[Elianna]

helene_t, on Jul 16 2007, 11:11 PM, said:

As for this rule that length in RHO's suit increases that chance of a fit: how strong is that correlation?

This is not the rule that people are claiming is true. The rule that people are claiming is "If I'm long in RHO's suit, there is a higher chance that partner is short in that suit".

There's this worry that if I have winner's/length in RHO's suit, maybe LHO has shortness and will be ruffing. But the hope is that partner (as dummy) will be overruffing.

[awm]

helene_t, on Jul 17 2007, 02:11 AM, said:

As for this rule that length in RHO's suit increases that chance of a fit: how strong is that correlation?

In fact I think it's a negative correlation.

If you give me the number of spades in my hand and RHO's hand, I can compute the probability distribution of partner's number of spades. The number of cards in other suits is not relevant.

So the only way my number of clubs effects partner's number of spades, is that it effects RHO's number of spades. The more clubs I have, the more likely RHO is balanced and has some spades. So more clubs for me implies more spades for RHO and less spades for partner. Note that this is only true because of the restriction that RHO has a 1♣ opening.

I agree with Elianna that the value in holding clubs when overcalling a four-card suit is that it increases the odds that partner has a useful ruffing value to go with his three trumps, not that it in any way makes partner more likely to have three trumps.

[goobers]

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

[joshs]

goobers, on Jul 17 2007, 06:50 PM, said:

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).

[joshs]

awm, on Jul 17 2007, 06:35 PM, said:

helene_t, on Jul 17 2007, 02:11 AM, said:

As for this rule that length in RHO's suit increases that chance of a fit: how strong is that correlation?

In fact I think it's a negative correlation.

If you give me the number of spades in my hand and RHO's hand, I can compute the probability distribution of partner's number of spades. The number of cards in other suits is not relevant.

So the only way my number of clubs effects partner's number of spades, is that it effects RHO's number of spades. The more clubs I have, the more likely RHO is balanced and has some spades. So more clubs for me implies more spades for RHO and less spades for partner. Note that this is only true because of the restriction that RHO has a 1♣ opening.

I agree with Elianna that the value in holding clubs when overcalling a four-card suit is that it increases the odds that partner has a useful ruffing value to go with his three trumps, not that it in any way makes partner more likely to have three trumps.

Its not just that partner will have shortage. Its that partner will have well placed shortage (shortage that can't be overruffed).

If you have 4C and 4S, AND RHO opens 1C, and partner raises there is an increased likelyhood that partner's shortage is Clubs (which probably can't be over ruffed). Furthermore, you had a doubelton in a red suit. Unless partner has 5 or more in that suit, its very unlikely that your LHO can over ruff you. This makes playing a 4-3 fit very effective. So when this is the situation, the opps almost have to lead trumps, and that surrenders the tempo to you, and means that they can't force you.

[goobers]

joshs, on Jul 17 2007, 07:06 PM, said:

goobers, on Jul 17 2007, 06:50 PM, said:

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]

goobers, on Jul 17 2007, 07:20 PM, said:

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.

[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]

SoTired, on Jul 18 2007, 05:47 AM, said:

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]

cherdano, on Jul 18 2007, 07:29 AM, said:

SoTired, on Jul 18 2007, 05:47 AM, said:

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?

[jtfanclub]

Jlall, on Jul 18 2007, 12:22 AM, said:

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]

Jlall, on Jul 18 2007, 08:47 AM, said:

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?

[helene_t]

Jlall, on Jul 18 2007, 03:47 PM, said:

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

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.

[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]

cherdano, on Jul 18 2007, 05:04 PM, said:

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.