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Thank you for your insightful and comprehensive coverage.

I will tell my partner (besides the results of the poll) that Michael Rosenberg said:

“With Qxx, I only want to unblock it when I KNOW partner has J10x. So, there has to be one card that SHOWS that holding - and it MUST be the jack. Therefore the 10 is either no J or J10 doubleton. This is ‘signaling’ by logic - it overrides the normal agreement.”

So, he (an ex-national player) was the one who gave J from JTx, and I could not drop the Q (at MPs). Afterwards, I told him that if he had given the T, I might have… He said: “Maybe, I am not sure”, and it stopped there.

NOTE: As for UDCA, it just adapts here very well. I am not in a position to claim that it is superior, although I find it slightly more fit with the natural.

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Thank you all for your votes and valuable comments, which have led to some interesting points out of this initially-simple-looking (!) combination.

Let me also express my views (to advocate my own vote), before giving the declarer's hand. NOTE: New votes are still appreciated.

(1) The main question is which ♦ North should insert from JTx: J or T?

As David Burn stated a few times, if pard has ♦JT doubleton, he should play J for sure (imo, too). Then, it seems that he should play T from JTx (since he cannot afford to play the x).

Now, if pard plays T, is it totally clear for me what to do? I am still not sure. As noted earlier, pard may be giving count with Tx, or trying to look like a person who has QT tight. Or is it a matter of how to split?

In addition, pard may think that giving the T denies the J; so, he may play the J from JTx to make me sure that he has the T…

Consequently, I would not know surely what to do if pard played the T.

(2) Declarer has either 4 or 5 (or even 6) ♦s, and it is very likely that the contract cannot be set if I do not unblock ♦Q.

As Bernard Yomtov pointed out, an exception may be that declarer has ♠Qx ♥AJ9 ♦Axxxx ♣Axx. In this case, pard holds ♥QTx, and we may expect him to give ♣s differently (like giving the 6 on the third round instead of 5, or leading a higher spot on the second round). I agree with Michael Hargreaves that partner’s playing all the small ♣s does not imply that he likes ♦s; however, if he had ♥QTx, I think he would not play the smallest ♣ at each turn.

(3) Assuming that it is a make without unblock, unblocking results in an overtrick only, which costs a 1-imp loss (in teams), but probably a lot of MPs (in pairs). Then, since I am not so sure what my partner is doing, I tend to drop ♦Q in IMPs regardless of partner’s ♦ card.

However, to risk that in MPs, I certainly need a better partnership understanding/experience.

(4) As Michael Askgaard mentioned, an UDCA’er has no problem at all. Play J from JTx, T from JT tight; and if you like, falsecard with T from Tx, at no risk.

(5) Here is the declarer's hand: (Michael Rosenberg got it almost right in his thread above.)

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Let me add this:

If we assume that W has 3 ♠'s and E has 6 ♠'s; then, the probability that W has 3 ♥'s (40.72 %) or 4 ♥'s (23.76 %) is 0.645.

Now, cashing ♥AK wins when W has ♥T65, T64, T54, 654 or T654. This makes 21.04 % of that 0.645.

Playing ♥J wins when W has ♥QT6, QT5, QT4, QT65, QT64, QT54, Q654; resulting in 31.22 %.

Dividing these by 0.645, we get, 32.6 % for the first line, and 48.4 % for the other one. (Michael Kopera got exactly the same result for Line 2 using ‘SuitPlay’ in an above thread.)

NOTE: These two are comparable with each other on the precondition that W has 3 ♠'s and 3 ♥'s or 4 ♥'s (this is the whole).

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SUCCESS CONDITIONS

For cashing ♥AK: W has ♥Qx or E has ♥Qx or stiff ♥Q.

NOTE: I assumed that W will always play ♥Q from ♥QT tight on ♥A (as a false-card), and we will NOT finesse vs. ♥T in E.

For starting with ♥J: W has ♥QT or QTx or QTxx or Q654 ——————————— SUCCESS PROBABILITIES

1. With no information about ♠'s distribution in opponents' hands; Cashing ♥AK: 29.94 % Starting with ♥J: 24.84 % (Robert Balas gave the same numbers earlier above.)

2. Assuming that W has 3 ♠'s, so E has 6 ♠'s; that is, W has 10 and E has 7 vacant places.

The calculation is as follows.

