[jdeegan]

2♣ seems clear to me. Other posters have made the point that partner is almost surely short in spades so that the odds of a club fit are excellent. With your HCP power and the spade hook a 10-1 favorite, you can handle almost any subsequent bidding sequence. 3NT, 4♥ or 5♣ are all possible depending on pard's hand.

[mikeh]

 han, on 2011-September-01, 07:10, said:

If we assume that RHO holds exactly 5 spades then the other two hands will hold exactly 3 spades, which means there are only 31 empty spaces, giving RHO 8/31 x 13 diamonds on average (which still rounds down to 3).

Of course your incorrect (?) simplification may well be more accurate than my correction (?) of your simplification.

Well, I mistakenly set out that my calculation was based on the assumption that rho held precisely 5 spades when I should have said that it was based on the assumption that he held at least 5 spades. Make that correction, which is appropriate, and there are 34 spaces into which one must fit not only 13 diamonds but 10 hearts, 8 clubs and 3 spades. So I think that your correction of my post resulted in a grammatically more accurate calculation but one that was nevertheless incorrect....mine was incorrectly described but correctly done, I think.

[whereagles]

 jdeegan, on 2011-September-01, 08:25, said:

Other posters have made the point that partner is almost surely short in spades so that the odds of a club fit are excellent.

Be careful with this sort of reasoning. While it is indeed true in general, I've noticed in practice that extreme length in (e.g. 11+) in RHO's + your suit tends to produce dummies with 55 or 65 in the other suits. This is why I'm normally cautious with this sort of hands.

[mikeh]

 whereagles, on 2011-September-01, 10:56, said:

Be careful with this sort of reasoning. While it is indeed true in general, I've noticed in practice that extreme length in (e.g. 11+) in RHO's + your suit tends to produce dummies with 55 or 65 in the other suits. This is why I'm normally cautious with this sort of hands.

You csn and in my view should do the sort of estimation I wrote about earlier.

In this case, granting rho at least 5 spades, we have 34 spaces in the other 3 hands into which to deal 8 clubs (and 13 diamonds, 10 hearts and 3 spades). Thus our expectation should be that partner will produce, on average, 8/34 X 13 clubs or 3 clubs. We can estimate that both partner and LHO will have distributions that cluster most closely around 1=4=5=3 shape. This isn't hard to do at the table. Allowing oneself to 'feel' that partner will usually be 5-5 or 4-6 reds is an error.

[VM1973]

 ArtK78, on 2011-August-31, 18:11, said:

I can see that you play a partnership game.

Might not partner, with significant length in diamonds and a fair hand, think that diamonds offers a better place to play this hand than notrump? After your 1NT overcall, you may find it difficult to restrain partner from reaching for slam in diamonds if he has something like:

x AKx AQTxxxxx x

He might be very disappointed to find out that there are two (or three!) trump losers opposite a 1NT overcall.

In any case, I am not likely to dissaude you from overcalling 1NT. I find that choice to be exceedingly weird.

Partner's proposed hand: 13 HCPs.
My hand: 16 HCPs.
Opener's hand: 13 HCPs.
Total: 42 HCPs.
Maximum in Deck: 40 Maybe we can reach 42 if we count 10s as 0.5 points?

I believe this scenario requires LHO to hold a Yarborough. Apparently the odds of that are 1827 to 1. Considering this is matchpoints, I'll take that chance.

[semeai]

 VM1973, on 2011-September-01, 11:55, said:

I believe this scenario requires LHO to hold a Yarborough. Apparently the odds of that are 1827 to 1. Considering this is matchpoints, I'll take that chance.

The odds LHO has a Yarborough given that you have 16 hcp and RHO has opened has got to be closer to 20 to 1. There's likely somewhere around 10-12 hcp left for partner and LHO, which is likely held in 3-5 cards. It's 1 in 2^3 = 8, 1 in 2^4 = 16, or 1 in 2^5 = 32 that partner has a Yarborough in each of the cases for 3, 4, or 5 high cards remaining.

