Help me understand deal generators and probability

[ralph23]

This is tending to back up what I learned from my teachers, that 3-1 split occurs less often than 2-2 split with a 6-3 declarer combination.

Couple quick comments

 

1. Here, once again, its very important to remember that there are two possible 3-1 splits to worry about and only one 2-2 split. A 3-1 split is MORE common than a 2-2 split. However, either one of the two specific 3-1 splits is less common.

A is less than B.

 

But A times 2 is greater than B??

 

Seriously, can that really happen?

[jtfanclub]

If you read my first post, I asked for no insults. If you have something positive or constructive to comment, please do so.

 

I am happy for everyone that can keep quoting the same things to me. Please test and results from your trials as I have done.

There is no testing and no trial. If you use the ACTUAL CASE and not some other configuration, I am very confident that you will come out with the correct result.

 

There is no testing and no trials necessary in math. Just set up your configuration EXACTLY for what is being tested for, and I am confident you will get the correct result.

 

If you do not believe this is not an accurate test, please post the configuration file for a test you believe to be accurate.

 

308912 hands with a 6 card spade suit in any hand

195462 hands with a 6-3 spade fit between any 2 hands

 

You're testing for a 6-3 fit BETWEEN ANY TWO HANDS.

 

Why are you testing any two hands? How does knowing what the split is when South has 6 spades and an opponent has 3 have anything whatsoever to do with when South has 6 spades an North has 3 spades?

 

The correct configuration file is....

 

1. SOUTH has exactly 6 spades.

2. NORTH has exactly 3 spades.

 

Case 1: East has exactly 0 spades (split 4-0)

Case 2: East has exactly 1 spade (split 3-1)

Case 3: East has exactly 2 spades (split 2-2)

Case 4: East has exactly 3 spades (split 1-3)

Case 5: East has exactly 4 spades (split 0-4)

 

Do you understand why THIS is an actual test of the probablilities for when South has 6 spades and North has 3 spades, and anything that disagrees with this result must be wrong?

 

-----------------------------------------------------------------------------------

 

 

So why is this configuration different from your configuation?

----------------------------------------------------------------------------------------

You have two dimes, and two nickels. You have three people that you pass a coin to randomly, keeping one for yourself.

 

1. If ANY coin you passed out is a nickel, what are the odds that the coin you kept is a nickel.

 

2. If THE FIRST coin you passed out is a nickel, what are the odds that you kept is a nickel?

 

3. Why are these odds different?

[hrothgar]

Any teacher who tells you that the odds of the opponent's split depends on how the cards split between you & dummy should be fired.  That assertion is easy to prove mathematically false.

Please do run a simulation and show results that match your hypothesis.

 

I've already gone over the mathematics of it. Probability says that 6-3-2-2 occurs more often than 6-3-3-1.

Here's the results of two very simple scripts that I created using Hans van Stavern's Dealer program

 

Script 1:

 

condition

 

spades(north) == 5 and spades(south) == 4

 

action

 

average spades(east) == 4,

average spades(east) == 3,

average spades(east) == 2,

average spades(east) == 1,

average spades(east) == 0

 

results:

 

0.0475299

0.2484

0.407608

0.248668

0.0477933

 

Script 2:

 

condition

 

spades(north) == 6 and spades(south) == 3

 

action

 

average spades(east) == 4,

average spades(east) == 3,

average spades(east) == 2,

average spades(east) == 1,

average spades(east) == 0

 

results

 

0.0478948

0.248502

0.407759

0.247195

0.0486479

[ralph23]

I can't believe you haven't done this yet, given the significance and novelty of your discoveries. Why waste any more time on this silly little forum? Get a real forum!

 

Otherwise, we're going to start thinking that you really don't believe your own findings. So go for it!!

Not my discovery. It is already known by many.

 

I'm only trying to enlighten others and help spread knowledge.

