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♣J. Since this is vacant space week I suggest the following argument: Comparing running the ♣J with ♣A then finesse, is to compare LHO's ♣Qxxx vs. ♣xxx. Therefore we may as well assume that the ♣x's are distributed 3=0. If the hearts are 3=6 then vacant spaces are equal; if hearts are 2=7 or worse then RHO is more likely to be void than ♣Q alone. But a stronger clue is that RHO's 3♥ bid with no A or K, especially vulnerable, is easier to envision with a void.

Again, depends on the hand. If the opponents have enough tricks ready to defeat the contract it's not so bright. Case in point: considering that hanp stated the problem is 5 tricks, it may be that the contract is 7NT. Low to the J looks really implausible in that context. Or, it could be that there are potentially lots of tricks in other suits but we are just testing this suit first hoping to thereby avoid a finesse for example. I've played that one over and over in my mind and can only come up with 4 tricks.

:lol: It costs nothing

I like this train of thought.

That may be two different questions. It surely makes a difference, and somewhat you can account for it via vacant spaces. Suppose ♠AKQ ♠72 ♠43 ♠J is revealed somehow -- perhaps through cashing 2 rounds of ♠. Under the assumption that LHO has denied the ♠6543 and RHO denied the ♠2 it seems a good approximation to deduct 1 and 4 vacant spaces from LHO and RHO respectively. I have a feeling that this kind of machination must be applied with care (especially worried about bias in the way the information is obtained) and tentatively. But it could also be an underexploited gold mine.

Just a program. I specify opposing cards of sq 4sx 5h 8d 8c, and a condition such as hε1 or (hε1)&(sx ε 3) and it computes a table of of the 73 or 17 or whatever possible distributions (counting sq as a suit) and their corresponding probabilities. Easy manipulations on the table give various results. Answers to simple problems often turn out to be nice fractions with small denominators -- which in retrospect can probably be seen logically such as via vacant spaces -- but 1/360 etc are just rounded versions of decimal numbers with a lot of arithmetic behind them.

Probability that LHO has the ♠Q 4:3 from the ♥ distribution alone 9:8 (4% less) given the further information that the ♠ pips are 3=1 -- per vacant spaces. This estimate technically Increases by 1/7400 knowing no club void. Increases by 1/360 knowing no club void or singleton. Cashing the ♦A tells you nothing further. Increases by 1/130 knowing no minor shortage. Summary: The number 0.537 from accurate calculation is over twice as close to the approximate 9:8 than it is to the "wrong answer" 5:4.

Oh. Then 6/11, 43/83, 51/98, 5001/9608 or 0.545 0.518 0.5204 0.5205.

Interesting case. Unlike Fred's original hand, we have not seen the entire spade-spot suit (the pseudo- or sub-suit consisting of small spades) so cannot accurately compute via vacant spaces. But we've seen a fair fraction of it, so how inaccurate is ignoring it altogether? Plugging into my probability calculator: First calculation: Using only that hearts are 1=3, the ♠Q is 6:5 to be on the left. That I think is the vacant spaces calculation. Second calculation: Using also the observation that LHO has at least 2 small spades, the ♠Q is 7:6 to be on the left. Third calculation: Taking into account as well that RHO has at least 1 small spade, the ♠Q is 33:28 to be on the left As decimal probabilities the above are 0.545, 0.538, 0.541. In principle the ♦ plays tell us something as well -- no ♦ void about. That increases the last probability to 0.5413, but of course making a computation like that while ignoring inferences from the bidding and play is entering the twilight zone. From the above computations it seems that the vacant space rule is quite accurate. Note that at the opposite extreme if you misapply vacant spaces by counting every card seen, then you would often, half-way through a trick, think it is k+1:k that the player yet to play has any given card. Sorry, don't understand that part. It can make big difference. Perhaps you could adjust the vacant space theory to account for it.

