overcalling

[goobers]

I don't understand why people are raising with 2 card support

Ok, LHO opens 1H

Partner overcalls 1S

3'rd hand passes

And you have Qx xxx Axxx KJxx

 

Sadly you can not make a responsive x of partners bid.

 

You also have a 10 count with a good fitting honor, but you lack a 5 card suit to bid and lack anything that resembles a stopper for NT. I think a 2S bid would be nearly unananimous in a bidding panel here (you have some extra values to make up for the lack of a 3rd trump).

Backed into a corner here with no real reasonable alternatives; it's just that the impression I got was that some people were routinely raising with 2

[jtfanclub]

Backed into a corner here with no real reasonable alternatives; it's just that the impression I got was that some people were routinely raising with 2

I do.

 

One issue is the LAW...it makes it much tougher for the opponents to bid if they don't know if you have an 8 card fit.

 

In addition, it's often a *really* bad idea to introduce your real suit. For example:

 

1♥ 1♠ P ?

 

xx

xx

QTxx

AJTxx

 

Your choices end up being...

1. Pass now, and then decide later when they limp into 2♥.

2. Bid 1NT, with your xx in their suit.

3. Bid 2 clubs, which even if you play it as nonforcing must show a better hand than this, I would think.

4. Bid 2 spades.

 

For me, this is a routine 2♠ call. You can argue 'no reasonable alternative', but really, you can argue about that about lots of calls. It's a doubleton in partner's suit, it has 'shape', and it does not contain a full or half stop in the opponent's suit. They all look like something like this.

[Guest Jlall]

FWIW I think raising with 2 is nonsense, with Joshs's example just bid 1N, with jtfanclubs just pass.

 

As far as the argument that "it makes it harder for the opponents," well unfortunately you have a partner to. He needs to be able to judge when to compete, whether to bid game, etc.

[SoTired]

1) Jlall said what I said in previous post of this thread: raising with 2-card support is usually wrong even if partner had a 5-card suit. It is just a worse disaster on a 4-2 fit.

 

2) You are only allowed 13 cards in your hand. If you are short in one suit, that means the odds are higher that you are longer in another suit. So if you have 4♣ with opener (playing better minor a 1C opener has 80% chance of 4+♣), that means partner is likely short, meaning there is more room in partner's hand for spades.

[cherdano]

2) You are only allowed 13 cards in your hand. If you are short in one suit, that means the odds are higher that you are longer in another suit. So if you have 4♣ with opener (playing better minor a 1C opener has 80% chance of 4+♣), that means partner is likely short, meaning there is more room in partner's hand for spades.

This argument is brought up so often, but it is still 100% wrong. If you overcall 1S over 1C, then by the same argument length in hearts makes it more likely that partner has more room for spades in his hand.

[SoTired]

2) You are only allowed 13 cards in your hand. If you are short in one suit, that means the odds are higher that you are longer in another suit. So if you have 4♣ with opener (playing better minor a 1C opener has 80% chance of 4+♣), that means partner is likely short, meaning there is more room in partner's hand for spades.

This argument is brought up so often, but it is still 100% wrong. If you overcall 1S over 1C, then by the same argument length in hearts makes it more likely that partner has more room for spades in his hand.

no... if they open 1C and your length in hearts does not affect partner's length in spades. But if you have length in clubs it does. How can you not understand this?

[Guest Jlall]

SoTired you are wrong, I would explain it but I'm sure our math pro cherdano can put it in more concrete terms than I :blink:

[jtfanclub]

He needs to be able to judge when to compete, whether to bid game, etc.

The raising with 2 card support and solid points is a natural extension of Larry Cohen's books on the LAW of total tricks.

 

Judging when to compete is easy. If your partnership never bids 3 over 3 with the raise showing 3 card support, you aren't going to do it with 2 card support. As for when to sacrifice over game, well, you're better off knowing about my HCP by my bidding than you will be placed by my passing.

 

Judging when to bid game when overcaller has 5? The raise gives us a good chance to find 3NT when it's there. Passing means we get to play 1 heart.

