junyi_zhu

that's why I never understand why so many play 1C to show 2 or more. It just puts way too many hand types to 1C and make 1D opening less frequent. Therefore, it would be very difficult to set up trumps in clubs, especially after preemptive bids or 2NT jump rebids by the opener. After a 1D 2H(invitational) sequence and seeking for shortness, the reach of 6D is indeed very possible.

Well, it's still quite possible. 1S 3D(solid) 4H(how many side aces?) 4S(1) 5C(we have all aces, CK) 5D(no extra length, no side king) 5H(HK) 5S(SQ) 6H(HQ, 6NT is safe, so let's try to 7NT) 7NT(now HJ gets the value)

I feel 3NT is an underbid. This hand looks too slamish.

What is the logic about this? (Before I read this I thought making a post to ask this & I was almost sure that you could do it always.) Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..? I can only remember this if I understand it. Because vacant space calculation is an approximation in some situations. For example, suppose opps have 8 spades, you draw two rounds, both follow. Therefore, spades can't be 8-0 or 7-1. However, 6-2, 5-3, 4-4 are all possible, so you just don't really know the exact distribution of that suit. In this sense, you can't include this suit into your calculation. All you know is that spades distribution is rather unknown and it's impossible to be 8-0 or 7-1, which is still a very small percentage comparing with all the possible spades distributions. Therefore, It's still fairly good approximation to assume that you don't know the spade distribution at all. In the other thread talking about the probabilities, one post was very well said: what's the average number of your heart suit if you open 1H showing 5 or more? I guess it can't be 6 and should be in the middle of 5-6, because frequency of 5, 6, 7 hearts holdings are very different. In this case, it's the same logic. When they hold 8 card, even if you know the distribution can't be 7-1 or 8-0, you still don't know the exact distribution and 7-1 8-0 are very rare events, so you can't really take those out of your approximation easily. Therefore, you'd just stick to what you have and consider the distribution of spades is unknown. The real percentage can be calculated by computer simulations in a quite accurate way, which should be slightly different from the vacant space calculation. However, at the table, vacant space counting is still your best friend. Still, in many situations, vacant space counting actually gives the accurate result. Thanks for the effort. I want to believe the above, but I wonder how far off both methods are: [hv=n=skjt9hkqjxdxxcxxx&s=saxhxxxxxdakqcakq]133|200|[/hv] ♠ is trumps. LHO leads a small ♥, RHO return a ♥ and LHO ruffs. LHO plays a random ♦ and RHO follows that suit. You cash ♠A and lead a second round. All follow with small ♠'s. Probability that LHO has ♠Q according to vacant space theory: 13-1-3:13-6-1=9:6=0.6 Probability if WRONGLY taking the ♦ play into account: 13-1-3-1:13-6-1-1=8:5=0.615385 If Playing ♣/♦ before the 2nd trump then: - for Vacant Spaces this will always be 0.6 - for the WRONG calculation this will be: 1 minor: 0.615385 2 minors: 0.636364 3 minors: 0.666667 4 minors: 0.714286 5 minors: 0.8 ==> Is anyone able to calculate the real probabilities? BTW: Does it make a difference for vacant space theory if opps always play their small cards from low to high? You made a mistake in your vacant space count, spade situation is still unknown. You can apply vacant space counting only when the situation of a suit is clear, or missing one key card in your key suit. Here, you just don't know the spade distribution yet. It could be 3-4, 4-3, 5-2, 6-1.

This is a clear 6D hand. Well, 7D can be possible, but you may not have good gadgets to explore. Partner's 3H shows doubt in 3NT, so he can't hold a lot of value in H. After you show both D and S, he shows no interest at all. Therefore, he must either have a club one suiter, or a hand with good hearts but too good to bid 3NT. In both cases, you should push to 6D, which should offer you a very reasonable play.

What is the logic about this? (Before I read this I thought making a post to ask this & I was almost sure that you could do it always.) Logically you would think that if LHO leads a Heart then vacant spaces are 12:13..? I can only remember this if I understand it. Because vacant space calculation is an approximation in some situations. For example, suppose opps have 8 spades, you draw two rounds, both follow. Therefore, spades can't be 8-0 or 7-1. However, 6-2, 5-3, 4-4 are all possible, so you just don't really know the exact distribution of that suit. In this sense, you can't include this suit into your calculation. All you know is that spades distribution is rather unknown and it's impossible to be 8-0 or 7-1, which is still a very small percentage comparing with all the possible spades distributions. Therefore, It's still fairly good approximation to assume that you don't know the spade distribution at all. In the other thread talking about the probabilities, one post was very well said: what's the average number of your heart suit if you open 1H showing 5 or more? I guess it can't be 6 and should be in the middle of 5-6, because frequency of 5, 6, 7 hearts holdings are very different. In this case, it's the same logic. When they hold 8 card, even if you know the distribution can't be 7-1 or 8-0, you still don't know the exact distribution and 7-1 8-0 are very rare events, so you can't really take those out of your approximation easily. Therefore, you'd just stick to what you have and consider the distribution of spades is unknown. The real percentage can be calculated by computer simulations in a quite accurate way, which should be slightly different from the vacant space calculation. However, at the table, vacant space counting is still your best friend. Still, in many situations, vacant space counting actually gives the accurate result.

