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My simulations strongly suggest that “just bid 3N” is very wrong at matchpoints with this responding hand.

I did a series of 5000 deal simulations.

Over all possible hands for opener (15-17 balanced), 3NT is right around 50%. Whether it is slightly above 50% or slightly below 50% depends on secondary factors such as whether one opens 1NT with all balanced hands containing a 5 card major (in which case 3NT made 51% of the time) or chooses 1M with some of these.

My “simplified” criteria for whether to open 1NT with 5332 hands containing a 5 card major are: * with 15, open 1NT * with 17, open 1M * with 16, open 1S with 5 ♠s but 1NT with 5 ♥s.

Obviously, in real life, this decision is more complex, but the above gives a reasonable approximation (for simulation purposes) of what I do without having to specify very complex criteria.

Anyway, with *those* assumptions, 3NT made 48.2% overall. Intuitively, the drop is due to 15 HCP hands becoming relatively more frequent since all the 15s with a 5 card major are included, while all the 17s and some of the 16s are removed.

Refining the simulations (now sticking with including some but not all hands with 5 card majors as described above), I found the following: Opener has 15 HCPs: 3NT made 35% Opener has 16-17 HCPs: 3NT made 61.2%

I use the 15 vs. 16/17 as an approximate surrogate for whether opener would accept a NT game invite or not. Again, real world criteria are more complex (if you have good hand evaluation skills) as some 15 HCP hands are as good as average 16s, while some 16s are as bad as average 15s. Still, statisitically, for simulation purposes, I find the 15 vs. 16/17 to give a good approximation for “accept” vs. “decline” decision after a NT game invite.

The relative frequencies of 15 vs. 16/17 openers opposite the given responding hand was: 15: 47.3%; 16/17: 52.7% (again, if you opened 1N on all balanced 15-17 hands with a 5 card major, the percentage for “15” would be somewhat lower).

Anyway, from the above data it should be obvious that inviting with 2N wins big AT MATCHPOINTS vs. blasting 3N.

When opener has 16/17, it doesn't matter which strategy you choose, because you will play 3NT either way.

But when opener has *15*, he will decline the invite. In those cases, 2NT beats 3NT 65% to 35%. Combining that result with the relative frequencies, shows that overall, inviting (2NT) beats blasting (3NT) by: 0.473 * 65% = 0.527 * 50% = 57.1% at matchpoints. That is a considerable win for the invite.

Since OP problem specified matchpoints, I will not include here the IMP analysis (which I have done), except to say the results are quite different (particularly when VUL).

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I picked 2♦ like almost everyone else.

But 2NT is IMO a reasonable enough alternative to merit significantly more than the 1 vote out of 40 that it has garnered so far.

There is a *lot* of ambiguity on this auction about East's strength and shape. 10-18 HCPs. He might even have only four ♦s and five ♣s (or the other way around, or 5=5, or maybe he rebid 2♣ on a 3-bagger with good 3 card ♠ support and extra values).

Because of all that ambiguity, “stretching” slightly for a 2NT rebid with this hand type can be the winning choice.

The “stretch” on this OP hand with 9 HCPs, two working 10 spots and that big ♠9 is not huge, I think.

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I'll have more comments later after I've finished some simulations.

But I must say I looked up this deal as I was interested in what sort of hand South might have held for his (to me) rather unusual vulnerable actions of first passing in 3rd chair and later volunteering 2♥ over a (usually game invitational or better) Stayman 2♣ from his RHO.

It turns out that South's actual hand was: ♠ 97 ♥ AQJT95 ♦ 97 ♣ 864

It seems to me that perhaps many players would have methods permitting some sort of 3rd seat opening with this hand, no?

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If you prefer a game in which some decisions are significantly more important than other (as I do), then IMP pairs is probably your choice.

If think every correct decision should be rewarded equally and every mistake equally expensive, then matchpoints (or, better, BAM) would be your preference.

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Wish you would have included question about opener's holding in responder's (major) suit.

