[fred]

---

West leads a club against your 7NT.

I think I found a 100% line for 13 tricks, but there may exist additional lines that are just as good.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[Hanoi5]

I was sent this hand today and I think the 'best' solution is to cash a spade from dummy, cash diamonds and hearts finishing in hand and then cash clubs throwing red cards from dummy. If nothing interesting has happened (no heart Queen, no diamond Jack, no spade discard or discard on the high spade) there's only one opponent's hand that could have held 4 spades originally and having:

♠KT9


♠A87

You just play accordingly (If West could have it you cash the Ace, if East could have it you cash the King)

I think what's interesting about the hand is that everybody thinks a squeeze can be tried but I think it usually fails.

[fred]

Hanoi5, on 2012-February-23, 19:16, said:

I was sent this hand today

My too - by another Venezuelan actually, my friend Steve Hamaoui (a great player who I think is now living somewhere in Europe).

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[ron_ron]

This hand (with irrelevant changes to some spots and honors) also appears on p66 of Rosenberg's book "Bridge, Zia... and Me".

[Hanoi5]

fred, on 2012-February-23, 19:25, said:

My too - by another Venezuelan actually, my friend Steve Hamaoui (a great player who I think is now living somewhere in Europe).

Italy, yes.

I just saw a flaw in my solution. Maybe it's a squeeze?

[bluecalm]

I have a feeling we need to cash some clubs and plan the play depending on to how many of them RHO follows but for now I don't see a solution.

[AyunuS]

You can do it using a finesse guaranteed. Cash all the clubs, while dropping the 3 9s. This lets you keep a finesse ready on 3 different suits. Now, cash the spade K and diamond K. They each have 7 cards left, and they still need a guard in each suit, or it's yours. To have a guard in each suit, the remaining spades much be 3/0, and they can't be anything else because they must start 4/1 or else you'll see the void upon cashing the K and be able to finesse the other side. The diamonds can't be 4/3/3/3, or else you'll just be able to take with the last one. Therefore, you will be guaranteed to notice someone out of diamonds the 3rd time you play it. Assume the player that does NOT have diamonds has the spade and draw the big spade you don't need in order to finesse that opponent. If the person without diamonds is indeed the one with the spades, just finesse that player. If not, then the same person has the diamonds and spades, and in fact, every single diamond and spade left that you don't have, and thus can't have the hearts. Draw the rest of the big cards you have on spades, then draw the big heart from the side not needed to finesse the heart, then finesse the heart. You only need the 1 extra trick from this finesse, and then it's all locked up and you make it.

Edit: I realized a flaw. Do the diamonds before the clubs and then drop the 9 of diamonds when you saw the diamonds were not 4/3/3/3, and if they were, then you just take with the 9 and it's easy from there.

[benlessard]

Im going down if East has Jxxx,Qx(x),Jxxx,xx(x) I would be very surprised if there is a 100% line.

[Hanoi5]

Ok, here goes.

You grab the first club and rattle all of them tossing hearts from dummy. Now you know how clubs break.

A heart to the singleton Ace on the table and a spade honour from it too.

KQ of diamonds and a diamond to the Ace in hand. Now you know how diamonds are divided and whether the Ten is cashable.

If you don't have thirteen tricks by now you play your heart king and throw the diamond ten (unless West had the J and threw it on the heart).

Now the only hand that threatened my counting-not-squeeze solution was x(???)-xx(???)-Jxxx-xxx with East but by playing all the clubs first you get him to give you the count in that suit first.

I think this is the 100% line. There's sort of a squeeze by leaving the ♦T on the table, but if West wasn't the one guarding diamonds you have counted the hand.

This made me wonder, is there a highest number of cards that have to be seen in order to get a complete count of a hand?

[fred]

Hanoi5, on 2012-February-24, 06:22, said:

Ok, here goes.

You grab the first club and rattle all of them tossing hearts from dummy. Now you know how clubs break.

A heart to the singleton Ace on the table and a spade honour from it too.

KQ of diamonds and a diamond to the Ace in hand. Now you know how diamonds are divided and whether the Ten is cashable.

If you don't have thirteen tricks by now you play your heart king and throw the diamond ten (unless West had the J and threw it on the heart).

Now the only hand that threatened my counting-not-squeeze solution was x(???)-xx(???)-Jxxx-xxx with East but by playing all the clubs first you get him to give you the count in that suit first.

I think this is the 100% line. There's sort of a squeeze by leaving the ♦T on the table, but if West wasn't the one guarding diamonds you have counted the hand.

This made me wonder, is there a highest number of cards that have to be seen in order to get a complete count of a hand?

