Suitplay

[kgr]

From suitplay help text:

When comparing two lines of play at matchpoints, it only matters how often one

line wins over the other, and how often it loses. It does not matter by how

many tricks one line wins over the other. One line is better than another line

if it wins more often than it loses. The "mp-best" line determined by SuitPlay

wins more often than it loses when it is compared with any other possible line

of play for the specified suit combination.

In rare cases there does not exist a best matchpoint line. This is the case

when every possible line of play can be improved. The improved line can be

improved again, and so on. Try for instance AJ98 - K32.

What is meant by this? I would think that one line is always best percentage?

[helene_t]

Yes, it's possible. Suppose there are three (approximately) equally likely splits, say X, Y and Z, and declarer can chose three lines, say A, B and C. Now suppose the number of tricks made in case of split X, Y and Z, respectively, are:

A: 7-8-9

B: 8-9-7

C: 9-7-8

 

You see that B has a 2/3 chance of winning over A, while C has the same odds against B and A the same against C.

[Guest Jlall]

Yes it's like rock-paper-scissors where I tell you to go first.

[jdonn]

I understand the concept, but I don't suppose someone would be kind enough to do the analysis on AJ98 opposite K32? I just want to see it without having to figure it out myself.

[han]

Do you read the bridge world Josh? There were several pages on it some months ago.

[gwnn]

can somebody maybe outline the basic ideas on why this has no optimal play? I'm really really curious here.

[brianshark]

Line A (Best for 4 tricks) = cash K finesse J, if it loses finesse 9

Line B (Best for 3 tricks) = cash A cash K lead to J

Line C = run J, if it loses cash K finesse 8 (if covered double finesse)

Line D = finesse J, if it loses run 9

Line E = finesse 9, if it loses run J

Line F = cash K finesse 9, if it loses cash A

Line G = run J, if it loses run 9 (if covered double finesse)

Line H = finesse J, if it loses run 9

 

It says for matchpoint analysis, A<E<F<G<H<C<D<A

[dburn]

Suitplay is a remarkable program that can help players at all levels to improve their handling of many common situations that arise at the table. It assumes, as did Roudinesco in his magnum opus the Dictionary of Suit Combinations that the defenders will play optimally, false-carding when necessary and presenting you with as many losing options as they can.

 

Certain apparently simple combinations "break" it by causing it to run out of memory before being able to "solve" a combination completely; interestingly, it does not "know" how to play

 

Q82

 

A1063

 

to best advantage, nor will it cope when the six is replaced by the five, but when the six is replaced by the four, it can just about manage. Exactly why this should be I do not yet know, and this is something of a humbling experience.

 

Only the other day, several very good players (much better than I) were heatedly debating the correct handling of this simple combination:

 

A102

 

Q6543

 

for four tricks. One was adamant that you should start with a low card to the ten; another equally convinced that you should start with a low card to the queen, then finesse the ten if the queen loses to the king. Two others were of the opinion that it did not matter; both lines were equally good.

 

Sizable wagers were struck, and I was appealed to for a verdict. It surprises me not a little that very many of the world's truly great players do not know the technically correct answer to questions such as this, nor do they care.

 

Of course, it may be they know that the percentage difference between various reasonable lines is often negligible, and they rely on their judgement and table presence rather than their knowledge of the odds against optimal defence when they decide how to play a suit or a hand. Certainly, one of the players involved in the discussion was convinced that he would unerringly select the winning line at the table, and "I don't care what the odds say."

 

But as an exercise for those whose judgement is not quite as good as Zia's, try these questions without artificial aids. Assuming that you have adequate entries to both hands, and that loss of tempo or control is not an issue:

 

What is actually the best line for four tricks with this combination, assuming you are playing against robots who will defend correctly in perfect tempo?

 

What is the best line for three tricks?

 

At matchpoints, why would you not follow any of these lines, and what line would you follow instead?

[xcurt]

Only the other day, several very good players (much better than I) were heatedly debating the correct handling of this simple combination:

 

A102

 

Q6543

 

for four tricks. One was adamant that you should start with a low card to the ten; another equally convinced that you should start with a low card to the queen, then finesse the ten if the queen loses to the king. Two others were of the opinion that it did not matter; both lines were equally good.