Traditionally, I believe the 3NT bid was supposed to have a singleton in responder's suit. I believe I have seen Eric Kokish in print advocate this as being quite important (virtually a requirement).

It seems that many players are (much) less strict about this, though.

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The most likely shape for 2♦ overcall followed by later TO double of (4♠) is 1=3=6=3 with maximum overcall values. Or, possibly, 0=3=6=4 .

Therefore, on this auction, I do not think 5♣ from south is more reasonable than 5♦, although you might argue for 4NT. Actually, among those three, I think I prefer 5♦, 4NT second, 5♣ last.

Of course, best choice by that point of the auction would be to pass (4♠X) and “take the money” since any game is far from certain.

Much better for North to start with 2NT to show is two strong minors. Maybe he will later get a chance to double (♠s) to suggest his 0=3=5=5 shape.

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Since this problem is largely about tactics and perhaps judging what your specific opponents are likely to do under pressure, I won't pretend that computer simulation can be of much value here.

Even generating random deals and looking at them “by hand” to try to determine what might happen if you open 1♦ vs. 5♦ is not very productive because on many deals it is very hard to say what the other three players will do (lots of shape around the table).

However, I thought it might be of some interest just to ask, statistically speaking, how many tricks North is likely to make playing in ♦s over a large number (I chose 5000) of random deals.

Here are the double-dummy results for tricks makeable in ♦s by North over 5,000 random deals with given North hand:

13 tricks: 295 (6% cumulative) 12 tricks: 989 (26% cumulative for 12+ tricks) 11 tricks: 1476 (55% cumulative for 11+ tricks) 10 tricks: 1259 (80% cumulative for 10+ tricks) 9 tricks: 710 (95% cumulative for 9+ tricks) 8 tricks: 266 (99.9% cumulative for 8+ tricks) 7 tricks: 5 (100% cumulative for 7+ tricks)

Those numbers make the 5♦ opening look pretty appealing since most often that is actually the best contract for our side, and it puts maximum pressure on the opponents.

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Your objections to OP's exact phrasing seem valid.

However, let's try to look past his perhaps less than ideal wording to the essential question:

When, after the initial reply to a keycard ask and a subsequent king showing reply to asker's “specific king(s) ask” follow-up ask, asker then continues with a suit bid below 6 of the agreed trump suit, what is the usual meaning of that bid?

If that is the question, I believe the *systemic* answer is that the bid in question is asking about the KING in the bid suit. This of course presumes that it is possible for replier to hold this king.

One could then debate what replier is supposed to do when he holds that king. Does he bid the grand in the agreed trump suit? I think that is the common, perhaps “traditional” style. It presumes that the specific king ask guarantees that the partnership holds all the keycards and the trump queen.

But, it has been pointed out that little harm is done if replier instead replies one step above 6 of the trump suit. This might ocassionally allow asker to try for 6NT with a missing key-card if possession of the king in question will suffice for 12 tricks in NT.

Another important question is “what should replier do if he doesn't have the king of the suit bid by asker but he *does* have another higher ranking king that could be shown below 6 of the trump suit?”

For example, suppose our suit is ♠s, and (after 4NT and reply), the reply to the 5NT specific king ask is 6♣ (showing the ♣ king). Now asker bids 6♦. Presumably, if replier *has* the ♦K, he bids above 6♠ (say 6NT). But what if he lacks the ♦K but has the ♥K?? In this case, he should bid 6♥ to show the ♥K in case that is enough for asker to commit to a contract higher than 6♠.

What should replier do in response to a specific king ask if he has *all three* side suit kings? I believe that cue-bidding a king (his cheapest) is supposed to show that king and *at most one higher* if that is possible. If he has all 3 side suit kings, he should probably reply with 6NT.

There has been other discussion about some non-standard agreements. One was that the reply should show *either* just the king in the bid suit *or* specifically the *other two* side suit kings. Maybe that agreement could work, but I do not believe it is standard and I think it could certainly sometimes suffer from the ambiguity.