 

Judging when to bid game when overcaller has 6? That's where you can lose out. Of course, you can also gain when the 6-2 is enough for game and you'd miss it if you passed, but I'd say overall it's a negative vs. knowing partner has 3.

 

The obstruction value of raising now and not competing later is gigantic, IMHO. It is so much easier for them to find game, correctly judge when to compete to the 3 level, or when to X if you pass now and limp in later.

 

But, well, to each their own. Just whatever you do, discuss it with partner first.

[SoTired]

SoTired you are wrong, I would explain it but I'm sure our math pro cherdano can put it in more concrete terms than I :blink:

So you having 4♣ with opener does not make the odds higher that partner has spade support. Is that right? Then why did Mike Lawrence say club length makes the 4-card overcall more attractive? Is it just the unlikelyhood of a club overruff?

[helene_t]

SoTired you are wrong, I would explain it but I'm sure our math pro cherdano can put it in more concrete terms than I :blink:

Let's try to simplify. Suppose there are only two cards of each suit and two cards in each player's hand. RHO promissed at least one club. You have a club and spade. What's the probability that p has the other spade? You know your own cards and one of RHO's. That leaves 5 cards for 5 slots. The probability that the missing spade is in one of partner's two slots is 2/5 = 40%

 

Now suppose you have a spade and a heart. Now there are 6 cards for 6 slots, that's 6! = 720 permuations but since we don't distinguish between the two slots by the same player, the number of combinations is 720/(2^3) = 90. But the restriction that RHO has at least one club rules out 6 hands he could have, each of which leaves 6 options for dividing the remaining 4 cards between p and LHO, leaving 90-6*6= 54 combinations.

 

Those 54 include 6 combinations with RHO having both clubs, of which 3 gives one spade to p. In 24 cases, RHO has one club and p has the other club, leaving 4 slots for the spade of which p has one, i.e. 24/4=6 combinations. Finally 24 cases with p not having the other club, i.e. p having 2 of the 4 slots for the spade, i.e. 24/2 =12 combinations.

 

In total 3+6+12 = 21 of the 54 combinations give p the spade, that is 38.888888 %.

 

So it seems that having the club actually inproves the chance that p has a spade fit, at least in this particular case. I'm sure someone can get the general picture (at least qualitatively) by plain logical reasoning. I'm getting a little dizzy by this kind of thinking so I'll have to make computer simulations.

[SoTired]

I am happy you simplified it. :)

[Guest Jlall]

Judging when to compete is easy. If your partnership never bids 3 over 3 with the raise showing 3 card support, you aren't going to do it with 2 card support.

I don't know what this means... you NEVER bid 3 over 3 in a competitive auction? I don't see how this can be true so I'm misunderstanding you clearly. My point is that it is much harder for partner to figure out whether to bid 3 over 3 when you have 2-3 trumps as opposed to 3 trumps. He cannot judge accurately.

 

As for when to sacrifice over game, well, you're better off knowing about my HCP by my bidding than you will be placed by my passing.

 

I don't think that knowing that you have a few HCP (something that he can usually infer anyways) matters at all when deciding whether to sacrifice, the degree of your fit is much more important.

 

 

Judging when to bid game when overcaller has 5?  The raise gives us a good chance to find 3NT when it's there.  Passing means we get to play 1 heart. 

 

Except that hands re-evaluate when you have a fit and when you do not. If you have something like a prime 5431 16 count you will usually have a game opposite a fit and some high cards. The hand is not nearly as good opposite no fit and you don't even have 3 level safety. 3N is likely not in the picture when you have a random 7 count and no stopper and partner cannot X the opening bid. However when you have a fit game could easily be in the picture.

[cherdano]

Helene's explanation shows how hard it is to calculate something like this mathematically: Her toy example neglects that RHO will always open 1♣ with two clubs, but some of the time with only one club RHO will open 1N (1100 is a balanced hand in 2-card bridge after all) or 1♦ (and note that I am assuming 2-card majors so he will never open 1M with one club).

