matchpoints declarer play

[vang]

[hv=d=n&v=e&n=s643hq10dkj52cj1032&s=saq9875hda4caq876]133|200|Scoring: MP

p - p - 1♠ - p

... final contract 4♠[/hv]

 

opponents silent, you reach 4♠ (6 seems a little too aggresive at matchpoints). lead ♥ K.

[helene_t]

Finese spades, if spades are 2-2 ♠6 is an entry for the club finese. Am I mising something?

[Finch]

<deleted>

[BebopKid]

Holding 6-3 in declarer/dummy:

 

2-2 split is 54.1%

3-1 split is 16.7%

1-3 split is 16.7%

4-0 split is 6.25%

0-4 split is 6.25%

 

3-1 split with Singleton K is 4.175%

 

Best line of play: finesse to QS, then lead AS

Works 58.275 % of the time.

[ralph23]

Holding 6-3 in declarer/dummy:

 

2-2 split is 54.1%

3-1 split is 16.7%

1-3 split is 16.7%

4-0 split is 6.25%

0-4 split is 6.25%

 

3-1 split with Singleton K is 4.175%

 

Best line of play: finesse to QS, then lead AS

Works 58.275 % of the time.

A singleton King on your right isn't good. If RHO has that holding, then LHO has JT2, and you will lose a trump trick to his Jack. Your only chance to avoid losing a trump trick is 2-2 split with King on your right.

 

Of course, just because you lose a trump trick doesn't mean you're going down. Maybe you won't lose a ♣.

 

So, your odds are calculated as:

 

Probability of NOT losing a trump trick = X%

Probability of losing a trump trick = (100 - X)% = Y

 

Probability of NOT losing a ♣ trick = Z

 

So, you will make your contract X% of the time "right away" on the trumps; the remaining Y% of the time, you will succeed in Z% of THOSE Y cases, unless you lost two trump tricks, which is certainly possible!!

 

This all assumes the probabilities are a priori. But you know that is not true. E.g. if your RHO turns up with ALL the trumps, or with 3 out of 4, then it is more likely that your LHO has any other given card, including the important club.

 

EDIT: LOL, I was playing for 6. B) . Still, it's MPs.....

Edited September 6, 2007 by ralph23

[Finch]

It's not as simple as that.

 

The best line of play in the spade suit, in isolation, depends on the number of tricks you want:

 

- For one loser, cash the ace and then lead towards the queen

- For no loser, finesse the queen then cash the ace. However this ends up with two losers when you finesse into singleton king.

 

Unfortunately, there are two new problems once you know this

 

i) Playing matchpoints, you don't know if you want to play the suit for 1 loser or no loser

ii) You would also like to take the club finesse, but you are a bit short of entries to dummy.

 

Here are some possible lines, after ruffing with the 7 at trick 1.

 

1. Cross in diamonds and take the spade finesse. If spades turn out to be 2-2, use the late spade entry to take the club finesse.

2. Cash the ace of spades (If East shows out, use your diamond entry to take the club finesse as you have 3 spade losers). Cross in diamonds and lead a spade up. If trumps are 2-2, use the late spade entry for the club finesse. If the SK is singleton, cash another trump and use the diamond entry for the club finesse.

3. Cash the ace of spades and play a spade from hand. Eventually take the club finesse.

4. Cash the ace of spades, cross in diamonds and run the jack of clubs. Whether or not covered, play a second club.

5. Cash the ace of spades, cross in diamonds and run the jack of clubs. Whatever happens to this, play a spade next.

[i think it must be technically wrong to select your line depending on whether the club is covered, otherwise an opponent can always make you do the wrong thing]

 

All other lines I can think of are (I believe) definitely worse than any of these.

 

It's hard to evaluate all of these lines, because there are so many possible layouts. The things that matter are

- how the spades break, and whether the king is singleton or not, and whether the king is onside or not

- how the clubs break, and whether the king is singleton or not, and whether the king is onside or not

- if clubs are 3-1, how many and which spades the singleton club holds

 

At IMPS, I would definitely cash the SA at trick 2.

 

My intuition (such as it is) is to cash the SA at trick 2 at matchpoints as well rather than cross in diamonds and take the spade finesse. But it's a real pain to try and justify whether it's right or not.

 

Now what? My intuition says line 2, but I'm not confident about this at all...

 

We've eliminated the the singleton SK hands and the 4-0 trump breaks.