The other issue was whether replier, lacking the requested king should ever reply higher than 6 of the trump suit because he believes his hand has some other feature(s) that will produce the needed 13th trick.

Sounds risky to me. We are in the middle of a conventional sequence (key card ask and specific king follow-up asks) with a well-defined protocol.

If partner is asking for a specific king, and you go ahead and bid a grand (or 6NT) without it, partner will probably forgive you if the resultant final contract makes. But if it doesn't, it will definitely be your fault.

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I agree that this would be a 1NT opener if playing weak NT openings.

But that doesn't mean it should rebid 2NT over a GF 2♣.

First, stoppers in unbid suits become more critical when choosing to rebid NT (especially at the 2 level, but applies to some extent even for 1NT rebids which is why opener often raises responder's 1M with only 3 card support).

When a balanced hand is in your 1NT (or 2NT) opening range, you pretty much (usually) have to choose that opening even with a small doubleton or unstopped suit somewhere because your system will (or might) bite you if you try to open something else.

So rebidding 2♦ after 1♦-2♣(GF) does not necessarily suggest a 6+ card suit and a one-suited hand, any more than rebidding 2M after 1M-2X suggests a 6 card suit and an unbalanced hand.

Rather, it is an efficient (space saving) rebid that shows a 5th ♦ and leaves room for further exploration.

Actually, in some styles, opener sometimes rebids 2♦ after 1♦-2♣ with only a *4* card suit in order to avoid mis-representing his stoppers and/or wasting space.

A 2NT rebid should at the very least promise stoppers in both majors.

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Quite astonished at the amount of support for 2NT here, particularly as this hand has a very good 5 card ♦ suit and an unstopped major.

In 2/1 auctions that start with a 1M openings, many (most?) play the style where the 2M rebid is the catch-all, i.e. not promising a 6 card suit. I know there are some modern “shift” rebid alternatives to this as well as older “Bergen” style of using 2NT as the catchall, but I think the 2M “catch-all” rebid is (still) quite mainstream.

Here, after 1♦-2♣ (GF), I would think the arguments in favor of (usually) rebidding 2♦ with *5* would be even stronger because (as constrasted with 1M-2x-2M) the original 1♦ opening has not promised 5 (or even 4 in most styles) ♦s.

So, by (usually) rebidding 2♦ with 5, we not only benefit from saving space (the main argument for 2M rebid as “catch-all” in 1M-2x auctions), but we are also giving partner some genuinely new information (that we hold 5+ ♦s).

After 2♦ rebid, there is plenty of room to discover a 4-4 majors suit fit and/or to explore stoppers for NT (and, hopefully, get 3NT played from the better side when it matters). And once in awhile, the knowledge that opener has 5 ♦s (especially when good ones as in OP), might be what responder needs to know to get to a superior ♦ (game or) slam.

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You seem to be arguing that this is not a forcing pass situation.

I've noticed that “forcing pass” does not seem to be enjoying its former popularity of late, but I think positing that FP does not apply on *this* auction is a very deep position! We have an opening bid with a splinter raise opposite a GF 2/1 response and a voluntarily bid slam. I think most would consider that “enough” for FP to be “on” at the 6 level!

I see no reason to presume from the auction that partner is VOID in ♥s. I mean it is *posible*, but I would say definitely not probable, much less certain.

If we pass (6♥), the normal meaning of that call is that we have first round ♥ control. Suppose partner has some hand like: Ax-x-AJxxx-KQJxx He might well think our “pass” justified his bidding 7♦ (perhaps he'd place us with x-Ax-KQxxx-Axxxx).

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I beg your pardon? “Awful hand to open” ??

I think not. I have quite high standards for my opening bids, yet consider this hand a clear opener.

Why not? It has: * 12 HCPs * 2 quick tricks * a *good* 5 card suit headed by AQJ * no particularly weak minor honors (J(x), Q(x), stiff K) * no rebid problems. * satisfies “rule of 20” and “rule of 22”.