What could be more in tune with your mission then, than letting everyone know -- by contacing the powers-that-be and getting them on board with your ideas? Then springing to a wider platform??

 

Oh well .... no sense arguing about it. I just think you owe it to the bridge world. You're just sitting on a gold mine and refuse to dig.

 

Everyone else here thinks a 22 division of four missing cards in a suit is less likely overall than a 31 division, because 31 can be 3 left, one right OR 3 right, one left, so while A < B, nevertheless 2A > B. Pavlicek included. He'll be grateful for the enlightenment, too !!!

[ralph23]

duplicate post

[Stephen Tu]

Okay, using Richard Pavlicek's deal generator with this RPD config

D 65000:1000000000

123WS=6

123WS=3

01234WS=4  (typo?)

0123WS=2

 

You are counting the wrong things, just as jtfanclub previously asserted! You are counting all 6322s, including those when one of the opponents has 3 and partner only has 2, rather than only the hands where partner has 3. In effect you are counting deals with a 6-2 fit & a 3-2 split between the opps along with the hands where there is 6-3 fit and 2-2 split between opps which are the only ones we are interested in. Out of the 6322 deals where you have 6, partner will only have the 3 1/3 of the time. Whereas for the 6331 hands, partner will have 3, 2/3 of the time. So your errors end up with you counting the 2-2s double in relationship to the 3-1s which explains the discrepancy between your stats & theoretical.

 

The test file you really want to use is something like: [edited from original since at first didn't understand how Pavlicek's "|" & "||" OR operators worked]

D 0:10000000

WS=6 & ES=3

NS<4

NS<3

NS<2

NS<1

 

which gives output:

Dealt: 10000000 22.08 sec D/s: 452899

Test 1: 115606 Percent: 1.1561 5236

Test 2: 110145 95.2762 1.1015 4988

Test 3: 81485 73.9798 0.8149 3690

Test 4: 34490 42.3268 0.3449 1562

Test 5: 5550 16.0916 0.0555 251

 

Translated, with the results from my test run:

deal 10,000,000 hands

115606 with west having 6sp & east 3sp

5550 with north having < 1 spades, so 5550 0-4 splits

34490 with north < 2 sp, so 34490 - 5550 = 28940 1-3 splits

81485 north < 3 spades, 81485 - 34490 = 46995 2-2 splits

110145 north < 4 spades, 110145 - 81485 = 28660 3-1 splits

115606 - 110145 = 5461 4-0 splits

 

So, in total:

2-2 split = 46995/115606 = 40.65%

3-1 split or 1-3 split = (28940+28660)/115606 = 49.82%

4-0 split or 0-4 split = (5550 + 5461)/115606 = 9.52%

 

as you see the percentages match up fairly closely with the theoretical 40.7, 49.75, 9.57

[jtfanclub]

Please don't be convinced by Richard and Stephen's trials and results. Do please be convinced by looking at the configurations they used.

 

Math is not of game of whoever does the most trials wins.

[Stephen Tu]

And here is another Pavlicek script, that gives the right answers, which replicates somewhat what I think you were trying to do originally:

 

D 0:10000000

WS = 6 & ES = 3 || ES = 3 & WS = 6 || NS = 6 & SS=3 || NS=3 & SS=6

0123 WS=4

0123 WS=2

 

 

I.e. select deal if either EW or NS have 6-3 fit either way, then exclude 4-0 breaks, then exclude 2-2 breaks, count them up.

 

The problem with your original script was that East got to be partners with North & South also, a strange game that doubles 4-0 & 2-2 relative to 3-1s. Your original numbers work out close to theory if you multiply by 1/3, 1/3, 2/3 respectively.

[han]

Here is my positive and constructive comment: go back to college. You are obviously interested in this stuff, there are people who teach probability theory at any level.

