semeai

East's odds did go up, from 13/26 = .50 to 13/25 = .52 when West's odds went down from 13/26 = .50 to 12/25 = .48. This is the same as the Monty Hall situation, or a variant rather. There's a goat, a sheep, and a car. You pick door #1, Monty has control of doors #2 and #3. By the vacant doors principle, your odds are 1/3 and his are 2/3 of having the car. You ask him point blank "do you have the goat behind one of your doors" (just like him having to bid 1♣ if he has ♣10 and pass otherwise) and he says, fortunately, yes. Now your odds are 1/2 and his are 1/2 since the vacant doors are 1 to 1 now. His odds decreased from 2/3 to 1/2 and yours increased from 1/3 to 1/2.

Good find. This would make the two lines quite close, with yours then likely getting the edge if the ♠J shows heart suit preference.

It used to be 13/26. Now it's 12/25 = 0.48. The denominator did change, as it does in Bayesian situations, just not as you're doing it here.

Yes, the 3-3 both minors comment was irrelevant. We both make on 2=4=3=4 (with my follow-up comment that I should switch to playing a spade at trick 5 if diamonds are 3-3), you make most 2=4=2=5's and I make most 2=3=4=4's. The latter are more common, aren't they? Added: Perhaps you're right about ♠J showing hearts. In any case, if diamonds are 4=2 (and then RHO is 2=4=2=5 or 2=3=2=6), I can still make by playing ♣K and shifting to spades if it wins, if LHO has J or 10 of hearts, I think.

This answer is correct only if LHO plays his heart randomly. If you read the previous post of mine, this is equivalent to assuming p_n is 1/n. As I said in the second footnote (egad!), LHO's motivations for playing ♥K are not likely to change much as the number of spot cards increases beyond 2, and even for 1 spot card, i.e. Kx, it's likely pretty close to 1, not 1/2.

Thanks for posting here, and for your helpful comments. I'm happy to know you took the care to worry about non-double-dummy play in the book.

All I said was it wasn't takeout. :) I hadn't even attempted to undertake defining the ones that weren't takeout as penalty, cards/values, DSIP, etc. That said, your point is good and this seems like a difficult undertaking. Do some professional partnerships have detailed lists like this that are basically comprehensive? Are they "trade secrets" if so? Thanks, good one.

This is not what inquiry was asking. The answer to this question is likely* very close to, but not exactly the same as, the answer to that question. This question is more complicated because LHO would not always play ♥K if he has it. Probability of LHO holding a specific club right now given that diamonds are 5-3 and all we've done is play a heart up and see ♥K pop: prob(hearts K-6)*p_1*(7/11) + prob(hearts K1-5)*p_2*(6/11) + prob(hearts K2-4)*p_3*(5/11) + prob(hearts K3-3)*p_4*(4/11) + prob(hearts K4-2)*p_5*(3/11) + prob(hearts K5-1)*p_6*(2/11) + prob(hearts K6-0)*p_7*(1/11) divided by prob(hearts K-6)*p_1 + ... + prob(hearts K6-0)*p_7 Here by prob(hearts K2-4) I mean the probability that hearts are Kxx - xxxx given diamonds 5-3 already. Here p_n is the probability that LHO plays the K when he has n of them. For example p_1 is 1 and p_2 is pretty close to 1. Let me say that the rest are all equal for simplicity.** Note I'm ignoring reasons other than length that may distinguish playing ♥K or not for LHO. I won't do the calculation out, but the first two terms (hearts K-6 or K&1-5) are not so likely and p_n is not too small, so the rest of the terms dominate. There the p_n's are assumed to be all the same, and they drop out of the numerator and denominator, giving us: prob(hearts K-6)*(7/11) + ... + prob(hearts K6-0)*(1/11) divided by prob(hearts K-6) + ... + prob(hearts K6-0) Which is the answer to the sort of question inquiry asked. It's also exactly equal to 7/17, the vacant spaces answer, which I invite you to check or prove. *This question has no definitive numerical answer because it depends on the probability that LHO will play ♥K from various holdings. **This is where the ♥K is different from say the ♥7. LHO's motivations for playing ♥K or not are not likely to change much as the number of spot cards increases beyond 2, but the likelihood he'd play a random spot like the 7 would decrease if he had more random spots to go with it.

2♣ now is enough.

No, the point is you have to place ♠K with RHO to make, so that does alter the vacant space calculations. This is exactly the same as placing the ♥K in the original hand.