 

 

Some edits thanks to later posters especially gnasher.

 

Interesting that none of your experts mentioned the third line, cashing the ace first.

 

If the suit is 5-0 you cannot take 4 tricks.

 

If the suit is 4-1, you can take 4 tricks if RHO has stiff K (cash the A, come to hand, lead to the T), or 1 4-1 break. That looks like the only way to take 4 tricks when the suit is 4-1 assuming we don't divine to finesse the T after (starting in dummy) x-K-x-x.

 

If the suit is 3-2, then we get the following.

 

Low to the T wins whenever the J is in LHO's hand, plus whenever the K is doubleton (Jx {3 cases}, Jxx {3 cases}, xxx {1 case}, KJ {3 cases}, KJx {3 cases}, Kx {3 cases}). Thats 14 3-2 breaks, or 47.5%.

 

Low to the Q wins whenever the K in in RHO's hand, plus whenever the J is in LHO's hand. Specifically

we pick up RHO holding (Kany {4 cases}, Kanyany {6 cases}, xxx {1 case} and xx {3 cases}). That's 14 3-2 breaks or 47.5%.

 

Cashing the A wins picks up the first 11 3-2 cases listed above for low to the Q, plus Jx {3 cases}, plus the one 4-1 case, totaling just over 50%. I'm getting more appreciation for the work J-M Roudinesco did!

 

I could well have missed a case or two here, I probably need my head examined for trying to work this out. I used to own a copy of Roudinesco but no more. Unfortunately it seems to be out of print. If someone knows where to come by a copy, please let me know.

 

By the way, such preference cycles ( eg A > B > C > A ...) are known as Condorcet cycles in the literature on voting.

 

curt

[brianshark]

Surely the fact that the program can't tell the best matchpoint line doesn't mean that one doesn't exist rather that the algorithm for computing it is flawed or that they are all equal.

 

I would have thought that any line yields:

min+1 tricks x% of the time

min+2 tricks y% of the time

...

max tricks z% of the time

 

And you can total this to get the trick taking expectency of this particular line and repeat for each line.

 

There must be some winner, even if there are just many joint winners.

 

I don't get it.

[dburn]

No one said it was easy. Whereas transitivity is intuitive (if A is taller than B is and B is taller than C, then A must be taller than C), non-transitivity is distinctly counter-intuitive (if A always beats B at tennis, and B always beats C at tennis, then A might always lose to C at tennis).

 

Yet Justin had it exactly right: rock always beats scissors, scissors always beat paper, and paper always beats rock - there are no "joint winners". The Marquis de Condorcet attempted to show that democracy was possible in spite of this; it was left to Kenneth Arrow to show that democracy was actually impossible in theory, and to George W. Bush to show that it was also impossible in practice.

[dicklont]

Only the other day, several very good players (much better than I) were heatedly debating the correct handling of this simple combination:

 

A102

 

Q6543

 

What is actually the best line for four tricks with this combination, assuming you are playing against robots who will defend correctly in perfect tempo?

What is the best line for three tricks?

I would most certainly not touch this suit for as long as possible. Meanwhile I would try to work out where the king is.

When West has the king I play Ace, then 2 and cover East's Jack or duck East's x.

When East has the king I play Ace, then 2 and play the Queen over East´s x.

 

If I needed only three tricks I would really be unhappy with this suit. I finesse the ten and realise that studying suit combinations is probably underrated.

 

Which book should I read to get this right more often? The encyclopedia?

[dburn]

Only the other day, several very good players (much better than I) were heatedly debating the correct handling of this simple combination:

 

A102

 

Q6543

 

What is actually the best line for four tricks with this combination, assuming you are playing against robots who will defend correctly in perfect tempo?

What is the best line for three tricks?

I would most certainly not touch this suit for as long as possible. Which book should I read to get this right more often? The encyclopedia?

It happened to be your trump suit in a small slam. In those circumstances, you do not really have the option to avoid touching it for very long at all.