Interesting case. Unlike Fred's original hand, we have not seen the entire spade-spot suit (the pseudo- or sub-suit consisting of small spades) so cannot accurately compute via vacant spaces. But we've seen a fair fraction of it, so how inaccurate is ignoring it altogether? Plugging into my probability calculator: First calculation: Using only that hearts are 1=3, the ♠Q is 6:5 to be on the left. That I think is the vacant spaces calculation. Second calculation: Using also the observation that LHO has at least 2 small spades, the ♠Q is 7:6 to be on the left. Third calculation: Taking into account as well that RHO has at least 1 small spade, the ♠Q is 33:28 to be on the left As decimal probabilities the above are 0.545, 0.538, 0.541. In principle the ♦ plays tell us something as well -- no ♦ void about. That increases the last probability to 0.5413, but of course making a computation like that while ignoring inferences from the bidding and play is entering the twilight zone. From the above computations it seems that the vacant space rule is quite accurate. Note that at the opposite extreme if you misapply vacant spaces by counting every card seen, then you would always, half-way through a trick, think it is k+1:k that the player yet to play has any given card. Sorry, don't understand that part. It can make big difference. Perhaps you could adjust the vacant space theory to account for it.

Which side is vulnerable, i.e. which side is NS?

It might not matter, but double dummy declarer can attack clubs and create a club trick if you allow two entries to the dummy, or even if you discard a ♠. ♣3 is the only sure defense. Edit -- ♠ discard is ok for now. Then win declarer's ♣ play, exit trump, win ♣, exit with three rounds of ♠, ♥ ruffed in dummy -- now discard ♣ and claim last two tricks.

It would be equally logical to argue the opposite -- that because other tables may be in an advantageous position that lets them try for the club break without risk, our best chance to catch up is to hope that the clubs don't work. The matchpoint math prefers whichever argument corresponds to the best chance. If the club play is say 50%+x to succeed, then comparing against the declarer who gets a free ride, risking the club ruff ties 50%+x and loses a match point 50%-x for a net negative expectation of 50%-x. Similarly, not risking the club ruff ties the 50%-x that the clubs lie badly and loses otherwise for a negative expectation of 50%+x. Hence what you should do depends only on whether x is positive or negative regardless of the lead at other tables. That said, if Kehela thinks x>0 it probably is.

very nice. I guess it falls under the category of entries.

From failing to draw trumps, declarer sees a possible benefit to ruffing twice in dummy. Therefore discarding and subsequently leading a round of trumps (or maybe over-ruffing the dummy's small trump) seems at least break-even compared to ruffing. It is of course nominally break-even: Ruff the Q and allow two ruffs in dummy, or allow the Q and one ruff in dummy. So for the discard to gain there has to be a matter of tempo or entries, such as stopping the 5th heart from scoring, say from ♥AQ976. Indeed if declarer is even so weak as Kx,AQ976,AKJx,xx (i.e. no ♣J), ruffing isn't a sufficient defense. But neither is discarding: next ♥9,10,♦x,? . So against what hand does it matter? Discarding does have one advantage. You don't have to decide how to attack in the black suits.

Can you spell this out? It looks to me like a free and routine play. The only cost I can dream up is the specific case of ♠QJx ♣xx and declarer next plays a club.

Technically speaking, it also gains if RHO has ♠QJ9 to begin with, along with 4 clubs, though I realize it is unlikely RHO plays a deep game and false card ♠Q. Good catch. I at first listed that case but for no reason that makes sense at the moment crossed it off. Big mistake. Given the inferences of red suit length with LHO, RHO is hugely likely to have been dealt 3♠ and 4♣. True, the holding should be discounted pro-rata based on RHO's probability to choose exactly the ♠Q to play from ♠QJx, but even if you figure 1/3 for each card it now seems to me that the chance of RHO holding ♠QJx and ♣xxxx exceeds either of the other probabilities that I computed and compared.

