[inquiry]

 ritong, on 2012-July-23, 04:06, said:

imho, answer is : we are 95% sure.. east might have concealed the spade 3 from the start, seeing the whole thing coming and simulating club 10xx- that would be a very deep play, for sure- now, it might be an answer to ben's question: rodwell finessed because it was possible that east was out of clubs.

Couldn't he win the jack in this case, and play a low club back to his six? I mean the ♣8 is gone, the ♣7 is gone, so the six forces the ten? I think the carding in clubs alone remove this possibility. Then there is the lack of the double of 3NT.

[inquiry]

 TWO4BRIDGE, on 2012-July-23, 08:29, said:

I too think the ♣3 discard by East was unusual ( at trick 7 ) when East had a "safe" discard of one of the ♥.
Of course the lead of the ♦K next ( trick 8 ) "forces" East to then discard a ♣ coming down to the 3 card ending .
So, Rodwell's thinking goes: Which discard scenario is more likely for East holding the stiff ♣10.

I think I remember watching this hand ( can't remember one day from the next ) and was amazed he found the ♣10.

I am thinking along these lines as well, but with an opposite view.

At trick 7, NUNES sees 8 tricks for Rodwell. 4♦ (opening lead marks south with aj109), 2♥, 2♣. If south has the ♣10, hand is over. So he mentally places that card with West. Now he has to create an image that he does hold the ♣10. How would you do it against a duffer? You would pitch winners to hold onto clubs. But then Rodwell will endplay you, but not a duffer. How would you play against an advanced player? You would discard the ♠Q, then a heart, finally a club (and if you were a coffeehouser, you would pitch that club verrrrrrry slowly.

But you are playing against one of the best card readers it becomes a game of psychology. If East had the ♣10, he would want to discard as if he didn't. But he knows that Rodwell knows, he is capable of discarding like he didn't have the club ten, so if he discarded like he didn't have it, he would have it. So.... well you get the idea that becomes a spy-vs-spy situation (forgive the MAD magazine reference).

So what I thought is you would go back to the "correct" line of play, and play for the drop (the math is simple, and favors the drop. There are three "slots" in Eas't clubs that can be the "ten", and only two in West's hand. Without fancy math, and ignoring the actual spot cards played (treating 8753 as true equals) here are the combinations...

West with the ten: KT3, KT5, KT7, KT8 (this is C(4,1) = 4)

East with the ten: K87, K85, K83, K75, K73, K53 (this is C(4,2) = 6).

As you can see, there are six combinations where East has the 10 to four where West has the ten: that is 6 to 4 or 3 to 2 (the number of vacant spaces. In a spy-vs-spy situation based upon carding, I would have thought Eric would go with the percentage play and play for the drop (so did all the commentator). So when he finessed the ten after a lot of thought, I knew there had to be something we all missed. I still don't know what it is, but I am certain it is there. Whatever the clue was, it doesn't seem to be related to West's carding. His one discard was an honest and sensible diamond two.

[keylime]

My gut feeling is that is was a table feel decision, based on the potential that he felt he was behind to try for any swing.

[quiddity]

 sfi, on 2012-July-23, 02:02, said:

But there are two ways that West can play their spots with K85 in the first instance, which is where restricted choice comes into it.

West is marked with the 8 after East signals with the 7 on the first round, so he is just playing the card he is known to hold.

[inquiry]

 keylime, on 2012-July-23, 10:09, said:

My gut feeling is that is was a table feel decision, based on the potential that he felt he was behind to try for any swing.

This was board 22 of a 64 board match (assuming no board thrown out like they did stupidly in the semi-finals). At this time, the match was very tight. So surely this was not a state-of-the-match play. The timing of the play (up to which club to play from dummy) seemed normal to me. It is true, both defenders would quickly determine the possession of the ♣10 was key to the hand, surely no later than when the ♥A won the fourth trick. I make some assumptions here... The ♦8 was top of nothing, the ♣7 was count, and the ♥T was probably s/p for spades, but south must have ♥Q for continuing hearts.

[keylime]

True but as you would agree, sometimes the score doesn't reflect the feeling at the table. Even tho I have the highest admiration for Rodwell in general, I can't quite put my finger on this being a technical play versus a feel one.

[inquiry]

 keylime, on 2012-July-23, 10:36, said:

True but as you would agree, sometimes the score doesn't reflect the feeling at the table. Even tho I have the highest admiration for Rodwell in general, I can't quite put my finger on this being a technical play versus a feel one.