Total number of distributions of the remaining 17 cards in vacant places in W and E: 17!/(10!x7!)=19448.

(A) Two ♥'s in W: 12!/(8!x4!) x 5!/(2!x3!) = 495x10=4950. Probability of 2 ♥'s in W=4950/19448=0.25452 (25.45 %).

(B) Three ♥'s in W: 12!/(7!x5!) x 5!/(3!x2!) = 792x10=7920. Probability=7920/19448=0.40724 (40.72 %).

(C) Four ♥'s in W: 12!/(6!x6!) x 5!/(4!x1!) = 924x5=4620. Probability=4620/19448=0.23756 (23.76 %).

Now, success probabilities:

Cashing ♥AK wins in 4 cases of total 10 in (A), 4 of 10 in (B), and 1 of 5 in (C); then,

25.45x4/10 + 40.72x4/10 + 23.76x1/5 = 31.22 %.

Starting with ♥J wins in 1 case of 10 in (A), 3 of 10 in (B), 4 of 5 in (C); then,

25.45x1/10 + 40.72x3/10 + 23.76x4/5 = 33.77 %. ——————————– Note that cashing ♥AK wins in 9 cases, while playing ♥J wins in 8 cases. However, when ♠'s are 3-6, it is quite more likely that W has 4 ♥'s, and the second line wins in 4 cases out of a total of 5 then. ——————————– The line suggested by Nick Hardy is worth considering (esp. when ♠'s are 3-6). That is, cash ♥A and if you see ♥T from E, finesse vs. ♥Q in W.

The probability for this: 25.45x4/10 + 40.72x3/10 + 23.76x2/5 = 31.90 %. ——————————– NOTE 1: Espen: In order to get accurately comparable results for the overall success probabilities of the two lines, it was necessary to assume that W has 3 ♠'s and E has 6♠'s (not that W has 3 or 4 ♥'s). In other words, I ignored W's 3♥ bid, to focus on the calculation with vacant places. With a precondition that W has 3 or 4 ♥'s, similar calculations can be made. But then, I don't think that all cases can be compared meaningfully.

NOTE 2: These (hand) results are consistent with the results obtainable through a web site by Theodore T. Triandaphyllopoulos, ( http://www.automaton.gr/tt/en/bridge.htm ), which can handle vacant spaces. I also intend to verify them using the “Dealer” software (Staveren's) when I have more time next week.

EDIT: Brackets in (A), (B), (C) above did not show up, so I had to rearrange them; and changed wording in NOTE 1.

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For me, too, (as I said) Deep Finesse has been an indispensable bridge companion. Yet, as with almost all technological advancements, not everything it brings inherently is plus, unfortunately.

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Certainly true that Deep Finesse (and similar software) caused double-dummy problems to lose a lot of interest. I have some of those UK bridge magazines, each with a continuing double-dummy corner, which contains magnificent problems. Of course, solving double-dummy problems is quite different from playing at the table, but it has its own merits, fun and attractiveness.

I think, except some actual aficionados, nobody intends to think over double-dummy problems anymore. And it is not much meaningful to ask/pose such problems, since one can get the exact solution just by feeding the hands to a program.

Obviously, this offers an advantage; but the downside of it should not be ignored.

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Say that declarer ducks a ♥, East overtakes his pard's ♥9 with ♥T and normally returns a ♦. Now declarer has to decide who has the ♣T. If he finesses vs. ♣T (I would), now he will go down TWO. That becomes more puzzling!

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That is certainly what I meant by ‘relativity’.

What actually happens at the table may significantly differ from what should happen in our opinion when looking at (analyzing) the hands.

NOTE: Besides, in practice, big errors may happen at the other table (or at our table) and we (or our teammates) may never know about some of those (if we are not in the view-graph). As an aside, it is usually better (for all the parties) not to know everything during the heat of the competition! Real life is far from being perfect.

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Yes; declarer cannot afford to lose a ♣ trick since he needs 5 tricks in ♣'s to reach 9. So, he won ♣K, cashed a ♠, (himself) played ♣9, lost to ♣T, and after a ♦ return, conceded for down 1.

It seems that the lead is much more likely to be from JTx(x) than from Jx(x).

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It is a good idea to put him in 7♠, but please make your pard bid it, so that the pattern would be complete: 1♠, 2♠, … through 7♠.

NOTE: These guys were young (and aggressive) that time, and probably they had the understanding that opening weak 2♠ with an outside A was unacceptable (too good).