Better thought experiment: ♠32 ♣AQ ♠AQ ♣32 We're missing ♠KJ109 and ♣KJ and ♦32, which were the opponents' only original spades and clubs were the only cards the opponents held that were not unintentionally faced and turned into penalty cards earlier (i.e. they don't matter for any calculations), and we know from the bidding that the diamond distribution is split (i.e. 1-1 now). We need the last 4 tricks and are in hand. We play a club toward the Q and LHO follows with the J. What do we do? (Playing BAM in 7NXX) What about missing ♠KJ10 ♣KJ ♦432, knowing diamonds are 1-2?

How do I know this? Did he fumble his cards and drop them on the ground and one flipped face up and it was ♣10? In this case, yes my odds for the finesse are now 12/25. Did he lead ♣10? In this case no. The information is (almost entirely)* irrelevant, cf the Monty Hall Problem. He led one of many irrelevant cards which he is happy to show me (the ♠K he is not going to show me). The actual situation is different. We can must (to make the contract) place the ♥K in West's hand and it was not irrelevantly volunteered. The probabilities for different heart distributions are different now because the heart suit is now effectively 12 cards. A simple thought experiment: In a three card ending, you need clubs to be 2-2. There are 4 clubs out, and the ♠AK. Left hand opponent's bid places him with both ♠AK. Is this irrelevant? Added: This example is not so related. I'll give a better one below. A slightly less simple thought experiment which is a bit more related: In a five card ending, we're missing ♣Q1098 and ♠KJ10987. Dummy has ♠AQ ♣432 and declarer has ♠32 ♣AKJ. We need all the tricks, and pitched ♣5-7 and ♠6 from dummy on the run of our diamonds and hearts. We play club A, spade to Q, spade A (drops K, mandatory falsecard possible). Now we play a club. All have followed throughout. What is the probability we'll succeed on the drop in clubs? On the hook in clubs? * Leading the ♣10 suggests certain club holdings, and also suggests certain holdings that are attractive to lead from were not as likely to exist in other suits, so this is not exactly correct.

I'm interested in good, fairly comprehensive defaults for doubles and when they should be takeout, cards, or penalty. There are of course multiple systems of defaults, maybe one called "modern aggressive" with most doubles takeout and one called "modern conservative" with many doubles takeout but also many cards or penalty. Here are a few auctions to test your agreements on (please add more controversial sequences if you like): 1. Berkowitz and Sontag disagreed (board 9) on this sequence: 1♠ P 1NT 2♣; 2♦ X. 2. There was some discussion of the following sequence starting here. 1♦ P 1♥ P; 2♣ 2♠ P P; X. If the pass-and-then-come-in 2♠ offends, maybe 1♦ 1♠ X P; 2♣ 2♠ P P; X is good enough to consider instead. Here is a poor attempt at the beginning of some defaults in the "modern aggressive" style, to show what I'm talking about: Double of an opponent's suit bid is takeout unless: 1) Our side has already made a strength showing double or redouble (maybe this is supposed to be more restrictive) 2) We have a known fit 3) All four suits have been shown 4) The opponent's bid is artificial 5) Pass is forcing 6) Our side preempted 7) The doubler passed up an opportunity to make a takeout double earlier and is not balancing at the 2-level and no new suit has been bid 8) The opponents have shown 3 separate suits 9) The opponents have bid game and doubler has passed before [suggested by wyman below:] 10) Partner has suggested the suit (even implicitly, as via a t/o X) as a place to play [note that this does not apply, for instance, to (1Y) 1N (2Y) X, where partner has shown cards in the suit but has not suggested it as a place for us to play] I haven't gotten into cards vs penalty or more detailed stuff here yet.

My description of the Berkowitz - Sontag auction (board 9) was close but not exactly correct (Sontag only bid once). Here it is: 1♠ P 1NT 2♣ 2♦ X Sontag bid 2♣, Berkowitz was the doubler. In fact Berkowitz had both! 7 KQ853 J8765 Q2

Not the same. A low card is one of many. The relevant passage you seem to be referring to:

This is equivalent in this case with the ♥K No, eliminating void with LHO doesn't go far enough. You also need to eliminate ♥1072 with West and ♥KJ54 with East, and many others. Eliminating all such holdings is equivalent to just placing ♥K with West and moving on. It's a well-defined single card.

I bid 3♣ if my partner is aware I might fake a jump-shift suit (especially this one). This allows me to get to hearts when it's right and to attempt some intelligent decision between notrump and spades otherwise. I imagine these artificial 3♣ agreements discussed above grew out of the tendency to fake jump-shift suits, especially the lowest one, just like 4th suit forcing as an actual agreement grew out of all sorts of people faking a natural 4th suit. If partner will take me too seriously, probably I should just rebid 3NT.