 

Helene is certainly right that the odds (for partner having the other spade) are 3/6=50% when RHO has 2 clubs, and 18/48 = 37.5% when he has one club, but it is not at all clear how to weigh these two cases.

 

I agree with Adam's intuition that there is a negative correlation between our club length and partner's spade length, but I wouldn't be sure without a simulation (or a spreadsheet calculation by Frances), and if someone claims to know the answer without even saying anything about the assumed NT range, he/she is certainly wrong.

 

Arend

[helene_t]

Helene is certainly right that the odds (for partner having the other spade) are 3/6=50% when RHO has 2 clubs, and 18/48 = 37.5% when he has one club, but it is not at all clear how to weigh these two cases.

Even the 37.5% may not be correct. Suppose opps' 1-card priorities are HDCS, then if we know that RHO has excactly one club we also know he has the remaining spade. OTOH, if they play SCDH we know that RHO doesn't have the remaining spade.

 

I made the assumption that RHO will always open 1♣ if he has at least one club, no matter how strong he is and no matter which suit his other card is.

[jtfanclub]

I think I can simplify your simplification:

 

8 cards: C1, C2, D1, D2, H1, H2, S1, S2.

 

Opener has C1, you have S1. If you also have C2, then there are 5 cards remaining: D1, D2, H1, H2, and S2. There is a 2/5 chance that opener has the spade. If you also have H1, then there are 5 cards remaining, and there is a 2/5 chance that opener has the spade.

 

But let's change the rules a bit. Give each person 3 cards, but opener must have at least one club, and cannot have two cards in another suit, because he would have opened the suit.

 

12 cards: C1-C3, D1-D3, H1-H3, and S1-S3.

 

You have S1 and S2.

Opener has C1.

 

Without the two card rule, and allowing you to have a third spade, your partner has a 3/8 chance of having S3.

 

So let's look at opener's hand a little closer.

The first card is C1.

The second card could be a club, diamond, heart or spade. There is a 1/9 chance that it's a spade.

If the second card is a club (2/9), then there is a 1/8 chance that the third card is a spade (2/72)

If the second card is a diamond (3/9), then the third card cannot be a diamond, so there is a 1/6 chance that the third card is a spade (1/18)

If the second card is a heart, the third card cannot be a heart, so it comes to 1/18.

 

The total ends up being 4.5/18, or 25%.

 

What if your third card is a club? Let's compute those odds again.

 

Well, there is a 1/8 chance that the second card is a spade.

If the second card is a club (1/8), then there is a 1/7 chance the third card is a spade (1/56)

If the second card is a diamond (3/8), then the third card cannot be a diamond, so there is a 1/5 chance that the third card is a spade (3/40)

The same applies for the hearts (3/40).

 

So it comes out to about 11.5/40, or 29.3% that opener has a spade.

 

And what if your third card is a diamond (same for a heart)? I'll compute the odds a last time.

 

Still a 1/8 chance that the second card is a spade.

If the second card is a club (2/8), then there is a 1/7 chance the third card is a spade (2/56).

If the second card is a heart (3/8), then the third card cannot be a heart, so there is a 1/5 chance chance that it is a spade (3/40).

If the second card is a diamond (2/8), then the third card cannot be a diamond, so there is a 1/6 chance that it is a spade (2/48).

 

Total? 27.7% that opener has a spade.

 

So what does it come down to? If you have more of an opponent's suit, the opponent is more likely to have more of your suit. And, of course, if an opponent has more of your suit, your partner is odds on to have less. So it's less likely that you have a fit if you have more of your opponent's suit.

 

Another way to think of it is...when an opponent opens a club, he (99% of the time) has either a long club suit or is balanced. When you have more clubs, it makes it more likely that your opponent is balanced, and he'll have more of your suit on the average when he's balanced than when he has long clubs.

[jtfanclub]

I'll reply in detail this afternoon- I don't want to misquote. But there was one thing you said I found interesting....

 

If you have something like a prime 5431 16 count you will usually have a game opposite a fit and some high cards. The hand is not nearly as good opposite no fit and you don't even have 3 level safety.