Well done! This is essentially the same solution that I came up with (though I played my tricks in a different order, cashing the Ace of hearts and 3 diamonds ending in my hand before playing the clubs, discarding hearts, and finally the King of hearts discarding dummy's last diamond).

And yes I agree it is some kind of weird non-material squeeze. Whichever defender has the diamond guard is squeezed into giving you the information you need to play spades for four tricks.

There may be some other 100% line available, but so far I have not come up with any (not that I have been trying very hard).

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[Trumpace]

Very interesting Hand.

I think the following works (didn't read what the others wrote, so pardon me if there is a repeat). (EDIT: I see Hanoi and Fred have a similar line).

Win the club, spade to K, cash heart A, Diamond K,Q and then A. Then play clubs throwing 3 hearts. Play heart K throwing the diamond (if not a winner already).
Basically, we have cashed our winners in clubs, diamonds and hearts, one spade for 10 tricks and are left with QT9 in North opposite Axx in hand.

At this point, I claim that we have enough information to play spades correctly.

This is how I tried to prove it without a (more)cumbersome case analysis:

There are 21 cards of clubs+diamonds+hearts outstanding, and so at least one of the opponents must have 10 or more cards in those three. That opponent cannot be the one who holds four spades.

Now, if we know how the clubs, diamonds and heart suits split, we will have a perfect count of the hand and know exactly which (second) spade honour to cash(first being the K or Q in North) and who to finesse.

Thus, let us assume for the moment that there is no singleton heart about, and each has at least two hearts. We play the heart AK, so we can confirm that.

The diamond and club suits are such that by playing 3 rounds of diamonds and 4 rounds of clubs, we will exactly know how they split. Now there are 14 clubs and diamonds outstanding. So, if any opponent has at least 8 or more of the minors, then that person has at least 10 cards in minors+hearts, and so we can guess spades.

So assume each person has exactly 7 cards in the minors.

Now consider the guy with 4 or more diamonds (if they split 3-3, the T will be a winner). Since the only problem case is when he has exactly 7 cards in the minors, he has 3 or less clubs.

On the last club, he cannot un-guard the diamond stopper (we haven't cashed diamonds yet) and is forced to throw a heart (if he throws a spade, we are done).
Now when we cash the heart K on trick 10, if he throws a diamond, then we know how the hearts split, and have an exact count.

So we can assume he follows to the heart K. Thus he will have at least 3 hearts and exactly 7 cards in the minors so no more than 3 spades. We know how to play spades in that case.

[phil_20686]

Clearly the contract is only ever in danger if east has four diamonds. If west has four diamonds I have a trivial double squeeze (positional).

I cannot squeeze east in the red suits so any 100% line must involve squeezing east in spades and diamonds. Suppose then I start with ace of spades K of spades, rho has four spades. So I cash the third spade two diamonds ending in hand and cash a couple of clubs pitching a spade and a heart. Now the hand looks like

---

and I am cold double dummy. If east has two diamonds, I simply cash two hearts and the club ace pitching a heart. If he has no diamonds I can cash the diamond K and cross back to the heart K to play the last club for a double squeeze.

I can combine my chances in lots of almost cold ways, but I do not see anyway to be 100% to read the position.

[benlessard]

Very nice.

[Fluffy]

benlessard, on 2012-February-24, 05:25, said:

I would be very surprised if there is a 100% line.

I would be very surprised if there wasn't one, I wonder if you just missread the OP or just ignored it:

fred, on 2012-February-23, 18:48, said:

I think I found a 100% line for 13 tricks, but there may exist additional lines that are just as good.

[phil_20686]

fred, on 2012-February-24, 09:45, said:

Well done! This is essentially the same solution that I came up with (though I played my tricks in a different order, cashing the Ace of hearts and 3 diamonds ending in my hand before playing the clubs, discarding hearts, and finally the King of hearts discarding dummy's last diamond).

And yes I agree it is some kind of weird non-material squeeze. Whichever defender has the diamond guard is squeezed into giving you the information you need to play spades for four tricks.

There may be some other 100% line available, but so far I have not come up with any (not that I have been trying very hard).

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

This is very clever - I didn't see how to tell apart 1525-4243, and 4225-1543, but now I see the discard on the fourth club forces the defense to reveal the heart count. Very clever.

I sent my dad Adventures in cardplay for christmas, was telling him on the phone the other day that I had never seen a non-material squeeze in a real hand.

[Flameous]

There is a similar deal in last deal of the week. They are readable on bbo so I'd suggest everyone to check it up if haven't already done so.

I was pretty sure this was a nonmaterial squeeze when I saw the deal but I couldn't figure it out. My head explodes when I have to count all suits in all combinations