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I didn't say double invites a grand. I said double denies 1st round ♥ control. Pass would promise 1st round ♥ control. Partner is captain of this hand. I've opened ♦s and made a splinter raise in support of ♣s (for better or worse). Now I tell whether I've got 1st round ♥ control or not. Partner does whatever he wants.

I suppose if I had 1st round ♥ control in some *Great* hand, I could take it upon myself to bid a grand myself.

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Rafael, Yes, if we are just discussing the apriori (i.e. no special inferences available from bidding and/or play to change the odds), then the probabilities for *ANY TWO CARDS* whatsoever being in the same opponent's hand vs. one in each opponent's hand are the same. It doesn't matter whether the two cards of interest are in the same suit or different suits. If in the same suit, it doesn't matter how many cards our side holds in that suit.

The calculation is simple (it can be done in other, more complicated ways, but the below is correct and easy):

1. Conceptually “place” one of the two cards (say ♠K in your example) in one opponent's hand– let's say your LHO's hand.

2. Now, consider the relative liklihood of the *other* card of interest (say the ♠Q in your example) being in LHO's hand (remember, we're assuming he already has the ♠K) vs. in your RHO's hand.

Since LHO has the ♠K, he has only 12 empty “slots” available which might receive the ♠Q.

But RHO, with no cards already placed, has *13* empty slots available which might receive the ♠Q.

Therefore, the odds are 13 to 12 (or 52 to 48) that RHO will have the other card (♠Q) rather than LHO (who already, by assumption) has the ♠K.

As you can see, the above analysis is very general and would apply equally to any two cards of interest.

BTW, this type of reasoning (known as “vacant spaces” analysis) can be generalized to handle situations where something is already known about the opponents' hands.

For example, suppose in your situation ♥s were trump and your side has 9 of them (♥s). Suppose you have already drawn trump and found that your LHO started with three ♥s while your RHO had only one.

Well, because LHO started with three ♥S, he has only *10* “vacant spaces” available for other cards (including the ♠K and ♠Q). RHO, on the other hand, having started with only one ♥ has *12* “vacant spaces” remaining.

So now, we place the first card of interest, say the ♠K. The chances LHO has it are 10 out of 22 because there are 22 “vacant spaces” left total between LHO and RHO and LHO has only 10 of them.

Given LHO *has* the ♠K, now he has only *9* vacant spaces left out of 21 remaining (RHO has the other 12 still). So the probability that LHO also has the ♠Q (given that he has the ♠K) is only 9 out of 21.

Therefore, the overcall probability that LHO (with three ♥s) has both the ♠K and ♠Q is (10 / 22) * (9 / 21) = 90 / 462 ~ 19.5%

Doing a similar calculation for *RHO* who starts with 12 “vacant spaces” since he had only one ♥, we compute the probability that *he* has both the ♠K and ♠Q as: (12 / 22) * (11 / 21) = 132 / 462 ~ 28.6%

Notice how LHO's having more ♥s than RHO has shifted the probability of either player having both ♠ honors from equal (24% each) to unequal (19.5% vs. 28.6%) with the opponent with fewer ♥s becoming more likely to have both ♠s.

To compute the probability (with ♥s 1=3) of the ♠ honors being split, we could just subtract the sum of the above from 100% giving: split ♠ honor probability = 1 - ((132 + 90) / 462) = (462 - 222)/462 = 240 / 462 =~ 51.9%

Notice this hasn't changed much from the apriori 52% when we have no knowledge of the ♥ distribution (it is not exactly the same, though).

Craig Zastera

I did a series of 5000 deal simulations.

Over all possible hands for opener (15-17 balanced), 3NT is right around 50%.

Whether it is slightly above 50% or slightly below 50% depends on secondary factors such as whether one opens 1NT with all balanced hands containing a 5 card major (in which case 3NT made 51% of the time) or chooses 1M with some of these.