Yes, include ♥K. Think of it this way: you're enumerating hands which are consistent with the bidding and allow you to make the contract. All such enumerations will have (at least) 5 diamonds and the heart King with West. In fact, the heart King is easier to deal with than the diamonds for the vacant space calculation. It's a unique card. In the diamond suit, the fact that West has QJ and at least 3 lower cards is slightly harder, as it could instead be QJ and 4 lower cards, etc. The proper way to do things would be to find the relative probabilities of 5 diamonds to the QJ with West and three with East, 6 diamonds to the QJ with West and two with East, etc, and then do your vacant space calculation for each (including ♥K with West) and weight by the relative probabilities. If you think for the hand to be consistent with the bidding West must have at least one spade honor, there should be a correction to the vacant space calculation. As with diamonds above, however, you can't just put one spade with West. Instead, you compute the relative probability of a 1-1 split (of which there are two) and a 2-0 split (given the diamonds and heart King!) and do your vacant space calculations for each case, and then weight the answers by the relative probabilities. To be completely proper, you'd also want to exclude hands with a 5 card major from West, as well as anything else that would change the bidding or play up to this point. This shouldn't have much of an effect on this hand, but such considerations are the key point when vacant space computations are flawed (e.g. when someone has led a suit and you use that suit for vacant space purposes, you're forgetting to exclude some percentage of the cases in which they have a longer or equal length side suit, and it does end up mattering).

1♣ for now. Things may be more interesting later. This hand is certainly good enough to open, and the conditions (2nd seat, vul) are very wrong for preempting. I did worry for a moment that the hand had 14 cards, but realized V stood for void. :D In some playing card sets, V, for Valet, is the Jack.

I understand your desire for sourced comments and rigid rules on doubles. A common modern style is to go the other way, with takeout being the default. Larry Cohen article: I tend not to have rigid rules (it would be better to, of course), but here's a try. I was looking for a sourced one on Cohen's site, but didn't find one. Double of an opponent's suit bid is takeout unless: 1) Our side has already made a strength showing double or redouble (even here some play exceptions to this) 2) We have a known fit 3) All four suits have been shown 4) The opponent's bid is artificial Admittedly, rule (3) leaves us with the above auction not being takeout if you include it. Here's a possible follow-up rule: If double is not takeout because of 1, 2, or 3, then it shows extra values if doubler sits under the bidder and is penalty if doubler sits over the bidder. These rules only really make sense through the 3-level. You'll want different rules for higher levels. Added: This isn't very complete. Maybe I'll start a thread asking for good defaults. The key point is that nowadays many play that the defaults are for when a bid is not takeout, not for when it is.

Please understand that there is more than one way to bid and to define your bids/doubles, and no one source has "the" way, especially a source that is 50 years old. This auction would differ based on my partner. With some, I agree that (essentially) all low level doubles, including this one, are takeout. With others, many more doubles, including this one, are "cards/values" and easily passable. Certainly this double would not be "penalty" with either, however. An example to illustrate disagreements and differences in style here: In the Spingold, Gordon vs Baldursson, I was watching Berkowitz and Sontag, evidently a new partnership. Three suits were bid, clubs by Sontag and spades and diamonds by the opponents and Sontag had made two calls and it was the second round of the auction. Berkowitz doubled a natural 2D, intending to show hearts and club tolerance ("takeout") and Sontag interpreted it as showing diamonds. I'll try to find the auction, but it's not in the vugraph archive yet. Berkowitz apparently said "please remember that all low-level doubles are takeout" (or something similar).

Right, I was thinking he could exit clubs, but he can't do that if diamonds were 3-3 or we can then cash our 6 minor suit winners and endplay RHO in hearts, so your line of a spade at trick four five is best then. If instead diamonds were 2-4, a club exit is fine for them, so I think you need my line of clearing clubs first. (I understand you were assuming a specific distribution.)

RHO should have all high cards except ♠A, excluding maybe ♥J. We have one loser to burn and seemingly nothing to gain from keeping diamond honors, so I'll hook the diamond and cash ♦AK unless someone shows out. My plan now is to lead a spade through West at a point when he has no more clubs. This should endplay him into giving me my second spade trick (and I hope to take 6 minor suit tricks and a heart trick) or into leading hearts, at which point I'll play him to have at least one of ♥J10. Accordingly, I lead ♣K from hand. Assuming East lets me win this, I lead low to ♣J. If East now wins and finds the spade exit and clubs are 3-3, West can win ♠A and play back a club and I'm now sunk if diamonds were 4-2, but if they were 3-3 I'm home by endplaying East in hearts. If East instead exits a club after winning ♣J (possibly after cashing a long diamond) I have to guess clubs and then lead a spade up.

Here's an article about this sort of squeeze, with various names suggested. The author calls it a squeeze-trim-endplay. At the bottom, other people's names for it are given.

This is a good question. I don't do so myself, but I've seen some play e.g. 1♦-1♠;2♠-4♦ as a splinter.