hand: AQ8752 A64 AK9 A I opened 1C (strong, 18+), partner responded 1N (= 7-10, balanced). On my 2S partner responded 4S, and I asked with 4 N (RKC, 1430) When partner respionded 5D showing that he did not have the SK, the slam seemed much safer in my mind. I was happy that partner did not have too many aces!
>>
>>>
>>
>>>
>>
>>>
>>
>>> I don't really follow - if partner is missing the SK, there could well be a loser.
>>
>>
>>
>> I think the idea is that, assuming partner has 4+ spades, there's a 50% chance of a loser in a suit--but given that partner's response shows a narrow HCP range, the missing king in spades would be (very roughly speaking) converted to three HCP in some other suit(s), and if it's the king in some other suit, that king would be a certain trick.  So following this logic, the hand might have a half-trick greater playing potential.  In reality I'm sure it's less than a half trick.
>>
>>
>>
>> This is something that might be studiable with a simulation.  Determine how many tricks are available, double-dummy, opposite balanced 7-10 hands with the spade king, and opposite balanced 7-10 hands without the spade king.
> 
> Sorry, I forgot to include the condition that the hand has to have four spades.  You could also do a simulation including hands with three spades but in that case I'm guessing you'd get the opposite result (i.e. the hands *with* the spade king will produce a higher number of average tricks).
> 
>                              -- Adam

I think we need a lot more description of the bidding then we've been
given if we want to try to make any real assumptions.  After 1C-1N;
2S-?, what is the difference between 3S and 4S and 2N followed by 4S.
What would 4C/4D/4H have meant[8 cards in the suits and concentrated
values seems logical, though possibly always 4-4].  Assuming partner
could still be the whole range of 7-10 on this auction seems horrible.

Travis