I see a better case for playing safe than risky. Compared to the safe play of cashing the other top ♠, a risky immediate ♣ ruff breaks even when ♣ are 4-3 and ♠ are 2-2. 1. Losses a. in the very unlikely case that LHO didn't lead a ♣ from ♣QJ10xx and RHO can ruff in from original ♠QJ b. in the unlikely case that LHO has ♣xx and RHO made a great falsecard from ♠Qx c. Therefore almost only when LHO has ♠J9x and ♣xx. 2. Gains -- only if LHO has 3♠s and 4♣s. LHO won't have ♥x, so would practically need to be 3244. Based on dealing probabilities the risky play loses by about 2:1. As against that maybe RHO would sometimes play a helpful club at trick 2 when holding ♣QJ10xx. (I realize I may be yelled at for raising such a dull idea in this forum.) The point that some of the field (especially if declaring by North) will get a different lead and be able to play safely is appealing but seems to me to have limited application. Compared to those tables, our risky play loses when the cards lie badly and our safe play loses when they lie well. Hence we have a negative expectation either way and if for example the bad lie is a 2:1 favorite, we expect -2/3 from the risky play and -1/3 from the safe play. Thus this does not seem to be one of those hands where bad luck at trick 1 justifies taking a chance to "catch up" with the field.

This is wrong because they can just adjust by never playing the K then J from KQJ. Yes, but maybe worth noting that the adjustment is a little intricate: KQJ can follow 5 ways: KQ QK QJ JQ JK KQ can follow 2 ways: KQ QK QJ can follow 2 ways: QJ JQ KJ can follow 1 way: JK So if KQJ chooses each play with probability 1/5, then JK would be 5:1 to be a doubleton and other plays would be 5:2. In order to get the same 3:1 always, KQJ must give 2 "shares" to JK, playing JK 1/3 and other combinations 1/6. (Alternatively you can (I think) make a rule to always play upwards and choose to hide each card from KQJ with probability 1/3. That means you select your first play with relative probabilities 4:2:0 for J, Q, K.) If considering all 5 carding orders, ideally select the first play with relative probabilities 3:2:1 for J, Q, K (then after the J, be 2/3 to next play the K). It may be unrealistic to imagine that defenders actually come close to this. In particular if you think your RHO opponent is even 1/3 to play the K first from KQJ, then when you see KQ played on the first two tricks it might be right, if there is a modest additional clue, to spurn the third round finesse.

LHO led the 2 of clubs playing 3rd and 5th, and RHO has already shown up with 5 clubs (he followed to the ace, pitched 2, won the king and exited one). I see. Watching the discards. Subtle clue.

Dissimilar. The Stayman doubler appears to be 5215 (♠QJ10x on lead is unlikely, double with only 4 ♣s nutty), whereas the other RHO can be 4414.

Sorry but this is nonsense. We can say the same thing about the heart 10, maybe the RHO always plays the jack from ♥J10x? Then we would be sure that the 10 is from ♥10x and we can certainly finesse. Or maybe RHO always plays the 10 from ♥J10x, so that it becomes much better to play RHO for ♥J10x instead of ♥10x? A good opponent will play the ♥J about half the time and the ♥10 also about half the time. That way he cannot be manipulated. For the same reason a good RHO will play (from ♣Q-6th) the club queen about half the time and the fifth small club about half the time. In that case we should play for the drop whether the queen appears or a small card appears, and it is 9:8 versus the finesse. It's a nice comparison -- the ♥J10 choice vs. the ♣Qx choice -- but are they really the same? I think it's facile to equate them and you're wrong on two counts, one psychological and one mathematical. I'll take them in that order. For the typical "quack" (or ♥J10) randomization, there is at most little scope in a reasonable A/E game for out-thinking what the opponent will do or expect, because we all know and know the opponent knows our way around this particular block, ad infinitum. Therefore it's just game theory, not mind games. But even if the ♣Q-x play were in principle the same thing, it does not follow that our merely expert (not cyborg) opponent will recognize it as such. For proof, simply note the varying reactions of several indubitably expert players here including even Fred. Yes, his uncertainty about the position was from the declarer side and perhaps as defender he would have seen it more easily. But anyway that asymmetry alone legitimizes hoping to out-guess the opposition here (e.g. Ken Rexford's discussion). Experts too can be had, under pressure and when there is something a little new. In fact, even in the tired old QJ randomization situation we can guess a bit better than 50% which card our opponent will play when holding both. But -- and here's the mathematical point -- in the classical case that doesn't help because the 2:1 restricted choice odds are so overwhelming that we quite properly and contemptuously dismiss mind reading the opponent unless we are Barry Crane against the rabbit. But the present case is different: The mere 8:9 nominal odds are quite likely to be a weaker reason for choosing a play than the psychological reason. The analogy between classical restricted choice randomization and this hand is a misleading one, fatally flawed.

