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They're not equivalent because after low to the 10 you have a guess on the next round. Only if you're a perfect guesser is low to the 10 equally good. Low to the jack, then ace, picks up Hx and KQx onside plus any KQ doubleton. Hence it wins in 6+3+2=11 cases. Suppose instead you play low to the 10 and 2nd hand is a weak player who never ducks from Hx. Then you'll finesse on the next round and end up winning against all Hx offside or KQ(x) onside, but that's still only 10 cases because you lose to KQ offside. If 2nd hand is a good defender who often ducks from Hx, then the best you can do is your birthright of playing for that holding, and incidentally picking up KQ onside as well. That's 7 cases. Reality may be between those extremes. Low to the jack is anyway clearly better. Edit -- I overlooked OP's idea about 2nd hand jumping in with Hxx. Yes, against such an opponent low to the 10 could be better. Cover what? I don't understand this comment.

I see. The chance of winning goes up for a while then comes down. However it plunges down then immediately reverses back up (especially if you count a tie as a 50% chance for a win, which I think makes sense), so it's more a sawtooth than a parabola. Too, the sawteeth continue. Plunging occurs at roughly 21 boards, 32 boards ... and in the long run of course the graph smooths and rises approaching 100% chance of winning. So I wouldn't bet that's what Jeff Rubens had in mind. Besides as you say it's unrealistic. The right point of view is what Stephen Tu said -- we only need to consider this choice arising once in a match, and we can ask the expected result over a long series of matches. Charles

For sure. Suppose team A is better by 1/4 IMP per board. Guessing that the variance is about the same, for a 16 board match the better team wants 22:1 odds to gamble 10 imps for 1, but the weaker team will take that gamble at 4:1. Over 64 boards the better team, though a mere 16 IMP favorite, needs over 1000:1 assurance to gamble for an overtrick. That conforms to the convention wisdom of don't waste your time. For the weaker team, 6:1 is enough -- interesting in that this line of thought begins to speak to the question of strategy for beating a better team. (I'm still using my simple-minded model of imp uncertainty because it's easier.) Charles

It's not unimodal. It's not necessarily symmetric either, although in the abstract case it is (for some combinations of teams, even of equal absolute standard, it isn't symmetric). I did say that none of this would necessarily change the conclusion, particularly for long matches (I started to get something resembling a normal distribution after about 10 boards). However, if you are (e.g.) playing a two-board tiebreak, it changes the odds on the first of the two boards. Ok, so the point is that we may have to consider the short run.

Thank you. That's nice to hear. David Burns is certainly right though that Jeff has a very sharp mind about these things, and he is a professional mathematician. There's a good chance that the resolution is misunderstanding of some kind.

When you say a 'normal probability distribution', do you mean a Normal probability distribution, or the distribution that you commonly see? I meant Normal, i.e. Gaussian. However, so long as the distribution is unimodal and symmetric I can't imagine the conclusion would be different. That is interesting. But regardless of the distribution for one board, the Central Limit Theorem seems to say it will be normal in the limit, i.e. arbitrarily close to normal for a sufficiently long match. Is your point that real matches aren't "sufficiently" long? Or that the CLT doesn't apply for some other reason?

I love a paradoxical result as much as anyone, but I don't see where such a result comes from. The behavior of the model that makes sense to me is not so intriguing. Rather -- In a 1-board match, I agree that a 51% chance merits playing for the 1 imp gain at the risk of a 10 imp loss. However, as the number of boards increases, the necessary odds simply increase monotonically asymptotically to the imp odds. That is, the longer the match, the closer to 90.9090...% must be your chance of success before it's worth risking the contract for an overtrick. I don't see any increase then decline, or the opposite. My model is as follows. The object is to end the match plus imps, no matter the margin. If the IMPs finish even, assume a coin-flip playoff. Other than the present board, assume the teams are even with a normal probability distribution assigned to the various IMP margins -- i.e. the largest probability for a tie, slightly less for + or - one imp, etc. The spread of this distribution -- i.e. the standard deviation -- is identically 0 for the zero other boards of a one-board match, and the longer the match the broader the spread. It doesn't require many boards before the threshold percentage to justify risking the contract is within a trivial margin of the imp odds. I.e. at a standard deviation of 10 imps, you need an 89.5% chance to justify gambling for an overtrick. Charles Brenner