 

i) If LHO has 3 spades and not the CK, we want top play line 3 or 4; line 5 also if the CK is singleton or doubleton.

ii) If LHO has 3 spades and the singleton CK we want to play line 2

iii) If RHO has 3 spades and not the CK we want to play line 2

iv) If RHO has 3 spades and exactly CKx play line 4

v) If spades are 2-2 and the CK is onside, we want to play line 2, 3 or 5; or 4 if RHO has Kxx clubs and LHO Kx of spades;

vi) If spades are 2-2 and the CK is offside, we want to do anything except 4 or 5 which lose to Kxx of clubs and Kx on our left.

 

help... too much to calculate

[bid_em_up]

Holding 6-3 in declarer/dummy:

 

2-2 split is 54.1%

3-1 split is 16.7%

1-3 split is 16.7%

4-0 split is 6.25%

0-4 split is 6.25%

 

3-1 split with Singleton K is 4.175%

Where did you get these percentages?

[jdonn]

Holding 6-3 in declarer/dummy:

 

2-2 split is 54.1%

3-1 split is 16.7%

1-3 split is 16.7%

4-0 split is 6.25%

0-4 split is 6.25%

 

3-1 split with Singleton K is 4.175%

Where did you get these percentages?

Outerspace. The true percentages (approximately, and formatted in the same odd way) are well known:

 

2-2 split is 40%

3-1 split is 25%

1-3 split is 25%

4-0 split is 5%

0-4 split is 5%

 

3-1 with singleton king is 6.25%

[bid_em_up]

Holding 6-3 in declarer/dummy:

 

2-2 split is 54.1%

3-1 split is 16.7%

1-3 split is 16.7%

4-0 split is 6.25%

0-4 split is 6.25%

 

3-1 split with Singleton K is 4.175%

Where did you get these percentages?

Outerspace. The true percentages (approximately, and formatted in the same odd way) are well known:

 

2-2 split is 40%

3-1 split is 25%

1-3 split is 25%

4-0 split is 5%

0-4 split is 5%

 

3-1 with singleton king is 6.25%

Well, to be honest, thats what I thought originally, but,

 

The percentages you list (which are the ones normally given) are for all combined holdings of 9 cards (missing 4 cards). A little further crazy thought says, the percentages for a 5-4, 6-3, 7-2, 8-1 and 9-0 breaks are likely different from each other though, no?

 

But I cannot find on the net the true odds opposite a 6-3 break, so I just wondered. For the moment, I will reserve judgement on where he got his odds from, hopefully a source can be provided. B)

[ralph23]

Holding 6-3 in declarer/dummy:

 

2-2 split is 54.1%

3-1 split is 16.7%

1-3 split is 16.7%

4-0 split is 6.25%

0-4 split is 6.25%

 

3-1 split with Singleton K is 4.175%

Where did you get these percentages?

Here's a link to Pavlicek.

 

http://www.rpbridge.net/xsb2.htm

 

The stated % are off. 4-0 is too hi, 3-1 too low.

[Stephen Tu]

The percentages you list (which are the ones normally given) are for all combined holdings of 9 cards (missing 4 cards). A little further crazy thought says, the percentages for a 5-4, 6-3, 7-2, 8-1 break, are likely different from each other though, no?

 

The applicable word there is "crazy". The percentages are the same, all that matters is the combined total. The percentages are calculated by taking the 26 unknown cards and giving 13 to each opponent randomly, there are 26!/(13!)(13!) specific distributions of the unknown cards. Trading cards between the known hands has absolutely no effect on the distribution of the unknown ones.

[BebopKid]

Where did you get these percentages?

I calculated them using probability and combinations.

 

6322 occurs 5.642490% of all hands (~54.2%)

6331 occurs 3.448188% of all hands (~33.1%)

6430 occurs 1.326226% of all hands (~12.7%)

 

jdonn has forgotten that you need to look at the whole suit and not just the 5 missing cards.

 

if the declarer dummy is 5-4 then the odds would be

5431 occurs 12.930705% of all hands (~52.2%)

5422 occurs 10.579668% of all hands (~42.8%)

5440 occurs 1.243337% of all hands (~5.0%)

 

I have no idea where jdonn came up with his percentages, but they are not correct.

 

[edit]

The combined total has no effect on the probable layout of the cards, it's the probability of the suits that is accurate.[/edit]

[edit again]I dealt over 130,000 hands to match the probability and came up with my predicted result.[/edit again]

[Trumpace]

jdonn has forgotten that you need to look at the whole suit and not just the 5 missing cards.

 

I have no idea where jdonn came up with his percentages, but they are not correct.

I am afraid jdonn is correct this time B)

 

You have forgotten to look at the whole hand, which is 52 cards.

 

You are considering only the "whole" suit which is just 13 cards. The percentages change based on how the other suits are distributed, which jdonn's method of calculation implicitly considers, while yours does not.

 

Please read Stephen Tu's post for an explanation why jdonn's method works.