[VM1973]

 mikeh, on 2011-August-31, 18:14, said:

1. assuming opener holds 5 spades (he may hold more but our length argues he will hold 5 most of the time), there are 34 empty spaces between his hand and the other two, into which we must fit 13 diamonds. Thus if he holds 5 spades he rates to hold 8/34 X 13 diamonds, or, rounding to integers....3, not the 2 you assume.

-snip-

I respect your opinion more than the others because you actually broke out a calculator instead of just making a knee-jerk rejection of the idea. In that, I salute you.

Although others have attempted to correct your assumptions, I think your assumptions are fine. Let's run with them.

We are assuming 34 empty spaces, 13 of which belong to my partner. Into those 13 spaces we must fit the remaining cards, so partner should hold:
1.15 spades.
3.82 hearts.
4.97 diamonds.
3.06 clubs.
Total cards: 13

Of course the same logic applies to LHO so he will hold (we are assuming) the same shape. So now we can calculate RHO's shape.

5.71 spades
2.35 hearts
3.06 diamonds
1.88 clubs
Total cards: 13

Assuming these numbers are accurate (which I don't claim, but I assume you are claiming) then:
A. Partner probably doesn't have an obscene number of diamonds.
B. If we get doubled, clubs will probably be a safe runout.

Of course you might argue that just because partner has on average 5 diamonds doesn't preclude him from having 6 (or 4), but the same argument could apply to hearts. Just because he has on average 4 doesn't mean he might not have 5 or 3.

Am I masterminding the hand a little by ruling out, perhaps unfairly, a contract of 5♣ or 6♣? Yes, maybe I am... but in matchpoints generally even with a minor suit fit you want to play NT.

[mikeh]

 VM1973, on 2011-September-01, 12:24, said:

I respect your opinion more than the others because you actually broke out a calculator instead of just making a knee-jerk rejection of the idea. In that, I salute you.

Although others have attempted to correct your assumptions, I think your assumptions are fine. Let's run with them.

We are assuming 34 empty spaces, 13 of which belong to my partner. Into those 13 spaces we must fit the remaining cards, so partner should hold:
1.15 spades.
3.82 hearts.
4.97 diamonds.
3.06 clubs.
Total cards: 13

Of course the same logic applies to LHO so he will hold (we are assuming) the same shape. So now we can calculate RHO's shape.

5.71 spades
2.35 hearts
3.06 diamonds
1.88 clubs
Total cards: 13

Assuming these numbers are accurate (which I don't claim, but I assume you are claiming) then:
A. Partner probably doesn't have an obscene number of diamonds.
B. If we get doubled, clubs will probably be a safe runout.

Of course you might argue that just because partner has on average 5 diamonds doesn't preclude him from having 6 (or 4), but the same argument could apply to hearts. Just because he has on average 4 doesn't mean he might not have 5 or 3.

Am I masterminding the hand a little by ruling out, perhaps unfairly, a contract of 5♣ or 6♣? Yes, maybe I am... but in matchpoints generally even with a minor suit fit you want to play NT.

This may shock you, but I didn't break out a calculator. Getting this sort of calculation to the nearest integer (since we never actually hold fractional cards) should be part of the toolkit of anyone who hopes to be a good bridge player....it assists in many aspects of the play of the hand as well as the bidding.

I am beginning to suspect that attempting to have a rational discussion with you about the bridge merits of your posts is an exercise in futility. You display the unfortunate habit of seeing posts, critical of your ideas, as invitations to repeat your ideas or even to expand upon them rather than as opportunities for you to learn from people who have far more experience and far greater accomplishments at the game.

Appeals to authority are poor substitutes for logical reasoning, and I do not for a moment suggest that you should simply accept as true anything I or any other real life expert says simply because it is we who say it. But I do suggest that you try to understand the points that others are making and that you accept at least the possibility (I would say the strong probability) that when several real experts say essentially the same thing to you, that they are right and you are wrong.

The realm of human knowledge and understanding is full of examples of lone voices whose opinions later became orthodoxy. People who cling to unpopular views often take comfort in that. But the reality is that while some lone voices were later proved to be correct, the vast majority of them were always wrong.

