matchpoints declarer play

[han]

That website lists the odds of any 6331 distribution, not the odds of a specific 6-3-3-1 distribution, that's why you got the wrong numbers.

[ralph23]

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 :)

Oh yes, silly silly silly. They didn't take moon phases into account either. :)

[Trumpace]

  <snip>

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

 

 

I don't think this is his method. (For instance, he has refered to a page which has only % of the suit split.)

 

What you are doing is using Bayes theorem (in a way) on top on what BebopKid is doing which was not even considered by him.

[ralph23]

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

Yep, I know bridge players never change their minds, no matter what the evidence.

 

But the contrary evidence in all published sources might want to make you reconsider, but I can already tell that it won't.

[han]

Details. I think that in spirit it is still his method.

[jtfanclub]

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

 

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

OK, sounds good, that should work fine. Luckily, I got a C in math in middle school.

 

For 9 cards known, there are five possible splits:

4-0

0-4

1-3

3-1

2-2

 

I can look up the odds of each of those on your web site.

 

6-3-4-0: .01326

6-3-0-4: .01326

6-3-3-1: .03448

6-3-1-3: .03448

6-3-2-2: .05642

 

I can add all of those together, and get .1519.

 

Now I can just divide each of those by .1519, and get the percentages.

 

6-3-4-0: 8.73%

6-3-0-4: 8.73%

6-3-3-1: 22.70%

6-3-1-3: 22.70%

6-3-2-2: 37.4%

 

Which jdonn rounded off to 40%, 25%, 25%, 5%, and 5%.

 

So your web site agrees with him, not you. I'm not going to check, but I'd guess that you forgot that 3-1 and 1-3 are different (as are 4-0 and 0-4), and therefore have to be included twice. That 2-2 has more permuations than 3-1 has already been included in the original probibilities.

[Trumpace]

Details. I think that in spirit it is still his method.

No really.

 

Given that he has "college math" training, and if this question was on a test and if I was grading him, he would not get any points for this.

 

In spirit, any "reasonable" method can be made correct.

 

The fact that he claims a 2-2 split for 6-3 holding vs 5-4 holding has different chances, make me want to give him negative marks then and there...

[han]

I don't think that the original problem is that complicated, and I think Frances and mikeh have it right in that line (2) is best. For example, when comparing line (2) with line (3) you see:

 

The lines are identical when spades are 2-2, 4-0 or when the king is singleton.

 

When spades are 3-1 onside line (2) will be superior when the club hook is off. When spades are 3-1 offside line (2) will be superior when the spade king is singleton offside, and line (3) is superior when the club hook is on.

 

So line (2) is clearly better than line (3).

 

Similarly you can compare line (2) with other lines.

[han]

Given that he has "college math" training, and if this question was on a test and if I was grading him, he would not get any points for this.

I would give partial credit. After all he remembers some of the method and I hate giving 0 credit for such an effort. I guess I have lower expectations.

[han]

For 9 cards known, there are five possible splits:

4-0

0-4

1-3

3-1

2-2

 

I can look up the odds of each of those on your web site.

 

6-3-4-0: .01326

6-3-0-4: .01326

6-3-3-1: .03448

6-3-1-3: .03448

6-3-2-2: .05642

 

I can add all of those together, and get .1519.

 

Now I can just divide each of those by .1519, and get the percentages.

 

6-3-4-0: 8.73%

6-3-0-4: 8.73%

6-3-3-1: 22.70%

6-3-1-3: 22.70%

6-3-2-2: 37.4%

FYI, this is nonsense, both the method and the outcomes.

[skjaeran]

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  :P .  It will verify the numbers I gave you.

As it happens, I also own a hard copy of the Encyclopedia of Bridge, 5th edition.

 

Under "Mathematical Tables, Table 4: Probability of Distribuiton of Cards In Two Hidden Hands on page 278 I find (4 outstanding cards):

3-1 49.74%

2-2 40.70%

4-0 9.57%

 

Any competent player knows that the percentages for 3-1/2-2/4-0 is roughly 50/40/10 respectively. (Btw, on the website you pointed to, you'll find this information too: http://www.durangobill.com/BrSplitStats.html)

 

The notion that the distribution of the 9 known cards between the two known hands should alter the odds is.....I haven't got words.

 

If you've got any understanding of probabilities you would know instinctively that this can't be true.

[Halo]

I would cross and finesse spades on the basis that if I cash the Ace I lose out to doubleton King on my right. That's a big loss and it doesn't feel there are nearly enough compensating distributions when I cash the Ace.

[matmat]

i like pascal's triangle :P

 

btw., what is this "math" you speak of?

[matmat]

Given that he has "college math" training, and if this question was on a test and if I was grading him, he would not get any points for this.

I would give partial credit. After all he remembers some of the method and I hate giving 0 credit for such an effort. I guess I have lower expectations.

your students must love you... do you teach basketweaving too? :P :lol:

[jtfanclub]

FYI, this is nonsense, both the method and the outcomes.

Maybe a C-. Sorry.

[jdonn]

FYI, this is nonsense, both the method and the outcomes.

Maybe a C-. Sorry.

But hey, on the bright side at least you get credit for realizing it when it is pointed out to you. And it only took one person, not 10 :P In fact, I now see people are proving Bebop wrong with his own quoted sources! Snicker.