Rodwell knows sooooooo much more than me about bridge and table feel and mathematics. When he finessed the nine, the knew the combination math. So something at the table tipped him off. To give up 3-to-2 odds, it was something major. Perhaps ritong is right, he factored in four clubs with west (making him 2-2-5-4) but then playing the Jack also works due the wasting of the ♣8. A mathematician would have gone down here. From the comments of the expert commentators, many (most) experts would have gone down. I would have gone down. I accept all of that. I still don't see how Rodwell correctly played the club suit.

There has to be something (magic?) that tipped him off. Maybe it is as simple as with T75 East probably would not have signaled correct count (at the time he played the club, the ownership of the ten of clubs might not have been as clear as critical). That is what I am leaning towards now as key.

[keylime]

Being strong in math myself I would have gone down here, albeit faster. Leads to a larger question now - how has Meckwell vs. Fantunes been in their history, overall? Might have influenced Eric in his line (under guises of knowing your customers).

[TWO4BRIDGE]

I got my trick #'s incorrect in my post #20 ... fixed now.

I think if Rodwell pegged East with the ♠Ace and East didn't discard it right after Rodwell discarded the ♠ K from dummy ( North ), then failure to do so, and previously discarding a ♣ instead, seems a "mistake" by East..... unless East doesn't know if Rodwell has the ♠3 or not in hand ????
If East had ♣ 10 x , then his final last 3 cards ( after discarding the ♠Ace ) could have been :
♥9
♣10 x

[aguahombre]

The answer obviously was not easy for Eric either. He took longer to decide than it normally takes him to bid and play two boards.

He certainly was planning this "read" way back at the beginning as he led the club Queen; so, anything extraneous happening way back then might have helped add to his decision at the end.

[benlessard]

Quote

With ♠AQJ93 East would probably double for the lead. It would have been a good double even on the actual deal.

I think making a lightning double with this is very dangerous. 3Nt xx & giving the contract away is just too likely.

The possibility S are 25 is IMO the only reason to finesse

[aguahombre]

 TWO4BRIDGE, on 2012-July-23, 08:29, said:

Now on trick 10, Declarer leading the good ♦9, and discarding the ♠K, East could safely
discard the ♠Ace ( if he has ♣10 x; but since he doesn't discard the ♠Ace but a ♣ instead, that seems to be the clue. [ End of EDIT ] .

Not safe at all; ref: Injuiry's post #3 and the heart exit by declarer for an endplay in clubs. North must have stiffed the club ten if he has it. That is the "restricted" choice , so Erick still had the same problem and only he can really answer whether it was a raw quess or an educated guess.

[TWO4BRIDGE]

 aguahombre, on 2012-July-23, 11:38, said:

Not safe at all; ref: Injuiry's post #3 and the heart exit by declarer for an endplay in clubs. North must have stiffed the club ten if he has it. That is the "restricted" choice , so Erick still had the same problem and only he can really answer whether it was a raw quess or an educated guess.

duh... that's right.... East is end-played in ♣ if he discards the ♠Ace:
♥9
♣ 10 x

So, as you say, he would have to blank the ♣10 ... IF he has it:
♠A
♥9
♣10

[SteveMoe]

 inquiry, on 2012-July-22, 22:28, said:

Well, I don't know how he got it right. I don't understand how restricted choice would apply here. Perhaps you could explain the implications of restricted choice in this situation, and if West play of the 8 has implications for the 5, what implications did East's discard of the 7 from possible 7 & 5 (or 10 & 7)?

Isolating on West:
K 10 8
K 8 5
for the purposes of this hand, 5=8.
Playing the 8 halves the likelihood that West holds the 5.

or

Perhaps R judged that 3N was likely at the other table, and needing a swing, decided to take the anti-percentage a prior odds choice to finesse (4 cases to 6 cases).

[aguahombre]

 inquiry, on 2012-July-23, 10:32, said:

This was board 22 of a 64 board match. At this time, the match was very tight. So surely this was not a state-of-the-match play.

 SteveMoe, on 2012-July-26, 00:15, said:

Perhaps R judged that 3N was likely at the other table, and needing a swing, decided to take the anti-percentage a prior odds choice to finesse (4 cases to 6 cases).

Maybe we would feel swingy --- seeded 70 out of 64 if we were there. I don't think Rodwell would have those thoughts in this match at that time.