This seems a very complicated way to get to the answer. "That means that" suggests a linear line of reasoning, but 2/3 and 4:3 seem to be from thin air here. You are probably right that I used a complicated way to get the answer. I do think that I have a very good understanding of the words "that means that". I assure you 2/3 and 4:3 did not come from think air. Once you convince yourself that 8:9 is the right answer, perhaps you can read my post again I tried to explain earlier why I think it is not correct to use that RHO played 5 little clubs. I am glad to see that the 8:9 also occurs when you compute the relative likelyhoods of ♥Hx ♣Qxxxxx and ♥J10x ♣?xxxx. I know they are not from thin air but they did not come from the immediately previous either, so I was complaining that I could not following your thinking. I understand that "seem to be from thin air [here]" can sound like "are from thin air" but such was not my intent. Yes, I followed your earlier argument and therefore tried to abort my post. Thought I had caught it too.

I think this part of what you say is not correct. The fact that RHO played small on the 11th trick gives no information because at that time (when we had already played the club jack from the dummy) he could play the queen or small from Qxxxxx. To get the correct answer, we should also consider Qxxxx of clubs on our right. This kind of argument can get a bit messy. Best is to ignore the card RHO plays. If RHO "falsecards" exactly the right amount of time from Qxxxxx there is no information to get from the play of the low club on our right. Let me try to phrase it in terms of strategies, I think it is easier. If we play for the drop no matter whether RHO plays low or the queen, we win if RHO started with J10x of hearts and Qxxxx or xxxxx of clubs, but lose if RHO started with Jx or 10x of hearts and Qxxxxx of clubs. ♥J10x and ♣xxxxx is exactly equally likely as ♥Hx and ♣Qxxxxx, but ♥J10x and ♣Qxxxx is a bit more likely than ♥J10x and ♣xxxxx, and thus also more likely than ♥Hx and ♣Qxxxxx. Therefore we should play for the drop. Thanks very much, Han. Yes, this makes sense and it is actually pretty similar to the theory I was afraid to post out of fear of further embarassing myself (though you definitely did a better job of explaining it to me than I did explaining it to myself!). Indidentally, the reason I found this problem interesting was because I could not really decide if the Queen of clubs should be treated of as a pure x (as it is in question 3), a pure non-x (like all the known cards in spades and diamonds), or as something in between. Fred Gitelman Bridge Base Inc. www.bridgebase.com Of course something in between. Ignoring what RHO actually plays is a cop-out -- it's saying that the actual problem is too hard so I'll instead consider how to play against computers. Hanp has reformulated the original problem to say that RHO's last discard is "a club" instead of "a small club". Under that stipulation here are the layouts in question (posterior probabilities in parentheses): xx Hxxx xxxx xxx (8/17) xxxx Hx x Qxxxxx xx xxx xxxx Qxxx (4/17) xxxx HHx x xxxxx xx xxx xxxx xxxx (5/17) xxxx HHx x Qxxxx giving 8:9 for finesse:drop. Assuming instead that you actually watch the spots, the 3rd possibility is eliminated but the 1st is counted pro-rata according to RHO's tendencies. If RHO last discards Q:x from the first case with ratio 5:4 in imitation of the Qxxxx:xxxxx prior probabilities, then yes 8:9 is correct. However in real life no one is that smart. At best they randomize 1:1 between the last two cards in which case the drop is 4:4 when the Q does not appear and 5:4 when it does. More likely you can judge RHO psychologically as either being a "clever" player who usually plays Q from Qx in order to "seem" longer in hearts -- in which case the drop approaches 4:0 when the Q does not appear and 5:8 when it does -- or a subtle (or oblivious) player who tends toward x from Qx -- in which case the drop approaches 4:8 when the Q does not appear and 5:0 when it does.