 

So on the auction 1♥-1♠-P, would you bid 2♠ with...

 

♠ 9xx

♥ xx

♦ Axxx

♣ 8xxx

 

I'm extrapolating here, so maybe I just misunderstand where you are coming from. But is it your position that a hand like this should raise in competition, and that it has 3 level safety across a 5431 16 count?

[Guest Jlall]

I'll reply in detail this afternoon- I don't want to misquote. But there was one thing you said I found interesting....

 

If you have something like a prime 5431 16 count you will usually have a game opposite a fit and some high cards. The hand is not nearly as good opposite no fit and you don't even have 3 level safety.

 

So on the auction 1♥-1♠-P, would you bid 2♠ with...

 

♠ 9xx

♥ xx

♦ Axxx

♣ 8xxx

 

I'm extrapolating here, so maybe I just misunderstand where you are coming from. But is it your position that a hand like this should raise in competition, and that it has 3 level safety across a 5431 16 count?

Please do not put words into my mouth. I never said I would raise with this hand on this auction. The fact that you cannot see the difference between the auction 1H 1S p ? and 1S 2C ? does not make it ok for you to put words into my mouth.

 

As far as 3 level safety, you might want to consider that even if you go down at the 3 level the opps will often make something.

 

You might want to also consider that in competitive auctions if 2 people make the most theoretically +EV bids they still may reach a contract that is -EV.

[jtfanclub]

I never said I would raise with this hand on this auction.

And I never said you would. I was careful to point out that I was extrapolating, and that I didn't know if this was your position or not. You did not, however, answer my question.

 

Logically, I figured that if you'd try for game with a 5431 16 count after

 

1♥ 1♠ P 2♠ P

 

you would also try for game with a 5431 16 count after

 

1♠ 2♣ 2♠ P

 

Am I wrong?

 

To me, the 2♠ on the first example auction needs to be wider than on the second auction, at least from a game try perspective. If I pass on the first auction, that may end it.

 

It seems odd to me that you'd criticize the raise with 2 card support as not safe for the 3 level across an invitational hand, while at the same time criticizing somebody else for not raising with 3 card support on a differenct auction, which looks to me to be not safe for the 3 level across an invitational hand.

 

For you, is a 2♠ call on these auctions primarily obstructive, and you don't really have any interest in game unless partner's got an 18 count or a 6 card suit? Or is it primarily constructive, and you want to give partner as much detail as possible in hopes that you have game across a 16-17 hcp hand? Or is one of the 2♠ bids constructive, and the other obstructive?

 

I don't understand where you're coming from.

[Fluffy]

SoTired you are wrong, I would explain it but I'm sure our math pro cherdano can put it in more concrete terms than I :)

So you having 4♣ with opener does not make the odds higher that partner has spade support. Is that right? Then why did Mike Lawrence say club length makes the 4-card overcall more attractive? Is it just the unlikelyhood of a club overruff?

I don't know if Mike said that, actually I am not even sure who Mike is. But if he did the reason is quite simple: You have no better way to enter the auction.

[awm]

Here are some simulation results. The constraints are:

 

East has at most four spades, at most four hearts, at least as many clubs as diamonds. If east has five or more diamonds, then clubs must be longer than diamonds. This gives East effectively a one club opening shape.

 

South has either 4-2-3-4 or 4-4-3-2 distribution (exactly in that order).

 

I measured the frequency of north holding various numbers of spades under these assumptions. Here are the results:

 

1. South has 4-2-3-4. Partner's number of spades:

 

0 spades: 0.94%

1 spades: 7.83%

2 spades: 23.74%

3 spades: 33.01%

4 spades: 23.13%

5 spades: 9.22%

6 spades: 1.90%

 

Expected (average) number of spades is 3.0595

 

2. South has 4-4-3-2. Partner's number of spades:

 

0 spades: 0.88%

1 spades: 7.57%

2 spades: 23.02%

3 spades: 32.49%

4 spades: 24.05%

5 spades: 9.72%

6 spades: 2.01%

 

Expected (average) number of spades is 3.0982

 

This is over a lot of trials. The result:

 

More clubs in my hand after a 1C opening on my right makes partner less likely to have a fit for me when I overcall a four-card spade suit. However, the effect is very slight.

