Join Bridge Winners
All comments by Neal Smith
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Here are the four hands:

1.
S - AKQJT32
H - 54
D - 98
C - 76

2.
S - 98
H - 76
D - AKQJT32
C - 54

3.
S - 76
H - AKQJT32
D - 54
C - 98

4.
S - 54
H - 98
D - 76
C - AKQJT32

Regardless of how the hands are distributed to N, S, E & W, the winning strategy is easy. Bid only to the three level. Double opponents if they bid to the four level. Works especially well with both Vul.

Aug. 14, 2013
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Law 6C: “A member of each side should be present during the shuffle and deal unless the Director instructs otherwise.”
Aug. 13, 2013
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As a less experienced player, I find this practice very intimidating. At the start of a match there's always a bit of nerves. My first major decision is not about bridge but whether to challenge the opponents' pre-deal? Please, guys, don't do this.
Aug. 13, 2013
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Well, what is their 1 NT range? I would check their card before making a decision.
Aug. 13, 2013
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My thought process:

If opponents have a 10 card spade fit, partner rates to have 4-5 hearts and 6-7 minor cards on average.

We don't need a lot for game. As little as xx xxxxx Qx Kxxx gives us a shot.

Slam is possible. Pard could have Ax xxxx xx KQxxx.

Opponents could have game. Give East a singleton D plus K of H and West a singleton C.

It's unlikely but possible that game is on for both sides. No way am I going to be the one to lose 15 IMPs on one hand at this point in the day.

I bid 5C. In tempo.
Aug. 13, 2013
Neal Smith edited this comment Aug. 13, 2013
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I think you mean “in tempo”, right?
Aug. 13, 2013
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Good call. But new partnership, we didn't have this in our arsenal.
Aug. 13, 2013
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I voted “Both equally” because the partnership system (apparently) does not include a 3S bid to right-side the contract, as pointed out by John T. I recognize that almost no one plays 3S that way anymore, but, still, it's a system flaw that responder cannot declare spades even when the hands call for it.
Aug. 8, 2013
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You're saying opponents bid to the 4 level with only 8 trumps, missing KJT. And only 14 HCP between them. Hmm.
Aug. 7, 2013
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OK, good stuff above, but here's another take on the math.

After 9 tricks have been played (3 spades, 4 hearts, 2 clubs), there are 8 EW cards outstanding, of which West is known to hold a spade. So the unknown cards are 2 clubs and 5 diamonds. If we deal West three of the seven cards randomly, the odds are:

1 D + 2 C: c(5,1) * c(2,2) / c(7,3) = 5*1/35 =..1/7
2 D + 1 C: c(5,2) * c(2,1) / c(7,3) = 10*2/35 = 4/7
3 D :….. c(5,3) / c(7,3) …….. = 10/35 =. 2/7

(where c(7,3) means number of combinations of three cards selected from 7)

If West has both clubs, then he started with 4-3-2-4 distribution and East 2-3-6-2, so West has only a 1/4 chance of holding the A D to go with his clubs.

If West has neither club, then the diamonds were 4-4 originally, so East has a 1/2 chance of holding the A D along with the long club.

To summarize:

Clubs (W/E)…Ace D….Odds…Matchpoints
…………………………Play it safe….Go For it
…3/3……….Any….4/7…….6…………..13
…2/4……….West…1/7…….7 1/2……….7 1/2
…2/4……….East…1/7…….9…………..1
…4/2……….West…1/28……9…………..1
…4/2……….East…3/28……7 1/2……….7 1/2

Expected matchpoints ——->….6.91………..9.48

So we have a statistical advantage of 2 1/2 matchpoints, pretty much confirming the consensus of this discussion.
Aug. 4, 2013
Neal Smith edited this comment Aug. 4, 2013
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I see it now. The key is unblocking the T8 on the 1st 2 rounds of D.
Aug. 3, 2013
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Well, I'm struggling to see how that would work. Could you be more specific?
Aug. 3, 2013
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On deal 2, call the director. North has only 12 cards.
July 20, 2013
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