Against Henri, it could be different. :D

You can count the vacant space only when the suit distribution is clear. Suppose for one certain suit, opps have 8 cards, you play two rounds, both follow, you can't count this suit yet, because the distribution of this suit is still unclear. vacant space count can only be applied when the suit distribution is clear(or one card missing in your key suit). So for your example, the correct way to count vacant space is: LHO: 13-1(singleton) -3(three trumps)=9 RHO: 13-5(5 in that side suit) -1 (one trump) = 7 So it is 9:7 better to play finesse vs. dropping. Wouldn't you have a complete guess since the difference is 2? No, this is the difference of total vacant space, which just gives your final answer. If The difference of your opps' total vacant space is zero, it's 50% to finesse or to drop. What Fred suggested is that vacant space difference other than the trump suit. When that number is 2, subtract the difference in trumps, which is 2, you get your total vacant space, which is 0, that gives even money to finesse or drop.

You can count the vacant space only when the suit distribution is clear. Suppose for one certain suit, opps have 8 cards, you play two rounds, both follow, you can't count this suit yet, because the distribution of this suit is still unclear. vacant space count can only be applied when the suit distribution is clear(or one card missing in your key suit). So for your example, the correct way to count vacant space is: LHO: 13-1(singleton) -3(three trumps)=9 RHO: 13-5(5 in that side suit) -1 (one trump) = 7 So it is 9:7 better to play finesse vs. dropping.

.... 3C 3D(5 or more D, not suitable for 3NT) 3H(two way cuebid, either C one suiter, or D fit, doubts in 3NT) 3S(S value) 4D(D fit and slam interest, without slam interest, responder can bid 5D) 4H(RKC, a little bit pushy, since partner denies a lot of H value, it looks not bad) 4N(3-0) 5C(Q?) 6C(CK and DQ) 7D.

I do not think Fred said anything about trump length having been shown. Fred, would your rule also apply if we had 9 trumps and RHO had, say, opened a weak 2 and marked the distribution in a suit that way instead? (ie 6 opposite 4 = guess) This is the correct way to count. His assumption is that 4 trumps are shown and trump Q is missing. The magic number of 2 is just due to the shown trump number difference, which is also 2. The other number 2 rule case is the 10 trumps case, suppose your LHO gets a ruff and plays low under your Q, should you drop or finesse K? This is also determined by the side suit shown cards difference. I'm a little confused. For deciding "odds" shouldn't it also matter how many cards remain to be played? Or does that simply determine how much of an advantage these observations give you? It is determined by the vacant space of your opp's unknown distributions. Suppose, LHO leads singleton and gets a ruff, you get the complete picture of that suit. So the vacant space for LHO is 12 (13-1) and the vacant space for RHO is 13- length of that suit. After you draw one round of trumps, play another card and LHO follows, your totally 4 trumps are shown with Q missing, now LHO's vacant space is 13-1(singleton side suit)-3(trumps) = 9. RHO's vacant space is 13- length of that side suit - 1(trump). Since the probability of Q is proportional to the vacant space, all you care is the difference of length of that side suit between RHO and LHO. If this number is 3 or more , you should finesse. If it is 1, you want to drop. If it is 2, it is 50% either way. Of course, it your opps show some distribution in other side suits (they could play out that suit, or they show it by bidding) the calculation can be different. The basic idea is still very simple, counting the vacant space.

yeah, just a simple matter of counting vacancies....

I do not think Fred said anything about trump length having been shown. Fred, would your rule also apply if we had 9 trumps and RHO had, say, opened a weak 2 and marked the distribution in a suit that way instead? (ie 6 opposite 4 = guess) This is the correct way to count. His assumption is that 4 trumps are shown and trump Q is missing. The magic number of 2 is just due to the shown trump number difference, which is also 2. The other number 2 rule case is the 10 trumps case, suppose your LHO gets a ruff and plays low under your Q, should you drop or finesse K? This is also determined by the side suit shown cards difference.

Any chance of an explanation for the somewhat idiots like myself who don't find this somewhat (or the least bit) obvious? The LHO shows two more trumps than RHO. Therefore, if the difference of side suit shown distribution of LHO and RHO < 2, RHO is more likely to hold the Q. (because the probability of holding Q is proportional to the unknown cards in one's hand) For example, LHO leads a doubleton to partner's AK and obtains a ruff and your side holds 8 card in that side suit, then LHO shows 2 cards and RHO shows 3 cards in that suit. Therefore, you should play to drop. If it is greater than 2, LHO is more likely to hold Q. For example, LHO leads his singleton to partner's ace and gets a ruff, his partner holds 5 card in that suit. You should take the finesse. If it equals 2, it's just a guess. For example, LHO leads a singleton, his partner holds Axx and gives him a ruff.

Good post. I actually never feel garbage stayman is very useful. Suppose you open 1NT with 2-2 in the majors, you just don't have a good place to play after garbage stayman. Therefore, a sign off gadget to 2D certainly looks better than garbage stayman IMO because it most likely may improve the contract.