 

There are, of course, advantages in the play because of likely shortage in partner's hand and who's ruffing what, which make 4-3 fits play better when I have a bunch of clubs than when I have few. But my holding a bunch of clubs does not make partner more likely to have spades with me -- in fact slightly the opposite.

[Guest Jlall]

Logically, I figured that if you'd try for game with a 5431 16 count after

 

1♥ 1♠ P 2♠ P

 

you would also try for game with a 5431 16 count after

 

1♠ 2♣ 2♠ P

 

Am I wrong?

 

No. I never said this was not the case. WTF is your point? Why would you bring 2 different auctions into play with the same hand and start discussing it, the other auction has no bearing on the merits of bidding or not bidding on this auction. Are you simply trying to discredit me, because if so you are doing a very poor job. In fact I have no idea how the fact that I would raise with xxx xx Axxx xxxx after 1S 2C 2S has any relevance to me advocating not raising with 2 trumps after 1H 1S p ? lol. This is really getting pretty far out.

 

It seems odd to me that you'd criticize the raise with 2 card support as not safe for the 3 level across an invitational hand, while at the same time criticizing somebody else for not raising with 3 card support on a differenct auction, which looks to me to be not safe for the 3 level across an invitational hand.

 

That is because you are equating 2 non similar auctions from different threads and looking at only one aspect of my argument. Please review my previous post:

 

You might want to also consider that in competitive auctions if 2 people make the most theoretically +EV bids they still may reach a contract that is -EV.

 

It seems odd to me that you'd criticize the raise with 2 card support as not safe for the 3 level across an invitational hand, while at the same time criticizing somebody else for not raising with 3 card support on a differenct auction, which looks to me to be not safe for the 3 level across an invitational hand.

 

OKKKKKKKKKKKKKKKKKKKKKKK thats nice for you. I will try one more time to make a non smartass/condescending reply to this assinine statement and then I am done.

 

1) When you have 2 trumps and a random 7 count then getting to the 3 level in the 5-2 fit is pretty much a disaster. The opponents are much less likely to make anything, you're still likely to go down, and opposite a distributional hand with 5 spades I'd take 3 trumps an ace and a ruffing value over the 2 trump hand type any day. You are underestimating how much the missing trump affects both your defense and your offense.

 

2) You cannot take arguments as a whole and then zone in on one aspect of that argument and compare it to the same aspect from another thread. For instance what if my view was raising with the 3 trump +ace+ruffing value hand type had 20 plus sides and one downside (getting too high when partner makes a game try). And what if my view was raising with 2 trumps has 1 upside and 20 downsides. You could easily focus on the one downside and ignore everything else, but that does not give an accurate representation of the merits of a bid.

 

I have stated already that I think raising with 2 trumps is bad because it will cause partner to misjudge. He may try for game assuming we have a fit and upgrading for that. He may misjudge a competitive auction. We may get to the wrong strain. These are all downsides. Again bringing up some non similar auction with a non similar hand from another thread serves no purpose towards discussing the merits of raising on your example hand that contained 2 trumps and 7 points. The only purpose it might serve would be to discredit me if you have as many holes in your logic as you seem to.

 

I don't understand where you're coming from.

 

If you still do not understand why I think raising with 2 trumps and random hands is a good idea, you never will.

[Guest Jlall]

Here are some simulation results. The constraints are:

 

East has at most four spades, at most four hearts, at least as many clubs as diamonds. If east has five or more diamonds, then clubs must be longer than diamonds. This gives East effectively a one club opening shape.

 

South has either 4-2-3-4 or 4-4-3-2 distribution (exactly in that order).