Simply on the question of IMP odds versus VP odds I've done some analysis. In summary, under typical assumptions there's very little difference but to my surprise VP does not encourage risk. Why is that? Doesn't the VP scale (like the IMP scale) compress large swings? As Helene said, it's sigmoid isn't it? I always assumed so. But a close look shows this isn't a very good approximation. Take for example the WBF 14-board scale. There are 5 imp differences (-2 to +2 imps difference) that correspond to a VP tie. And 5 imp differences (3-7 imps) that give one more VP. But the next bracket (8-10 imp) is smaller. The sequence of imp bracket sizes for winning 15-25 VP is (5 5 3 4 4 4 4 4 4 4 infinite). Hence the relation between imp and VP is better described as roughly linear but irregular, than concave upward. As a concrete example I assume the WBF 14-board VP scale, and that the imp consideration is risking 10 imps to gain 1. Hence in IMP terms it is break-even to try for the overtrick provided the chance of success is 10/11. A few comparisons of this number with the VP situation: Typical example: you estimate the match state as variously anywhere from even to plus or minus 40 imps with a standard deviation of imp uncertainty (and this includes both uncertainty in guessing the match state as well as volatility of future boards) of 17 imp. Break-even to try for the overtrick is 10.003/11 chance of success when behind to as much as 10.2/11 when far ahead -- you should be microscopically more conservative at VP. Extreme example: Suppose in the artificial extreme that your estimate of 10 imp lead has a standard deviation of only one imp. Then 7.3/11 chance is sufficient to justify taking the risk at VP.

"I don't know how often you play top level bridge" Of course not. That's why I told you what Hamman said. I don't know you either. Nonetheless, thanks for sharing your thoughts. Charles

I agree JLOL that it is not difficult to realize that playing slowly here isn't likely to help declarer. But that largely misses my point. I worry about revealing to my partner. Then we may have the kind of UI situation that no one ever complains about out loud, but that subtly and magically elevates the stodgy second tier pairs above their just station. As one who, at least on paper, aspires to be a class defender, I try to be inferior to Hamman because I think of less in the same time, not because I think of a little less in much more time. Do you recall the decisive hand in the World Championship a few years ago, where Hamman and Soloway stumbled through the defense of 5♦ x'd, only to luckily defeat it in the end when the Italian declarer carelessly played the wrong card from dummy? At trick one Hamman, third to play, gave a signal which on analysis was the wrong one. Why? He didn't think it good form to think too long even in that situation, so he played a card to send the message that was clear to him right away ("Don't underlead."). By the way, on the present hand, on reflection [EDIT] I'd consider the defense of taking the spade and playing a heart. Charles

"I surely didn't like my 2H opening ..." Me neither. As an indication, on hearing the double of 4S I would have a considerable sense of foreboding. "Note that not bidding 2H ... never get to 4H ..." which would make if ... There's always a justification of some sort. The other justification is that it pushed the opponents into a precarious contract, though as it happens it requires genius defense to defeat it. "Stiff club lead is IMO automatic. Ive played a discouraging card wich mean that i can tolerated a switch. Wich can only mean im ruffing D or have the Q." Third hand, especially at trick one, has time to think. Nonetheless the analysis here is garbled. Firstly, it implies that a large club would ask partner, who has just led a singleton, to lead more clubs. Secondly, having shortness OR the Q isn't the same has having shortness. "At trick 2 I think its not too tough to play A♠,A♦ followed by 2 ruffs." As second hand (such as partner here at trick 2) I always feel I should play in tempo. It takes a hell of a defender to take into account declarer's possible heart situation, judge to go counter to the instinct to duck this card combination, and correctly judge the best chance to defeat this contract -- especially all in tempo. Maybe the Garozzos of the world are entitled to expect their partners to think that crisply but from those of us whose own thinking is sometimes fuzzy, it's an unreasonable expectation. Charles