 

In your simulations did you try dealing 52 cards or just 13? If 52, can you please post the code?

[han]

Jdonn's method is just "remembering the well-known odds". He should receive no credit for this, even though he is obviously correct.

[han]

By the way, Bebob's method is also correct, but the numbers are wrong so he must have made a mistake somewhere.

[jdonn]

Where did you get these percentages?

I calculated them using probability and combinations.

 

6322 occurs 5.642490% of all hands (~54.2%)

6331 occurs 3.448188% of all hands (~33.1%)

6430 occurs 1.326226% of all hands (~12.7%)

 

jdonn has forgotten that you need to look at the whole suit and not just the 5 missing cards.

4 missing cards, and no you don't. Are you seriously suggesting the odds of a 2-2 break are different for a 5-4 fit than they are a 6-3 fit? lol.

 

if the declarer dummy is 5-4 then the odds would be

5431 occurs 12.930705% of all hands (~52.2%)

5422 occurs 10.579668% of all hands (~42.8%)

5440 occurs 1.243337% of all hands (~5.0%)

No way, you are!!! This is awesome. I am going to coin the term "Arkansas math" for being given one problem and taking the answer by solving a different problem :)

 

There are numerous ways to calculate this, but the most easily understood by people with little math background would be the following:

Say you are missing the AKQJ (taking cards with letters instead of numbers to avoid confusion).

 

2-2 break: There are 6 ways to distribute the cards

JQ KA

JK QA

JA QK

QK JA

QA JK

KA JQ

The probability of each is 13/26 * 12/25 * 13/24 * 12/23, which times 6 gives 40.7%

 

For 1-3

J QKA

Q JKA

K JQA

A JQK

4 * 13/26 * 13/25 * 12/24 * 11/23 = 24.9%

 

For 0-4

- JQKA

13/26 * 12/25 * 11/24 * 10/23 = 4.8%

 

I'm too slow to post, the peanut gallery has already been defending me. I <3 you guys.

[han]

I'm too slow to post, the peanut gallery has already been defending me. I <3 you guys.

Defending you, no way Josh! :)

 

But seriously, Bebob's method should also work if done carefully:

 

Calculate the odds of specific 6-3-0-4, 6-3-1-3, 6-3-2-2, 6-3-3-1 and 6-3-4-0 distributions, add them up and divide each by the total.

 

And, as Josh and others have pointed out, they are exactly the same answers that you would get for computing the odds for these distributions: 5-4-0-4, 5-4-1-3, 5-4-2-2, 5-4-3-1 and 5-4-4-0 (again, dividing each by the sum).

 

One reason Bebob's numbers could be wrong (I'm guessing now) is if you don't look at specific distributions, not all distributions get the same weight. For example, there are 24 different 5-4-3-1 distributions but only 12 different 5-4-2-2 distributions. Similarly, there are 12 6-3-3-1 and 6-3-2-2 distributions each, but 24 different 6-4-3-0 distributions.

 

But if you are careful enough then the following strategy should always work:

 

I calculated them using probability and combinations.

:)

[BebopKid]

There are numerous ways to calculate this, but the most easily understood by people with little math background would be the following:

Since I have over 20 hours of college math, I didn't use the non-math method.

 

I calculated the odds using probability and then dealt over 130,000 hands to validate my calculations. I may have made a typo or rounding error, but they are close.

 

This is also backed up by lessons with people such as Gordon Campbell, a member of the Canadian National Team that won in Salt Lake City a few years ago.

 

The layout of cards in two hands do affect the probability of the cards in the other two hands, because if you swap 63 to 54 you have to affect 2 other cards in each of the two hands, which affects the layout of other suits in the two hands.

 

After you've dealt 130,000 hands, please let me know your results.

 

You method is good for quick answers and intermediate players, but full knowledge of the probabilities helps later on. (and no, I'm not there yet, I'm still intermediate)

[ralph23]

I have no idea where jdonn came up with his percentages, but they are not correct.

 

He is indeed correct, albeit not precise to the umpteenth place, but he was not trying to be precise to the umpteenth place.

 

You can use the Pavlicek calculator if you are in doubt, and then argue with Pavlicek about it. But Pavlicek is correct as is jdonn (close enough for Govt work anyhow :) ).

 

The correct odds appear many times on the Internet though and they will in bridge books also.

 

http://www.prairienet.org/bridge/odds.htm

 

http://www.ronulrich.com/bridge/lessons/odds.html

 

http://www.himbuv.com/cpe.htm

Edited September 6, 2007 by ralph23

[jdonn]

There are numerous ways to calculate this, but the most easily understood by people with little math background would be the following:

Since I have over 20 hours of college math, I didn't use the non-math method.

