2022 August MSC

[veredk]

PROBLEM A: 4 Diamonds
PROBLEM B: Double
PROBLEM C: 3 Hearts
PROBLEM D: 2 Clubs
PROBLEM E: 4 Diamonds
PROBLEM F: 4 Spades
PROBLEM G: Double
PROBLEM H: Diamond 7

[ccr3]

PROBLEM A: 5 Diamonds
PROBLEM B: 3 Spades
PROBLEM C: 3 Clubs
PROBLEM D: 2 Hearts
PROBLEM E: 4 Diamonds
PROBLEM F: 4 Spades
PROBLEM G: 2 Notrump
PROBLEM H: Diamond 10

[blubayou]

SOLUTIONS FOR: Jock McQuade, 3 Bag End, Hobbiton OR 97030U.S.A.
PROBLEM A: 5 Diamonds  heavy sigh, for  "4 " but not "4 "
PROBLEM B: 5 Clubs 4-5 in his suits with  AK, AK  can claim,  and picking daisies won't make SLAM easier to bid than this picture-bid would.
PROBLEM C: 3 Clubs
PROBLEM D: Double
PROBLEM E: 4 Diamonds--endplaying pard into raising diamonds, THEN BIDDING  5    is not soft-peddaling this hand at all.  I could rant about playing leaping michaels as FORCING..and I will later--count on it!
PROBLEM F: 5 Hearts
PROBLEM G: 3 Diamonds
PROBLEM H: Diamond --9 no-brainer turned into  a fun problem  by your comments  wow

A copy of these solutions will be e-mailed to you for your records.

[jcreech]

SOLUTIONS FOR:
James Creech

Fredericksburg VA 22407
U.S.A.

Changes from my initial thoughts are discussed below:

PROBLEM A: 4 Spades  The ten-trick game is what finally tipped the decision.  I'm not sure that we don't have three losers, but I have no delusion that I am showing Ax in spades with terrific diamonds.  I still only promise two spades and less than 10 HCPs; I just think that 4 is better place to play than 5 or 3 NT.  There is reasonable play opposite ♠ KQxxx   ♥ xx   ♦Axxxx   ♣ x, and that is still short of what I expect from partner given his bidding.
PROBLEM B: 3 Spades
PROBLEM C: 3 Clubs
PROBLEM D: Double
PROBLEM E: 4 Hearts  I ended up spurning the hint for reasons akin to Todd's.  If the 7-5 had been reversed, I would have still taken the hint, but with a two-card difference in length favoring the major, I have to go with the major.
PROBLEM F: 5 Spades
PROBLEM G: 3 Diamonds
PROBLEM H: Diamond 9  I really don't think there will be a scoring difference for a diamond lead. regardless of choice.  However, the nine is a false-card that will not hurt partner and could talk declarer into taking a wrong line, so if there is a scoring difference, this may capture it. 

[Masse24]

In case I was unclear, I hate lead problems.

[kenberg]

I might not post this month. I shall return.

[jcreech]

In case I was unclear, I hate lead problems.

Todd,

we might be more inclined to believe you if you were less successful in your lead selections.  Last month was another 100, if I recall.

I might not post this month. I shall return.

Ken,

I always find your reasoning worth the read, even if I end up disagreeing your conclusion.  I hope that you will find time in these last few hours of this month's MSC to enter, even if your reasons may have to hold off until later.  But if you are not, I will still look forward to when you do return.  Thank you for all of your contributions, in MSC and throughout IAC.

[kenberg]

You have inspired me, I will give it a try. It might be sort of "choose first, think later".

[jcreech]

You have inspired me, I will give it a try. It might be sort of "choose first, think later".

Not a surprise, Ken.  After all you said:

If this set scores well I might decide to never think again.

Shortly after that, you finished 2nd in IAC and on the BW Honor Roll.  I am presuming that those results could be considered as a "... set (that) scores well ...".