 

If you don't want it to be your trump suit in a small slam very often, I would not recommend the Encyclopaedia. Instead, I would recommend almost all of the methods advocated here on an almost hourly basis by people who understand bidding theory, and would no doubt have been able to ask at the two level of the hand that had five to the queen whether or not it also had the jack. Regrettably, when the deal arose, such mehods were not in use.

[gnasher]

Instead, I would recommend almost all of the methods advocated here on an almost hourly basis by people who understand bidding theory, and would no doubt have been able to ask at the two level of the hand that had five to the queen whether or not it also had the jack.

But it's best to avoid the methods, also suggested quite frequently, of people for whom "bidding theory" is just a question of deciding at which moment to ask for keycards.

[gnasher]

I think this is right:

 

For four tricks, of the layouts where it's achievable:

 

- Low to the 10 loses to RHO's Jx, KJx and K (six 3-2s + one 4-1)

(xcurt - you missed out Kx-Jxx)

 

- Low to the Q then finesse on the way back loses to RHO's Jx, Jxx and K (six 3-2s and one 4-1)

(You could also pick up RHO's stiff K, but only if you don't mind losing to his KJ/KJx.)

 

- Ace and low to the queen loses to RHO's xx and Jxx (six 3-2s)

 

So, ace and another is best.

 

 

For three tricks, anything works against a 3-2 break, and nothing works against 5-0. Ace and low to the 10 picks up all 4-1 breaks, so must be best.

 

 

I'm sill thinking about the best matchpoint line.

[foo]

Only the other day, several very good players (much better than I) were heatedly debating the correct handling of this simple combination:

 

AT2

+

Q6543

 

What is actually the best line for four tricks with this combination, assuming you are playing against robots who will defend correctly in perfect tempo?

What is the best line for three tricks?

I would most certainly not touch this suit for as long as possible. Which book should I read to get this right more often? The encyclopedia?

It happened to be your trump suit in a small slam. In those circumstances, you do not really have the option to avoid touching it for very long at all.

Well, that constrains the problem in ways that things like SuitPlay can not yet take into account.

 

If you are in a good 6<mumble> with this as your trump suit, then you should have no problems in any other suits and plenty of communication.

 

4 tricks with this suit is just a smidge over 50%. 3 tricks is a virtual certainty.

You can't handle 50 breaks and the safety play for 3 tricks picks up KJxx+x

You start by playing the A.

 

None of the above lines is the play for Max tricks. Sometimes one has to play for Max tricks rather than to make one's contract when playing MP instead of IMPs or Total Points. (For instance if this slam is one that =everyone= rates to be in. Yeah right. That much Field Protection is rare.)

 

The Max Tricks lines begin by playing small to the T or small to the Q.

Kx+Jxx argues for small to the T.

xx+KJx argues for small to the Q.

Max tricks is ~3 2/5= ~3.4

 

Good luck figuring out which layout to play for opposite silent opponents.

[TimG]

- Ace and low to the queen loses to RHO's xx and Jxx (six 3-2s)

 

So, ace and another is best.

Ace and then low from the board, ducking when RHO does not play the Jack is just as good, isn't it? It loses to Jx-Kxx, but picks up Kx-Jxx.

[gnasher]

- Ace and low to the queen loses to RHO's xx and Jxx (six 3-2s)

 

So, ace and another is best.

Ace and then low from the board, ducking when RHO does not play the Jack is just as good, isn't it? It loses to Jx-Kxx, but picks up Kx-Jxx.

Good point. It does cost an extra undertrick when RHO has KJxx.

Edited March 18, 2008 by gnasher

[skjaeran]

I don't need Suitplay for this sort of problems. I just ask Helgemo, he'll give the correct answer in a couple of seconds.

[Cascade]

I don't need Suitplay for this sort of problems. I just ask Helgemo, he'll give the correct answer in a couple of seconds.

That reminds me of the chess computer was it called "Deep Blue" or something. There was a cartoon with a big mainframe with a door and Bobby Fischer walking inside the machine.