[gwnn]

There are 26 cards missing, so there are 26!/(13!*13!) possible ways of them splitting.

 

this is 10400600

 

In how many ways can a suit divide 4-0?

 

2(which opp is void?)*22!/(9!*13!) (which 13 cards does the opp with the void have out of the remaining 22?)

 

2*22!/(9!*13!)=994840

 

Exact percentage=9.5652173913043478260869565217391 (windows scientific calculator)

 

3-1?

 

2(which opp is singleton?)*4(which card is the stiff one?)*22!/(10!*12!) (which 12 cards does the opp with the singleton have?)

 

2*4*22!/(10!*12!)=5173168

 

Exact percentage=49.739130434782608695652173913043

 

2-2?

 

6(which 2 cards does LHO have?)*22!/(11!*11!) (how can 22 cards be dealt 11-11?)

 

6*22!/(11!*11!)=4232592

 

Exact percentage=40.695652173913043478260869565217

 

What does that say about the bridge hand? I have no idea. But the maths are correct.

[ralph23]

Using good ole' durango bills site:

 

A = No. of 6322 hands: 35,830,574,208

 

B = No. of 6331 hands: 21,896,462,016

C = B times 2 = 43,792,924,032 (because I can have 3 on left and 1 on right, or vice versa)

 

Ratio of C to A: 1.2222222

 

That is, for every time they divide 2-2 against me, they will divide 3-1 (or 1-3) against me 1.22222 times.

 

Compare against Pavlicek's calculator:Missing 4 cards:

 

E = Percentage Odds of EW dividing 2-2: 40.695652173913

F = Percentage Odds of EW dividing 3-1 or 1-3: 49.739130434783

 

Ratio of F to E: 1.2222222

[jtfanclub]

Darn it, Ralph, that's what I was trying to do. What did I do wrong?

[Guest Jlall]

A very interesting hand is posted and it turns into this.....

 

So what is the right line? My intuitive line is to cross to the DK and hook the spade. I recognize that this is effectively a math problem though so would like those who are good at these things to figure out what is the best line. Sorry if someoen did this and I missed it amongst the noise.

[ralph23]

Darn it, Ralph, that's what I was trying to do. What did I do wrong?

Well, I don't know.... I thought you very cogently pointed out to bebop that he had simply failed to count for both 3-1 and 1-3. But the point is, of course, the durango numbers on the web are correct; as is Pavlicek.

 

As is the other bridge calculator on the web that I cited and, I'm sure, he failed to even experiment with and wonder "Gee, I wonder why my numbers are different".

 

And as is the Encyclopedia of Bridge (and yes, I have one too). In fact all the published sources agree.

 

It's not the instrument that had a problem; it's the musician.

[Stephen Tu]

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

 

If you change 6-3 to 5-4, you are just trading cards in other suits from north to south & vice versa, you don't have to touch East/west's cards.

 

Say after each hand is shuffled & dealt, North & South examine their cards, and South always gives North all his spades, and North hands South the appropriate number of random non-spade cards in exchange. Why should the distribution probability of East/West change?

 

gwnn's calculations are the normal, direct way to do this in my mind. Jdonn calculated in a more time consuming way but also valid & the same answer. I don't quite comprehend how you are trying to do this.

[Guest Jlall]

I went through some crude analysis with han but it seemed like crossing and finessing was definitely a better MP line than cashing spade ace, crossing, and leading a spade up

[ralph23]

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

 

If you change 6-3 to 5-4, you are just trading cards in other suits from north to south & vice versa, you don't have to touch East/west's cards.

 

Say after each hand is shuffled & dealt, North & South examine their cards, and South always gives North all his spades, and North hands South the appropriate number of random non-spade cards in exchange. Why should the distribution probability of East/West change?

 

gwnn's calculations are the normal, direct way to do this in my mind. Jdonn calculated in a more time consuming way but also valid & the same answer. I don't quite comprehend how you are trying to do this.

Suppose after they are dealt (4 hands), NS just bunch their two hands in the middle. THey don't even look at their cards. This is the easiest way to see it, imo, but believe me, you're not going to convince him! :) :)

[joshs]

here are obviously 2 possible lines:

a. Cash trump A first and continue trumps (unless 4-0 onside), planning on hooking the club

b. use the entry to take the trump hook, and then take the other hook if trumps are 2-2

 

B Gains when:

Kx of trumps is onside

And when Kxx of trumps is onside and CK is offsides

 

A Gains when:

CK is on and [spade K is off or stiff SK is on]

 

Probabilities:

B: Kx in Spades on is 20%

Kxx on is 3/4 of 1/4=3/16

That AND CK off is thus 3/32

B gains 29.375% of the time

 

A:Spade K is off or stiff K is on=50%+1/4*1/4=9/16

So that and CK on is 9/32=28.125%

 

So it looks like B is slightly better. I probably need to think about the 4-0 breaks a little bit since you will not ALWAYS be able to repeat the club finesse as I have been assuming here, but that will only reduce the effectiveness of A a very little bit.

 

Anyway, I probably did my math wrong, so I will leave it to the peanut gallery to do 13,000 simulations :)