My “simplified” criteria for whether to open 1NT with 5332 hands containing a 5 card major are:

* with 15, open 1NT

* with 17, open 1M

* with 16, open 1S with 5 ♠s but 1NT with 5 ♥s.

Obviously, in real life, this decision is more complex, but the above gives a reasonable approximation (for simulation purposes) of what I do without having to specify very complex criteria.

Anyway, with *those* assumptions, 3NT made 48.2% overall.

Intuitively, the drop is due to 15 HCP hands becoming relatively more frequent since all the 15s with a 5 card major are included, while all the 17s and some of the 16s are removed.

Refining the simulations (now sticking with including some but not all hands with 5 card majors as described above), I found the following:

Opener has 15 HCPs: 3NT made 35%

Opener has 16-17 HCPs: 3NT made 61.2%

I use the 15 vs. 16/17 as an approximate surrogate for whether opener would accept a NT game invite or not.

Again, real world criteria are more complex (if you have good hand evaluation skills) as some 15 HCP hands are as good as average 16s, while some 16s are as bad as average 15s.

Still, statisitically, for simulation purposes, I find the 15 vs. 16/17 to give a good approximation for “accept” vs. “decline” decision after a NT game invite.

The relative frequencies of 15 vs. 16/17 openers opposite the given responding hand was:

15: 47.3%; 16/17: 52.7%

(again, if you opened 1N on all balanced 15-17 hands with a 5 card major, the percentage for “15” would be somewhat lower).

Anyway, from the above data it should be obvious that inviting with 2N wins big AT MATCHPOINTS vs. blasting 3N.

When opener has 16/17, it doesn't matter which strategy you choose, because you will play 3NT either way.

But when opener has *15*, he will decline the invite.

In those cases, 2NT beats 3NT 65% to 35%.

Combining that result with the relative frequencies, shows that overall, inviting (2NT) beats blasting (3NT) by:

0.473 * 65% = 0.527 * 50% = 57.1%

at matchpoints. That is a considerable win for the invite.

Since OP problem specified matchpoints, I will not include here the IMP analysis (which I have done), except to say the results are quite different (particularly when VUL).

Craig Zastera

But 2NT is IMO a reasonable enough alternative to merit significantly more than the 1 vote out of 40 that it has garnered so far.

There is a *lot* of ambiguity on this auction about East's strength and shape. 10-18 HCPs. He might even have only four ♦s and five ♣s (or the other way around, or 5=5, or maybe he rebid 2♣ on a 3-bagger with good 3 card ♠ support and extra values).

Because of all that ambiguity, “stretching” slightly for a 2NT rebid with this hand type can be the winning choice.

The “stretch” on this OP hand with 9 HCPs, two working 10 spots and that big ♠9 is not huge, I think.

Craig Zastera

But I must say I looked up this deal as I was interested in what sort of hand South might have held for his (to me) rather unusual vulnerable actions of first passing in 3rd chair and later volunteering 2♥ over a (usually game invitational or better) Stayman 2♣ from his RHO.

It turns out that South's actual hand was:

♠ 97 ♥ AQJT95 ♦ 97 ♣ 864

It seems to me that perhaps many players would have methods permitting some sort of 3rd seat opening with this hand, no?

Craig Zastera

If think every correct decision should be rewarded equally and every mistake equally expensive, then matchpoints (or, better, BAM) would be your preference.

Craig Zastera

Traditionally, I believe the 3NT bid was supposed to have a singleton in responder's suit. I believe I have seen Eric Kokish in print advocate this as being quite important (virtually a requirement).

It seems that many players are (much) less strict about this, though.

Craig Zastera

Therefore, on this auction, I do not think 5♣ from south is more reasonable than 5♦, although you might argue for 4NT.

Actually, among those three, I think I prefer 5♦, 4NT second, 5♣ last.

Of course, best choice by that point of the auction would be to pass (4♠X) and “take the money” since any game is far from certain.