[kenberg]

A: 4D
B: 5C
C: 4C
D: X
E: 3S
F: 5S
G: 2NT

H: What the ...?
I am still trying to understand the description.
I will take Blu's word for it that 2-1 points means 2 aces and one K but we still have the five in five 2-1 points to worry about. No, of course we do not all know what this means. It's a serious issue with relay systems, the explanations make sense only to those who have studied the system.

Anyway, I guess I am to assume that dummy will hit with a 5=1=3=4 shape, the diamond K, no other king, two aces, the club Q. It seems declarer must have two aces since otherwise he must have a void and somewhere along the line we should have been told that if they are playing it straight. Declarer can have at most six diamonds in his hand, so we count six diamond tricks in hand the spade A, the heart A and, perhaps, four club tricks. I doubt we can stop him from getting a heart ruff, so that would be 13 tricks. So we have to assume that either partner has the club J, holding declarer to three tricks in that suit, or that declarer has only five diamonds.

I really do not like this problem. In a typical online game I just relax when these relay systems arise but if I were playing something more seriously I would study their relay system beforehand and insist on some more detailed explanations. But time flies and I am leading the spade K. Nah, I am overthinking this. Some D.

H: D T

[kenberg]

MASTER SOLVERS CLUB SOLUTIONS RECEIVED


Your solutions have been received. This copy is for your records.

SOLVER: Ken Berg
       
Your Solutions for the August 2022 Contest
-------------------
PROBLEM A: 4 Diamonds
PROBLEM B: 5 Clubs
PROBLEM C: 4 Clubs
PROBLEM D: Double
PROBLEM E: 3 Spades
PROBLEM F: 5 Spades
PROBLEM G: 2 Notrump
PROBLEM H: Spade King

[Veeree]

PROBLEM A: 5 Diamonds
PROBLEM B: 5 Clubs
PROBLEM C: 3 Clubs
PROBLEM D: 2 Hearts
PROBLEM E: 4 Diamonds
PROBLEM F: 4 Spades
PROBLEM G: 3 Diamonds
PROBLEM H: Diamond 10

[yleexotee]

PROBLEM A: 4 Spades
PROBLEM B: 4 Clubs
PROBLEM C: 4 Hearts
PROBLEM D: 2 Hearts
PROBLEM E: 4 Diamonds
PROBLEM F: 6 Spades
PROBLEM G: Pass
PROBLEM H: Diamond 9

[Masse24]

August Results

Masse24 led the IAC solvers with 730, edging out both Hoki and Peuco, with 720 and 690 respectively.  Masse24 and Hoki also made the MSC Honor Roll!

Notably, of the eight Master Solver's problems, Hoki scored 100 on six of them.

NAME	BW-SCORE	RANK	MPs
Masse24	730	1	30
Hoki	720	2	25
Peuco	690	2	10
Also participating this month were:  BabsG, BluBayou, CCR3, DickHy, JCreech, KenBerg, VeeRee, VeredK, WackoJack, YleeXotee. Congratulations to all!

[jcreech]

It will be interesting to see how they justify the scoring on the lead problem. 

♠ K Q J 10   ♥ K Q J   ♦ 10 9 8 7   ♣ 10 9

The 100 went to 10/9 of , with a combined 7 votes.

The single top vote getter was the K also with 7 votes and a score of 30 (the Q/J had no votes and also got the same score).

The 10/9/8/7 of got a combined 9 votes with a score of 80

And the K/Q/J/10 had a combined 6 votes with a score of 90 (the Q/10 had no votes).

On the basis of single-card voting, the K the 10 and K should have gotten the 100, while on the basis of the block of co-equal cards with the most votes was a diamond with 9 votes, then any diamond lead should have gotten the 100.  However, neither criterion fit the top two scores.the rationale for the scoring is quite unclear to me.

edited because I accidently looked at the solver % and made it a Panel vote - oops