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4C was probably at least a mild slam try in Spades, though I am not totally sure. However, some (around here) play the next-higher suit of the transferred M after 2NT opng showing a slam interest in the M and not guaranteeing a ctrl in the suit bid. Even if North did not have a stiff C, he could still bid 4C. And with natural Clubs, he would bid 4H.

(And if it went 2NT-3D-3H- then 3S would show some slam interest in H, without promising a S ctrl. And with natural Spades, he would start with normal Stayman.)

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

I will tell my partner (besides the results of the poll) that Michael Rosenberg said:

“With Qxx, I only want to unblock it when I KNOW partner has J10x. So, there has to be one card that SHOWS that holding - and it MUST be the jack. Therefore the 10 is either no J or J10 doubleton.

This is ‘signaling’ by logic - it overrides the normal agreement.”

So, he (an ex-national player) was the one who gave J from JTx, and I could not drop the Q (at MPs). Afterwards, I told him that if he had given the T, I might have… He said: “Maybe, I am not sure”, and it stopped there.

NOTE: As for UDCA, it just adapts here very well. I am not in a position to claim that it is superior, although I find it slightly more fit with the natural.

Okan Zabunoglu

Let me also express my views (to advocate my own vote), before giving the declarer's hand. NOTE: New votes are still appreciated.

(1) The main question is which ♦ North should insert from JTx: J or T?

As David Burn stated a few times, if pard has ♦JT doubleton, he should play J for sure (imo, too). Then, it seems that he should play T from JTx (since he cannot afford to play the x).

Now, if pard plays T, is it totally clear for me what to do? I am still not sure. As noted earlier, pard may be giving count with Tx, or trying to look like a person who has QT tight. Or is it a matter of how to split?

In addition, pard may think that giving the T denies the J; so, he may play the J from JTx to make me sure that he has the T…

Consequently, I would not know surely what to do if pard played the T.

(2) Declarer has either 4 or 5 (or even 6) ♦s, and it is very likely that the contract cannot be set if I do not unblock ♦Q.

As Bernard Yomtov pointed out, an exception may be that declarer has ♠Qx ♥AJ9 ♦Axxxx ♣Axx. In this case, pard holds ♥QTx, and we may expect him to give ♣s differently (like giving the 6 on the third round instead of 5, or leading a higher spot on the second round). I agree with Michael Hargreaves that partner’s playing all the small ♣s does not imply that he likes ♦s; however, if he had ♥QTx, I think he would not play the smallest ♣ at each turn.

(3) Assuming that it is a make without unblock, unblocking results in an overtrick only, which costs a 1-imp loss (in teams), but probably a lot of MPs (in pairs). Then, since I am not so sure what my partner is doing, I tend to drop ♦Q in IMPs regardless of partner’s ♦ card.

However, to risk that in MPs, I certainly need a better partnership understanding/experience.

(4) As Michael Askgaard mentioned, an UDCA’er has no problem at all. Play J from JTx, T from JT tight; and if you like, falsecard with T from Tx, at no risk.

(5) Here is the declarer's hand: (Michael Rosenberg got it almost right in his thread above.)

♠J9 ♥AQT6 ♦A965 ♣A87

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

If we assume that W has 3 ♠'s and E has 6 ♠'s; then, the probability that W has 3 ♥'s (40.72 %) or 4 ♥'s (23.76 %) is 0.645.

Now, cashing ♥AK wins when W has ♥T65, T64, T54, 654 or T654. This makes 21.04 % of that 0.645.

Playing ♥J wins when W has ♥QT6, QT5, QT4, QT65, QT64, QT54, Q654; resulting in 31.22 %.

Dividing these by 0.645, we get,

32.6 % for the first line, and

48.4 % for the other one.

(Michael Kopera got exactly the same result for Line 2 using ‘SuitPlay’ in an above thread.)

NOTE: These two are comparable with each other on the precondition that W has 3 ♠'s and 3 ♥'s or 4 ♥'s (this is the whole).

Okan Zabunoglu

For cashing ♥AK:

W has ♥Qx or E has ♥Qx or stiff ♥Q.

NOTE: I assumed that W will always play ♥Q from ♥QT tight on ♥A (as a false-card), and we will NOT finesse vs. ♥T in E.

