[Trumpace]

BebopKid, on Sep 6 2007, 01:48 PM, said:

jdonn, on Sep 6 2007, 01:32 PM, said:

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]

ralph23, on Sep 6 2007, 01:51 PM, said:

http://www.prairiene...bridge/odds.htm

http://www.ronulrich...ssons/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]

ralph23, on Sep 6 2007, 01:51 PM, said:

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.durangobi...rSuitStats.html

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

[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]

jdonn, on Sep 6 2007, 03:12 PM, said:

ralph23, on Sep 6 2007, 01:51 PM, said:

http://www.prairiene...bridge/odds.htm

http://www.ronulrich...ssons/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]

Hannie, on Sep 6 2007, 02:15 PM, said:

  <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]

BebopKid, on Sep 6 2007, 03:16 PM, said:

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]

BebopKid, on Sep 6 2007, 02:16 PM, said:

http://www.durangobi...rSuitStats.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]

Hannie, on Sep 6 2007, 02:32 PM, said:

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]

Trumpace, on Sep 6 2007, 02:42 PM, said:

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]

jtfanclub, on Sep 6 2007, 02:39 PM, said:

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]

BebopKid, on Sep 6 2007, 09:16 PM, said:

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.

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.durangobi...SplitStats.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

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

[matmat]

Hannie, on Sep 6 2007, 02:46 PM, said:

Trumpace, on Sep 6 2007, 02:42 PM, said:

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?

[jtfanclub]

Hannie, on Sep 6 2007, 02:50 PM, said:

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

Maybe a C-. Sorry.