[lucky81]

Here are the results:

308912 hands with a 6 card spade suit in any hand

195462 hands with a 6-3 spade fit between any 2 hands

170625 hands without a 6-3-4-0 spade distribution

-> 24837 hand with a 6-3-4-0 spade distribution

65000 hands without a 6-3-2-2 spade distribution

-> 105625 hands with a 6-3-2-2 spade distribution

-> 65000 hands with a 6-3-3-1 spade distribution

 

So:

54.04% are 2-2 split

33.25% are 3-1 split

12.71% are 4-0 split

In your 6-3-3-1 hands there are TWO 6-3 fits that you are counting only once. If you correctly count them twice, you get:

2-2: 105625 or 40.55%

3-1: 130000 or 49.91%

4-0: 24837 or 9.54%

[Stephen Tu]

In your 6-3-3-1 hands there are TWO 6-3 fits that you are counting only once. If you correctly count them twice, you get:

 

Doubling the 3-1's "works" to correct the percentages, but the above is not what is really happening in his original sim. His problem was that he was 3x counting the 6322s (by including 6-2 fits with 3-2 breaks), and 3x counting 6340s (by including 6-0 fits with 4-3 breaks & 6-4 fits with 3-0 breaks), while 1.5x counting 6331s (including 6-1 fits with 3-3 breaks).

[lucky81]

Doubling the 3-1's "works" to correct the percentages, but the above is not what is really happening in his original sim.

Your explanation is a more convoluted way to say the same thing. I would even say my explanation is more correct, since he explicitly was trying to count 6-3 fits between ANY two hands, so including NE, NW, SE, SW. In other ways, he's not counting anything three times too many as you are implying but he's simply missing some of the spade fits in case there happen to be 2 in a single hand.

[Stephen Tu]

Your explanation is a more convoluted way to say the same thing.

No, it's a direct way to say the correct thing. Your explanation is simply erroneous.

 

since he explicitly was trying to count 6-3 fits between ANY two hands, so including NE, NW, SE, SW. In other ways, he's not counting anything three times too many

 

B.S. The original problem posited was how often a suit breaks 2-2, 3-1, 4-0 when someone has a 6-3 fit. To do a sim correctly you can count hands with any sort of combination where someone has 6 & his partner has 3. There are 4 different players who can have 6, the sim is OK if you count all, or any subset of the 4 cases, as long as you count the same subset for all the possible split cases. For the 2-2 split, if you want to count all possible 6-3 fits, you are supposed to count W6E3, E6W3, S6N3, N6S3. But Bebop's flawed script also threw in W6N3, W6S3, E6N3, E6S3, S6W3, S6E3, N6W3, N6E3, i.e. hands with 6-2 fits. That is 3x the proper number in total. Counting 6-3 fits between ANY two hands is wrong when the question is specifically how the cards will split when someone's side has the 6-3 fit.

 

 

he's simply missing some of the spade fits in case there happen to be 2 in a single hand

 

In his original stats, he did not "miss" any spade fits in the 3-1 case as you state. He in fact counted TOO MANY of them. He counted "fits" where the person with 6 found his partner with the singleton & the opponents with 3 as also belonging to the 3-1 split case with a 6-3 fit. He counted too many of each category, 3x 2-2, 3x 4-0, 1.5x 3-1s.

 

He didn't "miss" anything, he overcounted everything, some more than others.

[lucky81]

Counting 6-3 fits between ANY two hands is wrong when the question is specifically how the cards will split when someone's side has the 6-3 fit.

You are wrong by saying that the method is wrong, as it can be easily shown to be correct, i.e. to give the correct answer.

 

I believe in his latest simulation he was trying to count the following: assuming that given two hands (not necessarily a partnership) have a 6-3 spade fit, what is the distribution of the other two hands. He correctly assumed this would be the same distribution as for NS having a 6-3 fit. His method just gives more samples using the same set of deals than your method. He stated what he was doing by saying e.g. "195462 hands with a 6-3 spade fit between any 2 hands". ANY 2 hands. Not "any partnership".

