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Round 2, Board 11


inquiry

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[hv=d=s&v=n&n=saj73h5d7cakt9852&s=s8hkq642da964cq64]133|200|Scoring: IMP

West (here South) overcalls 4 after which opponents do not bid any more.

 

I thought this would be a double or bid 6 hand. It would be more complex for those who the double of 4 is takeout. I can't imagine 4 going down less than the value of 5, so 5 seems not be biddable to me. What do I know?[/hv]

 

6C=10, 4Sx=6, 5C=2, 4S=1, 5H=0

 

6CN gnasher/catch22

6CN hrothgar/Free

6CN jdonn/gib

6CN kfay/jchiu

6CN peachy/lg62

6CN rogerClee/cherdano

6CN tylere / bid_em_up

4SWx ant590 - crayzeejim

4SWx Flycycle/Wackojack

4SWx karlson/threenobob

4SWx lobowolf/bkjswan

4SWx mbodell - javabean

4Sxw olegru - driver733

4SWx Siegmund/MSchmahl

4Sxw tlgoodwin/timg

5CN bluecalm/redds

5CN East4Evil/sohcahtoa

5CN jlall/hanp

5CN kristen33/jillybean

5CN sallyally/joylson

5HS elianna/awm

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This is very tough.

We bid:

 

1 - 4 - 5 pass

pass pass

 

Initially I thought 6 from partner hand is better but now I don't think so. I did some simuls and slam doesn't make often enough.

Doubling would be clear if we were sure partner leaves it in. Unfortunately partner will be more than happy to bid his diamonds with 6-5 shape (or even 6-4/5-5?) as double just shows points I believe.

Overall I like partner's choice now. Opinions ?

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Obviously my score on this board is terrible and will not be discussed further.

 

However, I am not convinced that 4X is superior to 5, and certainly not by as much as the scoring indicates. Of course, 5 is virtually cold (the only question is +400 vs. +420, and it will usually but not always be the latter). So the question is our odds of taking six tricks on defense.

 

We have the A and two spade tricks pretty much for sure. North can also score a ruff for three spade tricks (a second ruff won't help, as it's with a natural trick). That's only four tricks, so to get the magic +500 we need to score both a heart trick and a club trick.

 

Now let's consider declarer's possible shapes. If he has only seven spades, then he won't be 7(321) or 7222 as these are not really 4 bids at equal... the most likely shapes would seem to be 7330 and 7141, neither of which allows us to get +500.

 

If declarer has eight spades, then we need him to have at least two hearts and at least one club. This means 8221 or 8311 allow +500, but 8131 and 8230 and 8320 do not.

 

Putting these together, I'm not convinced that we get +500 more than half the time. Even if we do, it's bound to be close.. and 5 > 4X any time we don't get +500 or more.

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Seems to me that 4Sx is down 800 or half the time, and 500 or less half the time. Since it loses to the slam, perhaps 6 is a tad high, but if we lower the double, we have to lower the 5C I think also.

 

So when 4Sx goes down 800, no game (5C or other) beats it. When it goes down only 500 it beats teh slams that go down (for instance, heart to ACE, spade ruff, big combo, but possible).

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Seems to me that 4Sx is down 800 or half the time, and 500 or less half the time. Since it loses to the slam, perhaps 6 is a tad high, but if we lower the double, we have to lower the 5C I think also.

 

So when 4Sx goes down 800, no game (5C or other) beats it. When it goes down only 500 it beats teh slams that go down (for instance, heart to ACE, spade ruff, big combo, but possible).

How is 4 down four for +800?

 

If declarer has eight spades, that's six tricks in spades and the A is only down three at most?

 

If declarer has seven spades, he will usually end up scoring a diamond trick (yes, I know he loses one to the ace and another gets ruffed away, but he will probably get the third round of diamonds after pulling trumps). He might also get a second diamond trick, or a second heart trick.

 

It seems like your statistics are as if the opponents were vulnerable, and -3 would score +800.

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This hand brings up an interesting bidding set piece of logic though. I don't think South can reopen if North passes at the table, but in a bidding contest letting them play in 4S white undoubled is completely absurd.

 

Maybe it's not in the spirit of the contest to take advantage of this, but if you do so you can easily pass the North hand and then partner has an almost mandatory reopening X regardless of what they have! Therefore almost all passes of 4 level preempts become penalty :P To take this a step further, then 5C directly has to be stronger than passing and bidding 5C when partner mandatorily reopens.

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Seems to me that 4Sx is down 800 or half the time, and 500 or less half the time. Since it loses to the slam, perhaps 6 is a tad high, but if we lower the double, we have to lower the 5C I think also.

 

So when 4Sx goes down 800, no game (5C or other) beats it. When it goes down only 500 it beats teh slams that go down (for instance, heart to ACE, spade ruff, big combo, but possible).

The only real question is which gives a higher score, 5 or 4X. The position of these contracts relative to other spots (4 undoubled, 6, 5 I guess) is identical.

 

Currently you have 5 scored as 2 and 4X as 6. Again, since the relative positions of other spots is identical, this means you think 4X will score better than 5 a substantial majority of the time.

 

My point is that I don't really think this is true. It's not clear to me that 4X goes down the necessary three tricks (to obtain +500 and beat 5 making) more than half the time. Even if it is more than half the time, I'm convinced it's not much more. Thus I think the score of 5 should be higher and the score of 4X should be lower; I would probably score them more like 5=5 and 4X=3.

 

Note that this has no effect on my own score, hopefully making me a reasonably neutral opinion on this.

 

It's possible to run a double-dummy simulation on this, and I think double-dummy is okay for this board (no obvious advantage to either side). The only trick is specifying hands for the 4 jump; I'd simply state that a jump to 4 shows one of:

 

(1) Any hand with all eight outstanding spades

(2) Seven of the remaining spades to the KQ, with either a void or two singletons.