Much better for North to start with 2NT to show is two strong minors. Maybe he will later get a chance to double (♠s) to suggest his 0=3=5=5 shape.

Craig Zastera

Even generating random deals and looking at them “by hand” to try to determine what might happen if you open 1♦ vs. 5♦ is not very productive because on many deals it is very hard to say what the other three players will do (lots of shape around the table).

However, I thought it might be of some interest just to ask, statistically speaking, how many tricks North is likely to make playing in ♦s over a large number (I chose 5000) of random deals.

Here are the double-dummy results for tricks makeable in ♦s by North over 5,000 random deals with given North hand:

13 tricks: 295 (6% cumulative)

12 tricks: 989 (26% cumulative for 12+ tricks)

11 tricks: 1476 (55% cumulative for 11+ tricks)

10 tricks: 1259 (80% cumulative for 10+ tricks)

9 tricks: 710 (95% cumulative for 9+ tricks)

8 tricks: 266 (99.9% cumulative for 8+ tricks)

7 tricks: 5 (100% cumulative for 7+ tricks)

Those numbers make the 5♦ opening look pretty appealing since most often that is actually the best contract for our side, and it puts maximum pressure on the opponents.

Craig Zastera

Craig Zastera

My view is that with 2-suiters it is usually better to show them both at once if you have a bid that does that.

Exception would be when the higher suit is longer because 2-suited bids usually suggest equal length suits or perhaps one card longer in lower suit.

South's 5♦ on the OP auction rather than just passing (4♠X) is pretty bad also, especially at IMPs.

Craig Zastera

However, let's try to look past his perhaps less than ideal wording to the essential question:

When, after the initial reply to a keycard ask and

a subsequent king showing reply to asker's

“specific king(s) ask” follow-up ask, asker then continues with

a suit bid below 6 of the agreed trump suit, what is

the usual meaning of that bid?

If that is the question, I believe the *systemic* answer is that the bid in question is asking about the KING in the bid suit. This of course presumes that it is possible for replier to hold this king.

One could then debate what replier is supposed to do when he holds that king. Does he bid the grand in the agreed trump suit?

I think that is the common, perhaps “traditional” style.

It presumes that the specific king ask guarantees that the partnership holds all the keycards and the trump queen.

But, it has been pointed out that little harm is done if replier instead replies one step above 6 of the trump suit. This might ocassionally allow asker to try for 6NT with a missing key-card if possession of the king in question will suffice for 12 tricks in NT.

Another important question is “what should replier do if he doesn't have the king of the suit bid by asker but he *does* have another higher ranking king that could be shown below 6 of the trump suit?”

For example, suppose our suit is ♠s, and (after 4NT and reply), the reply to the 5NT specific king ask is 6♣ (showing the ♣ king).

Now asker bids 6♦. Presumably, if replier *has* the ♦K, he bids above 6♠ (say 6NT).

But what if he lacks the ♦K but has the ♥K??

In this case, he should bid 6♥ to show the ♥K in case that is enough for asker to commit to a contract higher than 6♠.

What should replier do in response to a specific king ask if he has *all three* side suit kings?

I believe that cue-bidding a king (his cheapest) is supposed to show that king and *at most one higher* if that is possible.

If he has all 3 side suit kings, he should probably reply with 6NT.

There has been other discussion about some non-standard agreements. One was that the reply should show *either* just the king in the bid suit *or* specifically the *other two* side suit kings. Maybe that agreement could work, but I do not believe it is standard and I think it could certainly sometimes suffer from the ambiguity.

The other issue was whether replier, lacking the requested king should ever reply higher than 6 of the trump suit because he believes his hand has some other feature(s) that will produce the needed 13th trick.

Sounds risky to me. We are in the middle of a conventional sequence (key card ask and specific king follow-up asks) with a well-defined protocol.

If partner is asking for a specific king, and you go ahead and bid a grand (or 6NT) without it, partner will probably forgive you if the resultant final contract makes.

But if it doesn't, it will definitely be your fault.