♦ seems too obvious. Can't wait for reasons for other plays. The only contrary arguments I can dream up are -- 1. If pd not void in ♦ then presumably the ♥ lead would be 4th best. That's barely possible but depends on spots we are not told. 2. Is Qx, xx, AQ, AQ1098xx a 3♣ bid? If so, then maybe there's a case that the ♦ return tips a trump position declarer was about to get wrong.

If RHO makes non-vulnerable overcalls based on the (quite correct) criterion that it's fun to bid then obviously finesse through LHO. However it may be argued that there is much more point to overcalling when short in spades. Some hands with short clubs are good for a delayed t.o. double, or better suited to continuing, should partner (meaning LHO) bid spades, having initially passed. I hate trying to read overcaller's mind, but perhaps better players can comment. Anyway, assuming you buy this idea ... If RHO is known to have exactly a ♠xx it is a tossup which way to play clubs assuming you mark RHO with the ♦A. If you can convince yourself that the ♥ overcaller has on expectation 2 or fewer spades, cash the ♣K (and unblock). If LHO drops an honor, test the spades to gain a tiny edge. If and only if RHO has 3 or more spades, play the clubs to be 2-2. Charles

Also, if 3NT makes nearly half the time that already suggests that bidding 3NT must have an edge over passing: It gains when 3NT makes, and also there must be at least a few percent when it gains even though 3NT doesn't make, e.g. the opponents take a phantom save. A deeper analysis would admit that pass doesn't always end the auction. However when we hold 20hcp that may be a small point.

You're right -- it's certainly not a tossup. I forgot that there was bidding. RHO can hardly have both a singleton spade and the diamond king, so ♠A,Q makes no sense.

Gaining if it's a singleton and the clubs break; losing against ♠J9(x). ♠A, K succeeds against 4-1 onside; ♠A, Q against either 4-1 if the ♦ finesse works. Seems like a tossup.

I love these counterintuitive propositions so I hope you're right. If LHO has a singleton trump (22%) you have a chance but you will need to guess who has the ♦K and can guess right >50%. Do you therefore finesse if RHO ruffs & plays spade? If LHO has 2 trumps (40%) you are in reasonable shape if and only if also 3 clubs. If LHO has 3 trumps (26%) you have fair chances. Presumably if LHO ruffs first & plays a heart you will ruff & cash a trump. What happens if LHO has ♠Qxx and ♣x(x)? If LHO has 4 trumps (7%) not much chance single dummy. There are choice in both play and defense, but as I guess the play it adds to about 12% + 3% + 20-23% = 35-38% (larger number assumes they never beat you when LHO has ♠Qxx, ♣xx). For my line, about 0% + 28% + 10% =38%. How do you figure it? How do you plan to play?

I wrote (and calculated) that way because I was comparing Gnasher's line and it was his stipulation. I guess if RHO is ♠xxxxx ♥Jx ♦Kxx ♣Jxx we'd be ok. Are there additional chances? Any significant ones?

Surely the ♦ finesse won't "work" -- LHO will cover leaving you with a small diamond spot to worry about. If you then ruff a heart and cash a spade you'll need 3 rounds of clubs to live -- almost impossible since RHO can discard on the 3rd heart. Even without that discard problem it might be (depending on how you feel about down many) slightly better to finesse the spade regardless, rather than ruff a heart. Given the stipulation that LHO must have 2 or 3 spades to stand a chance: -- Diamond finesse line: The spade finesse is 60%, and you make in 2/3 of those cases (3 spades or long clubs on the right). -- My line: Unless there is some reason why LHO is particularly likely to bid 4♥ only when holding the ♦K, it's 2:1 (on empty spaces) that RHO has it. Playing top spades, top clubs will then succeed 3/4 of that time (losing mainly when RHO has ♠Qxx), plus when LHO has Qx, AKQxxxx, Kxx, x. I think it's nearly 15% better. == Spade, spade, spade might be better still if RHO can be relied on to miss the diamond return with Qxxx, Jx, Kxxx(x), Jx(x) after winning the 3rd spade -- quite an imaginative play.