1N here is a rejection of the motion of partnership bridge, and the rejection of the notion that one should, when contemplating various actions, consider how the auction may progress....you focus on the lucky ways the cards may lie and the auction may develop because those outcomes favour what you want to do. The good player focuses on both the good and the bad, and compares the resulting range of outcomes with the range of outcomes forseeable from alternative action. In this case, the good player would conclude (as all good players how have posted so far conclude) that 1N is silly.

[VM1973]

No, what I don't get is why you have a perverse desire to try to cram your opinion down my throat. As I said, "I'll try 1NT." Might it go horribly wrong? Sure it might. Might it be the wrong bid? Sure, it might. Did other people disagree with me, politely, and without trying to embroil themselves in a long discussion of the merits of one bid or another? Yes. Plenty of people made statements like: I've seen better diamond stoppers or: Do you think your partner will let you play NT if he has 6 diamonds?

Look, if you want to engage in a long discussion about the merits of 1NT I'll be happy to oblige. Send me a private message because I don't think the rest of the board is really that interested. Let's leave the party in peace and have our brawl outside if it really means that much to you.

[BunnyGo]

 VM1973, on 2011-September-01, 15:13, said:

No, what I don't get is why you have a perverse desire to try to teach me the right way to play bridge

FYP

Quote

If you want to engage in a long discussion about the merits of 1NT I'll be happy to oblige... I don't think the rest of the board is really that interested.

1) This was not a difficult problem.

2) When you got the answer wrong, Mike did you the courtesy of (IMO) clearly and patiently attempting to teach you (note that in my previous post, I did not do you the same courtesy). Note that he does not benefit from this, nor do I imagine your responses give him any satisfaction.

3) There are no merits to a 1NT bid here. There's the discussion.

4) At this point the rest of the board is only interested in the same manner that we occasionally enjoy rubbernecking accidents.

5) I strongly suggest that you consider attempting to play with better partners. If you feel that you cannot trust your current partners to bid well enough that you can afford to actually make an honest descriptive bid, then either you or they are the problem. Probably both. Bridge (and in particular bidding and defense) is a partnership effort. Regularly making bids that are far off of system and description (no matter the expected length of suits...that's a red herring that ignores both variance and more importantly partner's intelligence) ruin that effort. Even if you get lucky on a hand, it will ruin the trust necessary to have regular constructive auctions.

Sincerely,

Ben

[MrAce]

deleted

[Phil]

 VM1973, on 2011-September-01, 15:13, said:

As I said, "I'll try 1NT." Might it go horribly wrong? Sure it might. Might it be the wrong bid? Sure, it might.

This sounds like someone that really doesn't like 1N, but is choosing the option to simply be contrarian.

Speaking as someone who occasionally used to take views like this, I can tell you you rate to get more a constructive response if you present your position along the lines of "2♣ looks like the normal bid, but perhaps we should consider 1N for reasons x, y and z".

[JLOGIC]

Ok, I'm on board with the "VM is just a troll" thought process. I used to be a VM apologist, please BBF, forgive me for my sins.

[Free]

 VM1973, on 2011-September-01, 15:13, said:

Look, if you want to engage in a long discussion about the merits of 1NT I'll be happy to oblige. Send me a private message because I don't think the rest of the board is really that interested. Let's leave the party in peace and have our brawl outside if it really means that much to you.

This is a forum, it's purpose is to share knowledge and experiences, discuss thought processes behind certain decisions, learn from each other's successes, learn from each other's mistakes,..., and have fun.

When there are 2 clear choices (2♣ and pass) for 99% of all bridge players, you come up with a simple "I'll try 1NT". While your choice is appreciated and may have merit, people are interested in the thought process behind this at-first-sight crazy call. Maybe you're looking at the matter from an interesting point of view, who nows, maybe we can learn something. It's obvious that some sort of discussions will start when there are no valable arguments, what else did do you expect? Or do you prefer to be ignored by everyone?

It seems the only argument you've given is that you're allowed to misdescribe your hand completely because it's matchpoints. Matchpoints is still taking percentage actions, and doing so consistently will give you a good score most of the time. Personally I don't see the point in taking a huge risk for little reward. It's like playing in a casino: you'll win occasionally due to luck, but in the long run the house always wins.

[the hog]

2C for me and I would bid this at MPs and at Imps.
I am pleased to say that 1NT would never occur to me at any form of the game.