[ColdCrayon]

 inquiry, on 2012-July-22, 19:31, said:

Other relevant info is West had the ♣K (you can follow the early play in the diagram at the top of the post). It does seem like vacant space favors East having the ♣10, or does it?[/hv]

Vacant space suggests W has the card, not East. You can't really infer anything by east's showing out on the diamond lead, since W. is almost always going to lead his longest suit; so the fact that West has long diamonds means absolutely nothing. West is forced to lead. The strategy is for him to lead from his longest suit. Unless he's got a totally flat hand he's going to have at least one longish suit every time; this is why you generally disregard opening leads when figuring vacant space, otherwise you're always going to be putting missing cards in East's hand.

Anyway, West almost certainly has the T. You already know how the red cards are distributed, plus you know East has the A♠ (otherwise W gets into the bidding), so finessing is over 50%, while playing for the drop against 3 cards is 2-1 (I think).

My conclusion might be naive, but one thing I'm sure of is that restricted choice has no bearing here, since the 8 and 5 aren't equivalent: one can capture south's 6, and one can't.

full disclosure - I didn't see the hand. I missed the whole match, I've been teaching myself LISP.

[ColdCrayon]

To put it even more naively, there are equal slots in both hands. Why? Because West has 7 red cards, and east has six and the A of spades. So it's a play of standard technique.

E-W holds the second highest club plus two others; you're in a finessing position. It's analogous to a 10 card fit missing the K. How do you play that? You lead small to the queen. The percentages are extremely close, so Rodwell is definitely going to review the whole hand looking for clues, but unless he's got a reason to believe east is holding the ten, he'll finesse (West playing the King earlier doesn't make it more or less likely that he has it - in that spot, he'll cover the Queen if he can - if he can't, he'll play something else, but he'll never play the T. And neither will East.. I believe there's some over-thinking on this problem going on.

Was a similar position reached on the other board(s)? How did they play it?

[benlessard]

Quote

I believe there's some over-thinking on this problem going on.

I agree with this quote but i disagree with all the rest

If you know for sure that the distribution is 3=2=5=3 for West and 4=6=1=3 for East and WEST has the K of clubs , east going to have the T of clubs 60% and West 40% since hes got 3 available clubs to West 2 clubs. This is assuming that EAST would discard/play from T73 and 73 at random.

However note that if East has 5S hes going to discard exactly the same to induce declarer to go wrong. However if the hand are

Tx
A7
87xxx
KTxx

he might have led a club not a diamond (but probably not since 1D is catch all)

[ColdCrayon]

 benlessard, on 2012-July-27, 19:46, said:

I agree with this quote but i disagree with all the rest

If you know for sure that the distribution is 3=2=5=3 for West and 4=6=1=3 for East and WEST has the K of clubs , east going to have the T of clubs 60% and West 40% since hes got 3 available clubs to West 2 clubs. This is assuming that EAST would discard/play from T73 and 73 at random.

However note that if East has 5S hes going to discard exactly the same to induce declarer to go wrong. However if the hand are

Tx
A7
87xxx
KTxx

he might have led a club not a diamond (but probably not since 1D is catch all)

But you don't know that clubs are divided 3-3 (do you?) Also, West's play of the King doesn't make east more likely to have the T: if W doesn't duck, you can divide his clubs into two categories 1) K♣, and 2) everything else. If you put the T in his hand, he plays the King. Take the T out, he still plays the King. Put the T in east's hand, he plays a spot. Take it out, he plays a spot. So how can anything be inferred there? This is what I meant by overthinking: it's taking the principle of restricted choice and stretching it so far that Thomas Bayes is rolling over in his grave. Just make the standard "suit-combination" play, which is to finesse.

[edit] You could even make the opposite argument, that holding Txx, east has a non-zero chance of playing second-hand high, whereas after the queen is played, West will never put an honor UNDER an honor. Neither side played the ten, but since only east had a chance to play it, it follows that West is more likely to have it.
Nvm, forgot who was on lead on that hand.

[gnasher]

 ColdCrayon, on 2012-July-27, 20:00, said:

But you don't know that clubs are divided 3-3 (do you?)

The premise of Inquiry's post, and of almost all the comments in this thread, is that you do, in fact, know that clubs are 3-3.

If you think that the shapes might be 2254-5512, then yes that changes the odds. To determine whether it changes the odds sufficiently, you'd need to estimate the probability that the shapes were like this.

Quote

Just make the standard "suit-combination" play, which is to finesse.

The standard play of this suit assumes that there is, and was, no endplay possibility. On this occasion, we have two choices: play for a layout where the finesse will work, or play for a layout where the endplay would have worked.