With either Jx K10xxx Axx xxx or x K10xxx Axxx xxx West should have cashed ♦A. In either case, from his point of view, we could have five spade tricks, and he knows from his partner's discard that there are two diamond tricks to take. The only thing we can infer from West's heart continuation is that he's an idiot. Fair point -- although if West has ♠x declarer is an underdog to hold running spades. I was having trouble figuring out East's ♦ come-on, which is not a play one commonly makes with ♦K7632 over dummy's J108. Maybe it means that East figured out the spades are running if West has the ♦A, and that West will ignore the signal if holding ♠A.

Top spades, clubs. This requires RHO to have the ♦K, which should be a favorite just from the distribution, and LHO to have either ♠Qxx or a doubleton. In other words, it wins against nearly 50% of the 3-3 trump breaks and also most of the 2=4 breaks. If ♣ are 4-0 it still seems a reasonable start. Three rounds of spades (a) slightly increases the success rate against 3-3 spades to 50%, but at the expense of down 4 when LHO has ♠Qxx, and (b ) risks RHO finding the best defense of exiting ♦ rather than club after the 3rd spade when they are 2=4. (b ) is imponderable but even if it is zero I think (a) has a net negative expectation.

Peter Weichsel finessed (correctly) against me in this position once. My partner had defended in a certain fashion. I asked Weichsel how he guessed. "The man's wearing a hat," he said. "A man's wearing a hat, you've got to figure he knows what he's doing." After RHO discards on the 3rd heart, West can see that at least four tricks will be accounted for by hearts, at least three by the AQJ of clubs. If West has 2 or 3 spades then he is playing an awfully deep game if he doesn't also reckon on five rounds of spades. That leaves only one possible trick in diamonds, so why didn't West simply cash his ace? By not doing so he took some risk it could disappear and some risk of the actual lie, that the second defensive diamond trick could disappear. So if you think West is capable of counting tricks you should finesse the spade, and the answer to the question in the subject line is No, this is not a percentage play situation. (Actually the deep game idea doesn't hold water, because even if West holds ♠Jx he cannot know that we have a finessing option.)

Pass. I can see 1♠ but not 1NT which has so little upside without a double stopper (i.e. 3NT won't work out) and a minimum hand besides.

This hand from the Las Vegas Nationals was reported because it came before an appeals committee. You might enjoy deciding whether East can make 4♥ double dummy.[hv=d=e&v=n&n=sj103h4dqj864ca632&w=sq94ha96d10932c1074&e=sk72hkqj108753dk7c&s=sa865h2da5ckqj985]399|300|Scoring: BAM --------- 1♥ dbl 2♥ dbl 4♥ dbl pass 5♦ 5♥ dbl [/hv]

Yes, it helps. Win the SJ and knock out the ace of hearts. West has to play something back... if it's e.g. a club we put the ace up and cash hearts, and now if West started with a 'normal' takeout double we don't need to guess diamonds, instead we play them from the top and if West was originally 4432 have 9 tricks, whilst if he was 4441 he will have to give us a spade trick You're right. The jack can gain by obviating a guess. Assuming that the normal guess would be a diamond finesse, the J would gain when the doubler has AQxx, Axxx, xxx, Kx or AQxx, Axxx, xxxx, x. As against that, I still don't understand the argument that LHO would overcall 1S on Axxxx and subsequently double with Axxxx, Axxx, Jxxx, -- (when the J play loses). Nonetheless, it may be that hands of that sort are less likely than the AQxx sort. (I thought I had posted this yesterday.)