 

I measured the frequency of north holding various numbers of spades under these assumptions. Here are the results:

 

1. South has 4-2-3-4. Partner's number of spades:

 

0 spades: 0.94%

1 spades: 7.83%

2 spades: 23.74%

3 spades: 33.01%

4 spades: 23.13%

5 spades: 9.22%

6 spades: 1.90%

 

Expected (average) number of spades is 3.0595

 

2. South has 4-4-3-2. Partner's number of spades:

 

0 spades: 0.88%

1 spades: 7.57%

2 spades: 23.02%

3 spades: 32.49%

4 spades: 24.05%

5 spades: 9.72%

6 spades: 2.01%

 

Expected (average) number of spades is 3.0982

 

This is over a lot of trials. The result:

 

More clubs in my hand after a 1C opening on my right makes partner less likely to have a fit for me when I overcall a four-card spade suit. However, the effect is very slight.

 

There are, of course, advantages in the play because of likely shortage in partner's hand and who's ruffing what, which make 4-3 fits play better when I have a bunch of clubs than when I have few. But my holding a bunch of clubs does not make partner more likely to have spades with me -- in fact slightly the opposite.

Interesting, I thought it was probably going to be equal.

[han]

To see that it won't be equal you could take it to an extreme. If you have 4 spades and 9 clubs then you know opener has at most 4 clubs and likely has 2-4 spades (unless he has exactly 1444 shape). But if you have 4 spades and 0 clubs then opener likely has club length and therefore is more likely to have short spades. The non-spade cards in our hand are the same so don't influence the likelyhood for partner to have spades, but the more spades opener has the less likely it is that partner has spade length.

 

So the more clubs we have, the fewer support we should expect from partner, just like Adam has said all along (and I suspect for the same reason). I didn't read all the post in this thread in great detail. Apologies if someone has given a similar argument before.

[jtfanclub]

Logically, I figured that if you'd try for game with a 5431 16 count after

 

1♥ 1♠ P 2♠ P

 

you would also try for game with a 5431 16 count after

 

1♠ 2♣ 2♠ P

 

Am I wrong?

 

No. I never said this was not the case. WTF is your point?

My point is that I see both auctions as primarily obstructive. The object here is to make the opponents make the last guess. the objective is not to find find game, unless partner has an extreme hand.

 

My point is that I think that a 3+244 4 count with one ace and a 2+254 well supported 7 count with one ace are about equal in offense. If you don't think that's true, then up the point value to where they're equal, there's nothing magical about 7. While the auctions are not the same, partner's going to invite with the same hand in both cases. I think that if anything, partner should expect a weaker hand from 1♥ 1♠ P 2♠ because the bid may be the best of a bad set of choices, while after 1♠ 2♣ 2♠ you had both pass and X available to you.

 

My point was that if you can see the parallel strengths of the hands, and the parallel objectives in bidding the two hands, then hopefully you can see where I'm coming from. I obviously was not successful.

 

On the auction 1♥ 1♠ P, would you bid 2♠ with 9xx xx Axxx 8xxx?

 

If the answer is yes, then the parallels between the two bids should be obvious. If the answer is no, then the parallels don't immediately apply, but it doesn't make a lot of sense to me.

 

That is because you are equating 2 non similar auctions from different threads and looking at only one aspect of my argument.

 

That the best bid combined with the best bid can get you to a bad place? I thought that was self-evident.

 

1) When you have 2 trumps and a random 7 count then getting to the 3 level in the 5-2 fit is pretty much a disaster.

 

Cohen's rule is 'eight never, nine ever'. I was going to grab more quotes to explain where I was coming from, but I never got the chance.

 

I agree that getting to the three level in a 5-2 to fit would be a disaster. However, if you follow Cohen's rule, the original overcaller will never bid 3♠ without 6 cards in the suit. If the raise can be two cards, then it's extra sure that the original overcaller will never bid 3♠ without 6 cards in the suit. So you may end up in a 6-2, which violates Cohen's rule, but never a 5-2.

 

Take the 5431 16 count across the 2254 7 count. If the singleton is in opener's suit, then to try for game the overcaller bids 3 of his longer minor, and now you're playing in a 4-4 or a 5-4 fit. If the overcaller instead has the singleton in a minor and the opponent's suit well stopped, he bids 2NT, and with a 23 count and a fit in the minors that's hardly a horror. If the overcaller doesn't have shortness in opener's suit but doesn't have the stoppers for NT, he can pass or bid the minor. Game tries, in other words, are easily handled without any real loss. What else were you using those bids for?