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It's possible to run a double-dummy simulation on this, and I think double-dummy is okay for this board (no obvious advantage to either side). The only trick is specifying hands for the 4♠ jump; I'd simply state that a jump to 4♠ shows one of:

 

(1) Any hand with all eight outstanding spades

(2) Seven of the remaining spades to the KQ, with either a void or two singletons.

 

According to my simulation based on your assumptions 4 makes:

 

<7 tricks: 8.9%

7 tricks: 24.8%

8 tricks: 62.1%

9 tricks: 4.2%

10+ tricks: 0%

 

Also in 90.4% of hands the best lead is natural A which seems to confirm that double dummy simul doesn't skew the results too much.

 

I run 6 from N hand on the same set of hands and the results are:

6 makes: 68%

6 doesn't make: 32%

 

So assuming 5 always makes we have the following:

 

4x scores 32% vs slams and 43% against 5

5 scores 66% against 4x and 32% against slams

6 scores 68% against everything.

 

Assuming field is not so often in slam and equally often in 5/4 (say 40-40-20) scores should look like:

 

6 - 64.4%

5 - 52.8%

4x - 43.6%

 

Scalling it to make top score a 10 and rounding the other scores, we have:

(this is based on pure mp expectancy)

 

6 - 10

5 - 8 (it was 8.2)

4x - 7 (it was 6.7)

 

If you want to somehow reward the top spot the score could look like this for example:

 

6 - 10

5 - 7

4x - 6

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The only trick is specifying hands for the 4 jump; I'd simply state that a jump to 4 shows one of:

 

(1) Any hand with all eight outstanding spades

(2) Seven of the remaining spades to the KQ, with either a void or two singletons.

I think that is too restrictive for a jump to 4. All white at IMPs I think a lot of people would jump to 4 (especially over a 1 opener) with KQ-seventh of spades in a 7321 pattern and some with a 7222 (especially if it included either the A or K).

 

 

All white IMPs 1 to me with KQT9xxx Ax x Jxx and I'd bid 4 for sure. Preempting based on the rule of 2/3/4 suggests that you'd need one of the A or non-stiff K when you are 7321 or 7222.

 

How much would including 7321 and 7222 hands effect the simulation results? Not sure.

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The simulations I run on this one is less restrictive than bluecalms. I require WEST to have 7 or 8 spades. I did not require him to have KQ, nor a singleton, nor a void. There was a reason for this selection, the actual hand WEST held at the table when this hand was bid was the hand held when it was played (it was 7222 with only one spade honor).

 

Now, most of might not bid that way, and so perhaps if we restrict the opponents to being sane, one might reasonable conclude that this hand is not possible. For what it is worth, the majority of the field bid 3 or 4 spades with the shown hand. Guess that speaks the sanity of the field.

 

Be that as it may, when you run such simulations, you discover 5C is cold, 6C has excellent chances (just below 88%), and for 4Sx I get it takes 7 tricks or less 66.5% of the time, 8 or 9 tricks 33.5% of the time.

 

So both 4Sx and 6C should score higher than 5C, but clearly 5C is scored too low. To estimate 5C's score, I have figured that 1/3 of the time 4Sx will score less and nearly 1/3 of the time 6C will go down, using the contracts the panel bid here. Also all the 5C tie themselves.

 

I decided to matchpoint the bidding in this contest and normalize to 12 using the math for this type of results. The 6C guys get 0.5 point for the other 6C's (3MP), 1 MP versus 5H, plus 88% of the remaining spots (3+.88*13+1)=15.44/20=.77% = 9.2 MP on a 10 point scale.

 

4S doublers tie themselves (3.5 Mp), beat the 5H contract for 1MP, they beat 6C 12% of the time for (.12*7) and beat 5C 66.5% of the time (.665*5), this comes to 8.665 out of 20 mp or .433%, converting to 12 points scale this is 5.19 MP

 

The 5C bidders get 1MP for 5H, and 2.5 mp for the other five club bidders. They get 0.12 MP for each 6C bidder (.12*7) and 0.335% for each 4Sx (.335x8) which comes to 7.02 out 20, or .351%. Converting to a 12 point scale, this is 4.2

 

So calculated values are

6C = 9.2

4Sx = 5.2

5C = 4.2

 

So my plan is to round the scores UP, so 6C stays a 10; 4Sx stays a 6, but 5C goes up from 2 to 5.

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Looks sensible. I am not going to run more simuls on this one as they are quite time consuming. It's obvious that if you add more flat shapes for 4 bidder then 6 is more attractive and 4x is more attractive too.

 

the majority of the field bid 3 or 4 spades with the shown hand

 

Bidding 3 and bidding 4 are completely different thing though.

There aren't many people who would even thought of bidding with:

 

KQxxxxx xx xx xx

 

While it's true that many people would bid 4 with 7-2-2-2 or 7-3-2-1 they usually need quite a strong hand for that.

Maybe mine (awm's in fact) requirements were too restrictive but just saying 7+spades is too loose for sure.

More reasonable would be one of the following:

 

1) 8+spades

2) 7 spades to KQ and either void or 2singletons

3) 7spades and 12+hcp

 

Anyway, it's all close anyway. I respect your scores on this one. Interesting board for sure. In my opinion the one where most pairs are doomed to be in 5 :)

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Well, I can't say bidding 4S is retarded on a 7222 distribution because wereagles will be all over me, saying that I'm insulding all retards and him in particular. I think it's really bad though and I wouldn't expect that shape for a 4S bid.

 

Having said that, I have no problem with these scores.

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