[xcurt]

Problem 1 from this months "The Bidding Box" is, I think, missing a score.

Scoring: MP

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I could easilly see a few pairs stumbling into 6NT on this. How would you play 6NT on a red lead? How many matchpoints would you expect to get on a 12 top if you got to 6NT?

The scoring given was 6♥ 10, 6♣ 6, 5♥ 5, 5NT 4, 6♠ 3, 5♣ 2, 4♠ 1

[kenrexford]

The play:

"On a red lead?" These are wildly different. On a heart lead, I win the Queen, play a club, and claim. I cannot do that on a diamond lead.

On a diamond lead, I'm fairly boxed in. I win the diamond perforce. I run seven hearts just because (might as well play spades first). I then try spades. I need 109(x) somewhere.

With a diamond lead, I'm not playing one person for 4-4-1-4 pattern and his partner with 3-1-9-0 pattern,. If I see a void on the opening diamond lead, now I can play a club; the odds favor the guy without ten diamonds having the Ace. Although, I'd suspect that the guy with ten diamonds actually has the club Ace.

The Bidding (to 6NT?):

Sure. Could happen. Probably would happen a few times.

[CSGibson]

kenrexford, on May 23 2008, 07:21 PM, said:

The play:

"On a red lead?"  These are wildly different.  On a heart lead, I win the Queen, play a club, and claim.  I cannot do that on a diamond lead.

On a diamond lead, I'm fairly boxed in.  I win the diamond perforce.  I run seven hearts just because (might as well play spades first).  I then try spades.  I need 109(x) somewhere.

Same line of play as Ken, but note that you have an additional chance if a player with 4 spades also has the club Ace by pitching all your minor suit cards on the run of hearts. This leads to a stepping stone-squeeze.

[xcurt]

kenrexford, on May 23 2008, 10:21 PM, said:

"On a red lead?"  These are wildly different.  On a heart lead, I win the Queen, play a club, and claim.  I cannot do that on a diamond lead.

On a diamond lead, I'm fairly boxed in.  I win the diamond perforce.  I run seven hearts just because (might as well play spades first).  I then try spades.  I need 109(x) somewhere.

I make ST9 tight anywhere and ST9x anywhere 10.33% (1/21 5/2 breaks on each side plus 5/35 4/3 breaks on each side = 0.6217*5/35 + 0.3052*1/21).

Even if the opponents always lead diamonds and the field never goes set, namely 6H down on a club ruff or 6C down on two trump losers, that's worth 1.2 on a 12 top. I think 6NT is better than that, though. I'll post more tomorrow.

I mentioned both red leads only because there's nothing to the play on a black lead. On a club lead they will take the ace at T1 or you have a claim. On a spade lead you are just going to see if spades are 4-3 (or T9 tight) for 13 tricks.

Edit -- csgibson found the solution (which works regardless of who has the CJ if we read the ending and come down to AKQ8, -, -, K vs J, -, -, QTx)

Curt

[kenrexford]

CSGibson, on May 23 2008, 10:30 PM, said:

kenrexford, on May 23 2008, 07:21 PM, said:

The play:

"On a red lead?"  These are wildly different.  On a heart lead, I win the Queen, play a club, and claim.  I cannot do that on a diamond lead.

On a diamond lead, I'm fairly boxed in.  I win the diamond perforce.  I run seven hearts just because (might as well play spades first).  I then try spades.  I need 109(x) somewhere.

Same line of play as Ken, but note that you have an additional chance if a player with 4 spades also has the club Ace by pitching all your minor suit cards on the run of hearts. This leads to a stepping stone-squeeze.

Sexy!

I'm going for that line.

[xcurt]

Since I didn't add this yesterday, I would score 6NT as either 4 or 5.

The relevant variables are

1) how often the opponents lead diamonds
2) how often we read the endposition and how often the endposition allows us to make
3) how often other pairs go -50 on our cards

It's 45.2% (1) that the stepping-stone position exists. However, the opponent with the spade guard and {Cl]A might pitch down to xxx, -, x, A and give us a nasty surprise. Therefore we need to adopt a mixed strategy of sometimes overtaking the ♠J and sometimes cashing it and then playing a club in the 5-card ending. We can certainly do no worse than go right half the time is we randomly choose from the two lines. Of course then the opponent could always keep a diamond ensuring that when we go wrong, we go down many not halving the board with pairs to go -1 in 6♣ or 6♥.

Assuming that the opponents lead diamonds 90% of the time, we go right in the endposition 5/8 of the time, but half the time we go wrong we go down many (2), and that that 10% of the other results are -50 (3), I get a matchpoint expectation of between 4 and 5. I won't reproduce the algebra here.

(1) p(4-3 spades) * 8/17 + p(5-2 spades)*7/17 + p(6-1 spades * 6/17) + p(spades zip-split) * 5/17.
(2) Assuming that we get information 1/4 of the time that allows us to go right with certainty, and the remainder of the time we jchoose randomly from between the lines, we will go right 1/4 + 1/2*3/4 = 5/8 of the time.
(3) The -50's are from two club losers in club slams, or a club ruff in heart slams. These results are uncorreted with our chances of making 6NT (they are correlated with each other however) so we can ignore any correlation terms.

Finally, it's quite likely that there are Condorcet cycles among the various contracts, making the calculation of EV at BAM near-impossible.

[CSGibson]

board a match I would pitch down to J - - QTxx opposite AKQ8x - - -. This gives opponents an extra worry of an overtrick if they don't discard down to 4 spades and the ace of clubs. I feel fairly confident that I would be able to read the end postition, however, based on the lead and discards.

[nige1]

CSGibson, on May 24 2008, 02:04 PM, said:

board a match I would pitch down to J - - QTxx opposite AKQ8x - - -.  This gives opponents an extra worry of an overtrick if they don't discard down to 4 spades and the ace of clubs.  I feel fairly confident that I would be able to read the end postition, however, based on the lead and discards.

If West is declarer in 6N and North leads a ♦, then CSGibson is almost certain to make if the opponent with ♣A has 4 or more ♠ to the T or 9 .
That defender is almost certain to keep 4 spades. as he will dread his partner's postmortem comments when declarer has a small ♠ instead of the knave

Paradoxically, on a ♦ lead, CSGibson is more likely to go down when a less skilled declarer wraps up a carefree 13 tricks That is when when ♠T9 are doubleton or tripleton