[whereagles]

 mikeh, on 2011-September-01, 11:04, said:

You csn and in my view should do the sort of estimation I wrote about earlier.

In this case, granting rho at least 5 spades, we have 34 spaces in the other 3 hands into which to deal 8 clubs (and 13 diamonds, 10 hearts and 3 spades). Thus our expectation should be that partner will produce, on average, 8/34 X 13 clubs or 3 clubs. We can estimate that both partner and LHO will have distributions that cluster most closely around 1=4=5=3 shape. This isn't hard to do at the table. Allowing oneself to 'feel' that partner will usually be 5-5 or 4-6 reds is an error.

Well, regardless of the exact % figures, it's just my experience that stuff happens on this sort of hands. I learned to tread warily in such situations. Note that even if pard does have a 1-4-5-3, the known bad breaks make the play get complicated.

I'll try and run a sim later on this week-end. Maybe I'm being too cautious, who knows. But hey, I bid a lot... I think I'm entitled to pass once in a while LOL.

[han]

 mikeh, on 2011-September-01, 09:17, said:

Well, I mistakenly set out that my calculation was based on the assumption that rho held precisely 5 spades when I should have said that it was based on the assumption that he held at least 5 spades. Make that correction, which is appropriate, and there are 34 spaces into which one must fit not only 13 diamonds but 10 hearts, 8 clubs and 3 spades. So I think that your correction of my post resulted in a grammatically more accurate calculation but one that was nevertheless incorrect....mine was incorrectly described but correctly done, I think.

No, in that case your calculation was incorrectly described and incorrectly done.

Consider a bag with 2 red balls and 2 white balls. Suppose that our RHO randomly takes two of the 4 balls out of this bag, and we learn that he must have at least 1 red ball. What is his expected number of white balls?

Using your empty space method, you would get 2/3. After all, he has 1 "empty space", and 2 of the "remaining" 3 balls.

If we number the red balls A and B, and the white balls C and D, then we see that there is 1 way to draw two red balls (A+B) and there are 4 ways to draw a red ball and a white ball (AC, AD, BC, BD). Hence of the 5 possible drawings with at least one red ball, 4 have one white ball and 1 has no white ball, for an average of 4/5 white balls.

Where is your mistake? The answer you get would be correct if you knew that RHO had a specific red ball. For example, if we knew that RHO has ball A, then the expected 2/3 white balls is correct. But we don't know this, we know he has at least 1 red ball, a very different statement.

So what's the exact expected number of diamonds for RHO? I don't know, it is a very difficult computation. To get the right answer we would need to take HCP into consideration as well, which makes the question impossible to answer without knowing RHO's opening tendecies. Rounded to the nearest integer I am pretty sure that the answer is still 3 though. I don't know why this is actually interesting, but when somebody starts posting incorrect mathematics I feel a strong urge to correct them before anybody gets harmed.

[semeai]

To paraphrase for Han, minus the probability lecture:

If you're going to assume opener has five spades, do so. Then the vacant spaces are 31, and the rest of your calculation is basically fine using 31 instead of 34. (Han goes on to note that hcp considerations will matter slightly as well, hence the "basically.")

If you don't want to assume opener has five spades, then you have some more work to do. It's still not too bad (though not something you want to compute at the table), just more complicated, at least ignoring the confounding factor of hcp.

[whereagles]

These theoretical calculations have some subtleties... be sure you get them right because the bias in case of a mix-up is considerable.

[helene_t]

We had a long discussion a couple of years ago about the persistent belief in the idea that holding length in opps' suit increases the chance of a fit. AWM did some nice simulations.

It is indeed a very complicated thing if one wants to get the math right. And there is probably no simple answer.

Other than that for practical purposes it doesn't really matter. Just assume p and LHO each have on average a little more then 1/3 of the remaining cards in your suit. After all, RHO must on average hold more than 1/3 of the outstanding cards in his opening suit.

Btw rounding expected numbers of cards to nearest integer is silly. There is a hell of a difference between an expectation of 2.51 and 3.49. If the average number is not sufficient information then calculate the probability distribution across all possible integers (0..8 in this case).