Craig Zastera

But that doesn't mean it should rebid 2NT over a GF 2♣.

First, stoppers in unbid suits become more critical when choosing to rebid NT (especially at the 2 level, but applies to some extent even for 1NT rebids which is why opener often raises responder's 1M with only 3 card support).

When a balanced hand is in your 1NT (or 2NT) opening range, you pretty much (usually) have to choose that opening even with a small doubleton or unstopped suit somewhere because your system will (or might) bite you if you try to open something else.

So rebidding 2♦ after 1♦-2♣(GF) does not necessarily suggest a 6+ card suit and a one-suited hand, any more than rebidding 2M after 1M-2X suggests a 6 card suit and an unbalanced hand.

Rather, it is an efficient (space saving) rebid that shows a 5th ♦ and leaves room for further exploration.

Actually, in some styles, opener sometimes rebids 2♦ after 1♦-2♣ with only a *4* card suit in order to avoid mis-representing his stoppers and/or wasting space.

A 2NT rebid should at the very least promise stoppers in both majors.

Craig Zastera

In 2/1 auctions that start with a 1M openings, many (most?) play the style where the 2M rebid is the catch-all, i.e. not promising a 6 card suit. I know there are some modern “shift” rebid alternatives to this as well as older “Bergen” style of using 2NT as the catchall, but I think the 2M “catch-all” rebid is (still) quite mainstream.

Here, after 1♦-2♣ (GF), I would think the arguments in favor of (usually) rebidding 2♦ with *5* would be even stronger because (as constrasted with 1M-2x-2M) the original 1♦ opening has not promised 5 (or even 4 in most styles) ♦s.

So, by (usually) rebidding 2♦ with 5, we not only benefit from saving space (the main argument for 2M rebid as “catch-all” in 1M-2x auctions), but we are also giving partner some genuinely new information (that we hold 5+ ♦s).

After 2♦ rebid, there is plenty of room to discover a 4-4 majors suit fit and/or to explore stoppers for NT (and, hopefully, get 3NT played from the better side when it matters).

And once in awhile, the knowledge that opener has 5 ♦s (especially when good ones as in OP), might be what responder needs to know to get to a superior ♦ (game or) slam.

Craig Zastera

For example, suppose opener were 4=4=3=2 (perhaps also 4=4=4=1) with not unusual honor dispersion.

What to rebid for those cases after 1♦-2♣ ?

For me, those would be 2NT rebids.

Craig Zastera

Craig Zastera

I've noticed that “forcing pass” does not seem to be enjoying its former popularity of late, but I think positing that FP does not apply on *this* auction is a very deep position! We have an opening bid with a splinter raise opposite a GF 2/1 response and a voluntarily bid slam.

I think most would consider that “enough” for FP to be “on” at the 6 level!

I see no reason to presume from the auction that partner is VOID in ♥s. I mean it is *posible*, but I would say definitely not probable, much less certain.

If we pass (6♥), the normal meaning of that call is that we have first round ♥ control.

Suppose partner has some hand like:

Ax-x-AJxxx-KQJxx

He might well think our “pass” justified his bidding 7♦

(perhaps he'd place us with x-Ax-KQxxx-Axxxx).

Craig Zastera

I think not. I have quite high standards for my opening bids, yet consider this hand a clear opener.

Why not?

It has:

* 12 HCPs

* 2 quick tricks

* a *good* 5 card suit headed by AQJ

* no particularly weak minor honors (J(x), Q(x), stiff K)

* no rebid problems.

* satisfies “rule of 20” and “rule of 22”.

Craig Zastera

I said double denies 1st round ♥ control.

Pass would promise 1st round ♥ control.

Partner is captain of this hand.

I've opened ♦s and made a splinter raise in support of ♣s (for better or worse). Now I tell whether I've got 1st round ♥ control or not. Partner does whatever he wants.

I suppose if I had 1st round ♥ control in some *Great* hand, I could take it upon myself to bid a grand myself.