 

I calculated the odds using probability and then dealt over 130,000 hands to validate my calculations. I may have made a typo or rounding error, but they are close.

 

This is also backed up by lessons with people such as Gordon Campbell, a member of the Canadian National Team that won in Salt Lake City a few years ago.

 

The layout of cards in two hands do affect the probability of the cards in the other two hands, because if you swap 63 to 54 you have to affect 2 other cards in each of the two hands, which affects the layout of other suits in the two hands.

 

After you've dealt 130,000 hands, please let me know your results.

 

You method is good for quick answers and intermediate players, but full knowledge of the probabilities helps later on. (and no, I'm not there yet, I'm still intermediate)

Sorry I'm crying here. Where do I even start.

 

"Since I have over 20 hours of college math."

Sorry it's my fault, I didn't realize you were nearly at a full day! :) If you must know, I got 800 on my SATs in math, have a BA in math (and economics), and have worked as an actuary (which includes passing the first actuarial exam, in calculus and PROBABILITY) and in accounting the last 3 years, as well as being a tutor in probability before that. AND I went to math camp as a youth! *FLEX*

Edit: Wait wait! I forgot, 5 in AP Calc AB, 5 in AP Stat, 4 in AP Calc BC :) I think I had a bad day.

 

"I didn't use the non-math method."

Uh, neither of us did. I used the simple math method. You used the wrong math method.

 

"This is also backed up by lessons with people such as Gordon Campbell, a member of the Canadian National Team that won in Salt Lake City a few years ago."

Well mine is back up by lessons with people such as Han Peters, a member of....AAA? Who won...an online BBO team game recently? Damn you win.

 

"After you've dealt 130,000 hands, please let me know your results."

See this is the beauty of using math to solve your problems, presuming you do it correctly. It prevents methods which could lead to carpal tunnel syndrome at an early age.

 

"You method is good for quick answers and intermediate players, but full knowledge of the probabilities helps later on. "

By quick answers and intermediate players, I'm sure you meant correct answers and intermediate mathematicians.

 

Sorry I just have to repeat it for posterity, it's too priceless on many levels.

"Since I have over 20 hours of college math, I didn't use the non-math method."

[Trumpace]

There are numerous ways to calculate this, but the most easily understood by people with little math background would be the following:

Since I have over 20 hours of college math, I didn't use the non-math method.

What university, who was the professor and what grade did you get? Did the math include probability theory?

 

Sorry, this statement of yours is quite annoying. You don't have to answer the questions above, but making statements like these to try and support your argument only shows that you don't have a convincing logical argument which is pertinent to the discussion at hand.

[jdonn]

http://www.prairienet.org/bridge/odds.htm

 

http://www.ronulrich.com/bridge/lessons/odds.html

 

http://www.himbuv.com/cpe.htm

Sorry Ralph, they don't count. Those silly websites forgot to consider how the cards are split between declarer and dummy :)

[mikeh]

deleted post: deleted due to temporary (I hope) brain-freeze

[han]

To try to end this painful discussion I will try to do the calculation suggested by Bebob correctly. Hopefully this will show Bebobkid that his numbers were way off and the rest that his method was actually correct.

 

By a-b-c-d I will denote deals where we have a spades, partner b, LHO c and RHO d. The odds of having each spade distribution with 6-3 spades in our hands are:

 

6-3-4-0 = 0.055259424731 %

 

6-3-3-1 = 0.287349008602 %

 

6-3-2-2 = 0.470207468622 %

 

6-3-1-3 = 0.287349008602 %

 

6-3-0-4 = 0.055259424731 %

 

Total = 1.155424335288 %

 

Now divide each of the percentages by the total to get:

 

each 4-0 split = 4.78260869563%

 

each 3-1 split = 24.86956521738%

 

the 2-2 split = 40.69565217395%

 

Indeed, these are quite close to the percentages that jdonn gave, Bebob's method is correct.

[BebopKid]

You can use the Pavlicek calculator if you are in doubt, and then argue with Pavlicek about it. But Pavlicek is correct as is jdonn (close enough for Govt work anyhow :) ).

 

The correct odds appear many times on the Internet though and they will in bridge books also.

 

Those odds are not taking into account the actual layouts of 52 cards. They are using a simplistic method which may be close you for, but not close enough for me or the experts and world class players that I've taken lessons from.

 

Check the Encyclopedia of Bridge. We have a hard copy at our Bridge House, come by and I'll show you :) . It will verify the numbers I gave you.

 

But there are also many places you can find the statistics on the Internet if you search for them, but here are is one such list of calculations.

 

http://www.durangobill.com/BrSuitStats.html

 

So I'll stand by by original answer of the probability of the suit break.