For starting with ♥J:

W has ♥QT or QTx or QTxx or Q654

———————————

SUCCESS PROBABILITIES

1. With no information about ♠'s distribution in opponents' hands;

Cashing ♥AK: 29.94 %

Starting with ♥J: 24.84 %

(Robert Balas gave the same numbers earlier above.)

2. Assuming that W has 3 ♠'s, so E has 6 ♠'s;

that is, W has 10 and E has 7 vacant places.

The calculation is as follows.

Total number of distributions of the remaining 17 cards in vacant places in W and E: 17!/(10!x7!)=19448.

(A) Two ♥'s in W: 12!/(8!x4!) x 5!/(2!x3!) = 495x10=4950.

Probability of 2 ♥'s in W=4950/19448=0.25452 (25.45 %).

(B) Three ♥'s in W: 12!/(7!x5!) x 5!/(3!x2!) = 792x10=7920.

Probability=7920/19448=0.40724 (40.72 %).

(C) Four ♥'s in W: 12!/(6!x6!) x 5!/(4!x1!) = 924x5=4620.

Probability=4620/19448=0.23756 (23.76 %).

Now, success probabilities:

Cashing ♥AK wins in 4 cases of total 10 in (A), 4 of 10 in (B), and 1 of 5 in (C); then,

25.45x4/10 + 40.72x4/10 + 23.76x1/5 = 31.22 %.

Starting with ♥J wins in 1 case of 10 in (A), 3 of 10 in (B), 4 of 5 in (C); then,

25.45x1/10 + 40.72x3/10 + 23.76x4/5 = 33.77 %.

——————————–

Note that cashing ♥AK wins in 9 cases, while playing ♥J wins in 8 cases. However, when ♠'s are 3-6, it is quite more likely that W has 4 ♥'s, and the second line wins in 4 cases out of a total of 5 then.

——————————–

The line suggested by Nick Hardy is worth considering (esp. when ♠'s are 3-6). That is, cash ♥A and if you see ♥T from E, finesse vs. ♥Q in W.

The probability for this:

25.45x4/10 + 40.72x3/10 + 23.76x2/5 = 31.90 %.

——————————–

NOTE 1: Espen: In order to get accurately comparable results for the overall success probabilities of the two lines, it was necessary to assume that W has 3 ♠'s and E has 6♠'s (not that W has 3 or 4 ♥'s). In other words, I ignored W's 3♥ bid, to focus on the calculation with vacant places. With a precondition that W has 3 or 4 ♥'s, similar calculations can be made. But then, I don't think that all cases can be compared meaningfully.

NOTE 2: These (hand) results are consistent with the results obtainable through a web site by Theodore T. Triandaphyllopoulos,

( http://www.automaton.gr/tt/en/bridge.htm ),

which can handle vacant spaces. I also intend to verify them using the “Dealer” software (Staveren's) when I have more time next week.

EDIT: Brackets in (A), (B), (C) above did not show up, so I had to rearrange them; and changed wording in NOTE 1.

Okan Zabunoglu

Okan Zabunoglu

I have some of those UK bridge magazines, each with a continuing double-dummy corner, which contains magnificent problems. Of course, solving double-dummy problems is quite different from playing at the table, but it has its own merits, fun and attractiveness.

I think, except some actual aficionados, nobody intends to think over double-dummy problems anymore. And it is not much meaningful to ask/pose such problems, since one can get the exact solution just by feeding the hands to a program.

Obviously, this offers an advantage; but the downside of it should not be ignored.

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

What actually happens at the table may significantly differ from what should happen in our opinion when looking at (analyzing) the hands.

NOTE: Besides, in practice, big errors may happen at the other table (or at our table) and we (or our teammates) may never know about some of those (if we are not in the view-graph). As an aside, it is usually better (for all the parties) not to know everything during the heat of the competition! Real life is far from being perfect.

Okan Zabunoglu

It seems that the lead is much more likely to be from JTx(x) than from Jx(x).

Okan Zabunoglu

NOTE: They were my teammates, and 6♠ made. Yet, I did not note the full deal, since I found the South's bidding more interesting.

Okan Zabunoglu

NOTE: These guys were young (and aggressive) that time, and probably they had the understanding that opening weak 2♠ with an outside A was unacceptable (too good).

Okan Zabunoglu

(And if it went 2NT-3D-3H- then 3S would show some slam interest in H, without promising a S ctrl. And with natural Spades, he would start with normal Stayman.)