 

He is not counting anything 3 times - the numbers are not even divisible by 3. He is trying to count something else than you think he was trying to count.

[Trumpace]

Counting 6-3 fits between ANY two hands is wrong when the question is specifically how the cards will split when someone's side has the 6-3 fit.

You are wrong by saying that the method is wrong, as it can be easily shown to be correct, i.e. to give the correct answer.

Sorry, but in mathematics, ends don't justify the means.

 

Getting the right answer does not automatically imply the correctness of the method.

 

You just got lucky, lucky. (Sorry could not resist :) )

[lucky81]

A method is correct if and only if it always gives a correct answer. What is your definition of a correct method?

 

Let me state a more general proposition: the method will work for all types of fits (not just 6-3). I leave the (easy) proof as an exercise.

[Trumpace]

A method is correct if and only if it always gives a correct answer. What is your definition of a correct method?

 

Let me state a more general proposition: the method will work for all types of fits (not just 6-3). I leave the proof as an exercise.

A correct method is one which can be justified by the proper axioms at each step.

 

Just because you got the right answer does not mean there is no hole in your steps.

 

If you are a bridge player you should know this. Just because you make a contract does not mean you took the best play... does it? Not exactly a perfect analogy, but hope you see the point.

[lucky81]

Justified to do what? To give the correct answer. The method used can be justified to give the correct answer, as I have said. Please think about it for a while so that I do not have to explain more. You will probably see WHY the method is correct. Or else you will see a counterexample for the method (you won't), proving the method incorrect. If you really do not see it AFTER you have really thought about the method (which I think you have not done at all) - then I will be more than happy to justify the method in detail.

[Trumpace]

I won't waste anymore of your time. Sorry.

Please don't bother trying to explain it to me.

[Stephen Tu]

I believe in his latest simulation he was trying to count the following: assuming that given two hands (not necessarily a partnership) have a 6-3 spade fit, what is the distribution of the other two hands. He correctly assumed this would be the same distribution as for NS having a 6-3 fit.

 

No, he *incorrectly* assumed this. That is why his numbers came out wrong for the original question. His numbers are accurate if you phrase a different question. The question "if you have a 6-3 fit, how often will the opponent's cards divide 2-2?", is a different question, with a different answer, than "if you have a 6 cd suit, and someone else at the table has 3 of the suit, how often will the remaining cards be split 2-2 between the remaining hands?".

 

His method just gives more samples using the same set of deals than your method.

 

It doesn't "just give more samples", it gives samples that don't apply to the original question, samples where there are 6-2, 6-0, 6-1, 6-4 fits, not 6-3 fits. He overcounts all the splits, it so happens that his method of counting gets 3x the 2-2s and 1.5x the 3-1s.

 

He is not counting anything 3 times - the numbers are not even divisible by 3. He is trying to count something else than you think he was trying to count.

 

This is nonsense. I am not saying his script was counting the same deal 3 times. He is counting bogus deals, in such a way that his statistics (for the 2-2 & 4-0 split cases) will in the long run approach 3x the theoretically expected number. Since this is a statistical sample, of course the numbers do not have to be divisible by 3, that would only happen by pure chance. Only if one programmed the computer to exhaustively generate every possible bridge deal would you be able to guarantee divisibility by 3.

[helene_t]

Sorry, but in mathematics, ends don't justify the means.

 

Getting the right answer does not automatically imply the correctness of the method.

Lol. I was recently asked about my opinion about a machine learning method that had been proposed for some bioinformathics problem. I said that it seemed completely ad-hoc and I could see no reason why it would work. A colleague replied that I should not reject such methods solely on theoretical grounds, if they work fine in practice it is good enough. I replied that I'm a mathematician so it's my blody job to reject stuff solely on theoretical grounds.

 

As a mathematician I take the word "correct" as refering to its absolute, mathematical meaning. If that is not what is meant, a word like "useful" may be better.