How did Rodwell find the club Ten?

[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]

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]

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]

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 :rolleyes: , so Erick still had the same problem and only he can really answer whether it was a raw quess or an educated guess.

[TWO4BRIDGE]

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 :rolleyes: , 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]

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]

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.

 

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]

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]

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]

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]

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.

 

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.