Craig Zastera

While I'm not 100% opposed to rebidding 2NT with 5 ♦s, I do not think this hand is “special” enough to choose that somewhat non-standard action.

If you had KJx in *both* majors or even KJx-QJx, then I think 2NT would have more appeal.

But here, you have good ♦s and not even a real stopper in ♥s, hence no reason to deviate from “normal” 2♦ rebid.

Craig Zastera

It just says that I do not have 1st round ♥ control.

Period. Not a penalty double.

In fact, having opened ♦s and splintered in support of ♣s, partner might be encouraged to learn I don't have the (possibly wasted) ♥A.

Craig Zastera

Yes, if we are just discussing the apriori (i.e. no special inferences available from bidding and/or play to change the odds), then the probabilities for *ANY TWO CARDS* whatsoever being in the same opponent's hand vs. one in each opponent's hand are the same. It doesn't matter whether the two cards of interest are in the same suit or different suits.

If in the same suit, it doesn't matter how many cards our side holds in that suit.

The calculation is simple (it can be done in other, more complicated ways, but the below is correct and easy):

1. Conceptually “place” one of the two cards

(say ♠K in your example) in one opponent's hand–

let's say your LHO's hand.

2. Now, consider the relative liklihood of the *other*

card of interest (say the ♠Q in your example) being

in LHO's hand (remember, we're assuming he already

has the ♠K) vs. in your RHO's hand.

Since LHO has the ♠K, he has only 12 empty “slots”

available which might receive the ♠Q.

But RHO, with no cards already placed, has *13* empty

slots available which might receive the ♠Q.

Therefore, the odds are 13 to 12 (or 52 to 48) that

RHO will have the other card (♠Q) rather than LHO

(who already, by assumption) has the ♠K.

As you can see, the above analysis is very general and would apply equally to any two cards of interest.

BTW, this type of reasoning (known as “vacant spaces” analysis) can be generalized to handle situations where something is already known about the opponents' hands.

For example, suppose in your situation ♥s were trump and your side has 9 of them (♥s).

Suppose you have already drawn trump and found that your LHO started with three ♥s while your RHO had only one.

Now, what are the relative probabilities of the missing ♠K and ♠Q being (a) both with LHO vs. (b) both with RHO vs. © split between LHO and RHO ?

Well, because LHO started with three ♥S, he has only *10* “vacant spaces” available for other cards (including the ♠K and ♠Q).

RHO, on the other hand, having started with only one ♥ has *12* “vacant spaces” remaining.

So now, we place the first card of interest, say the ♠K.

The chances LHO has it are 10 out of 22 because there are 22 “vacant spaces” left total between LHO and RHO and LHO has only 10 of them.

Given LHO *has* the ♠K, now he has only *9* vacant spaces left out of 21 remaining (RHO has the other 12 still).

So the probability that LHO also has the ♠Q (given that he has the ♠K) is only 9 out of 21.

Therefore, the overcall probability that LHO (with three ♥s) has both the ♠K and ♠Q is

(10 / 22) * (9 / 21) = 90 / 462 ~ 19.5%

Doing a similar calculation for *RHO* who starts with 12 “vacant spaces” since he had only one ♥, we compute the probability that *he* has both the ♠K and ♠Q as:

(12 / 22) * (11 / 21) = 132 / 462 ~ 28.6%

Notice how LHO's having more ♥s than RHO has shifted the probability of either player having both ♠ honors from equal (24% each) to unequal (19.5% vs. 28.6%) with the opponent with fewer ♥s becoming more likely to have both ♠s.

To compute the probability (with ♥s 1=3) of the ♠ honors being split, we could just subtract the sum of the above from 100% giving:

split ♠ honor probability =

1 - ((132 + 90) / 462) = (462 - 222)/462 = 240 / 462

=~ 51.9%

Notice this hasn't changed much from the apriori 52% when we have no knowledge of the ♥ distribution (it is not exactly the same, though).