 

2) You cannot take arguments as a whole and then zone in on one aspect of that argument and compare it to the same aspect from another thread.

 

The others aspects I think I can argue through pretty easily. This is the only negative aspect I'm worried about.

 

I have stated already that I think raising with 2 trumps is bad because it will cause partner to misjudge. He may try for game assuming we have a fit and upgrading for that. He may misjudge a competitive auction. We may get to the wrong strain. These are all downsides.

 

Let's start with the idea that you're not fooling partner. Partner's going to assume you have three cards, but he's aware that you might have 2 and will factor that into the auction when it doesn't cost anything. That you've discussed this possibility before the game. Surprising partner is a bad thing regardless.

 

A) Trying for game: I don't think that partner trying for game assuming a fit is a problem as long as the minimum for two card support is signifanctly stronger than the minimum for 3 card support. If partner thinks we have 21+ hcp and a 5-3 fit, I don't think it'll be a disaster when we turn up with 24+ hcp and a 5-2 fit. Obviously, partner can't just bid 3♠ with 5, but there are far cheaper ways to try for game.

 

B) Misjudging a competitive auction: I think this is far more likely to happen to the opponents. Overcaller has already shown 5, there is no reason for him to show it again. If you follow the eight never principle, overcaller shouldn't be tempted to compete further with 5. If partner things we're 6-3 and we're actually 6-2, this can be an issue, sure. But the opponents don't even have that much information. If opener has 6, can he be sure partner has 2? If responder has 3, can he be sure partner has 6? 6-2 is not a bad fit.

 

C) Wrong strain: I think it's actually tougher to get to another strain without the raise. Obviously, if it gets passed out, you're not finding another strain. After the raise, you can afford to show an alternate strain if you also have interest in game. After 1♥ 1♠ P P 2♥, can you even find 3NT when it's right? I doubt I could.

 

If you still do not understand why I think raising with 2 trumps and random hands is a good idea, you never will.

 

Yeah, I get it. It's the same old argument used for everything from not opening a weak 2 without 2 of the top 3 honors to 1♦-1♠-2♠ with 3 card support to overcalling with 4 card suits to your call. How many times can you name where you've gotten that same argument from somebody else?

 

But if you start thinking about the LOTT and eight never, you can see why raising with 2 card support with certain hands fits with those rules and doesn't cause major problems. These hands are not random- that's why you don't do it with 2344 shape, or with the same minimum strength that you'd bid it with 3 card support.

[awm]

Bidding needs to be based on offense-to-defense ratio (ODR) and not just offensive value. While it's true that a hand with three-card support and lousy values may be equivalent offensively to a hand with two-card support and an extra ace, the second hand is much better defensively than the first.

 

Take the following three hands:

 

Jxx xx Axxx xxxx

 

Jx xxx AKxx xxxx

 

Jxx xx AKxx xxxx

 

It's certainly true that the first two hands will be worth close to the same number of tricks in a spade contract (the extra diamond king in the second hand compensates for the missing trump). The third hand is worth about one trick more than either of the first two. But the issue is, the first hand is much worse defensively than the second and third. If the auction continues with a competitive bid by opponents and partner has to decide whether to bid 3♠, he will be right to bid on either the first hand (3♠ likely down one, but their three level contract probably making) or the third hand (3♠ probably makes, their three level contract probably down one). He will be wrong to bid on the second hand (3♠ and their contract both probably down).

 

There is an additional issue when partner has "game try" values. On the first hand, it's generally right to reject a game try, and on the third hand it's generally right to accept. Easy enough. But the second hand is a pain -- if partner is bidding simply on power than you can probably make 3NT (but 4♠ can easily be hopeless). If partner is bidding on shape then 3♠ could easily be the limit (partner's shape not worth as much opposite only two trumps).