[helene_t]

Yeah, but that's a different issue. I am talking about people saying that there are patterns like South getting the better hand all evening, or West holding a 7-card suit three times in a row, etc.

BTW I don't believe the "well-known fact" of bad human shuffling either, but I could be wrong. It's very difficult to get good data on human shuffling so to some extent we will have to use our gut feelings. My gut feelings is that at most clubs, the difference between human and computer shuffling is too small to diagnose without rigorous analysis of thousands of hands. But I could be wrong.

[Vampyr]

helene_t, on Dec 9 2009, 03:02 PM, said:

BTW I don't believe the "well-known fact" of bad human shuffling either, but I could be wrong.

This applies less in duplicate bridge, I should have thought. Especially if the individual hands are shuffled before being returned to the board, prior to taking them out at the next occasion and shuffling and dealing.

But human shuffling does not generate hand records. In my youth I could remember all of the hands of a session -- thank goodness the possession by most clubs of a duplicating machine coincides with my mental decline.

[Grazy69]

Interesting thread.
As a scorer for many years with a computer scoring program that dealt with this type of problem when a redeal is required part way through the evening....
I would let scores stand for folks already played the board as stanza 1
I would have the board redealt ( when discovered ) and all scores from then on would be stanza 2.

As an aside ...... Many moons ago, on one of the forums or suchlike, someone came up with a very nice definition of manually well shuffled cards.
It was simple ...you should try it.

Suit the whole pack A23456 etc in all suits ( like a new deck ).
Use whatever type of shuffle you choose, and there are many.
Definition: A good shuffle is that no more than 3 pairs of cards remain together after the shuffle.
Eh Voila

[blackshoe]

cherdanno, on Dec 9 2009, 04:42 AM, said:

Btw, I wouldn't let the earlier results stand. Even if these players didn't notice it explicitly, they may have benefitted without realizing it (e.g. "learning" how to play a certain suit combination on the previous board).

I might buy that if the hands had not been rotated. They were, I don't.

[pran]

Grazy69, on Dec 9 2009, 06:17 PM, said:

As an aside ...... Many moons ago, on one of the forums or suchlike, someone came up with a very nice definition of manually well shuffled cards.
It was simple ...you should try it.

Suit the whole pack A23456 etc in all suits ( like a new deck ).
Use whatever type of shuffle you choose, and there are many.
Definition: A good shuffle is that no more than 3 pairs of cards remain together after the shuffle.
Eh Voila

I ran a quick simulation (very reliable) and found that in the long run one shuffle out of every five in a perfectly random shuffled pack of cards will exhibit more than 4 pairs in the pack.

(A "pair" then defined at two neighbouring cards in the pack being of the same rank)

[helene_t]

Yeah but that is not what was meant.

The number of retained pairs after a perfect shuffle is appr. Poison distributed with parameter 1. In that approximation, the probability of more than 3 pairs is 1.8%.

[pran]

helene_t, on Dec 9 2009, 11:54 PM, said:

Yeah but that is not what was meant.

The number of retained pairs after a perfect shuffle is appr. Poison distributed with parameter 1. In that approximation, the probability of more than 3 pairs is 1.8%.

Sorry - what is a "retained pair" then?

[nige1]

I think seven crude riffle shuffles usually suffice to randomise the deck. (Although eight perfect riffle shuffles return the original order).
A single riffle shuffle is ineffectual; and a simple magic trick depends on that for its effect: You (surreptitiously) arrange a full deck of fifty-two cards. You give it to any member of your audience. He cuts it and performs a single riffle shuffle (perfect or crude, it does not matter). Without disturbing the order of the deck, you deal it off in pairs, fours, and thirteens ...

• Each pair comprises a red card and a black card.
  
• Each set of four cards comprises one of each suit.
  
• Each set of thirteen cards comprises one of each rank.

provided you choose the right starting position

[helene_t]

pran, on Dec 10 2009, 12:19 AM, said:

helene_t, on Dec 9 2009, 11:54 PM, said:

Yeah but that is not what was meant.

The number of retained pairs after a perfect shuffle is appr. Poison distributed with parameter 1. In that approximation, the probability of more than 3 pairs is 1.8%.

Sorry - what is a "retained pair" then?

Say if ♠2 is followed by ♠3 in the original order, that pair is retained if ♠2 is also followed byt ♠3 after shuffling.

[pran]

helene_t, on Dec 10 2009, 12:53 AM, said:

pran, on Dec 10 2009, 12:19 AM, said:

helene_t, on Dec 9 2009, 11:54 PM, said:

Yeah but that is not what was meant.

The number of retained pairs after a perfect shuffle is appr. Poison distributed with parameter 1. In that approximation, the probability of more than 3 pairs is 1.8%.

Sorry - what is a "retained pair" then?

Say if ♠2 is followed by ♠3 in the original order, that pair is retained if ♠2 is also followed byt ♠3 after shuffling.

And this should be a test for a perfect shuffle?

Split the pack in two exact halves and perform one single "perfect riffle shuffle". Then there are no retained pairs, but perfect shuffle?????

[helene_t]

pran, on Dec 10 2009, 09:27 AM, said:

helene_t, on Dec 10 2009, 12:53 AM, said:

pran, on Dec 10 2009, 12:19 AM, said:

helene_t, on Dec 9 2009, 11:54 PM, said:

Yeah but that is not what was meant.

The number of retained pairs after a perfect shuffle is appr. Poison distributed with parameter 1. In that approximation, the probability of more than 3 pairs is 1.8%.

Sorry - what is a "retained pair" then?

Say if ♠2 is followed by ♠3 in the original order, that pair is retained if ♠2 is also followed byt ♠3 after shuffling.

And this should be a test for a perfect shuffle?

Split the pack in two exact halves and perform one single "perfect riffle shuffle". Then there are no retained pairs, but perfect shuffle?????

Lol, it wasn't my claim that it's a test for a perfect shuffle

Obviously you can't establish on the basis of a single sample that the cards have been shuffled well.

You may be able to establish that they haven't been shuffled well, though.

[jdonn]

helene_t, on Dec 10 2009, 07:26 AM, said:

Obviously you can't establish on the basis of a single sample that the cards have been shuffled well.

You may be able to establish that they haven't been shuffled well, though.

This may be the first time I have ever thought you were wrong about a science or statistics question, but I disagree with your second statement here.

[aguahombre]

jdonn, on Dec 10 2009, 09:42 AM, said:

helene_t, on Dec 10 2009, 07:26 AM, said:

Obviously you can't establish on the basis of a single sample that the cards have been shuffled well.

You may be able to establish that they haven't been shuffled well, though.

This may be the first time I have ever thought you were wrong about a science or statistics question, but I disagree with your second statement here.

Yes, I agree with your disagree. This thread started with a hand that was an exact duplicate except for the rotation one step counter-clock wise. This would be equivalent to taking the cards out to shuffle, getting sidetracked, and putting the cards back into the board. I would call that "not shuffled well".

[helene_t]

jdonn, on Dec 10 2009, 05:42 PM, said:

helene_t, on Dec 10 2009, 07:26 AM, said:

Obviously you can't establish on the basis of a single sample that the cards have been shuffled well.

You may be able to establish that they haven't been shuffled well, though.

This may be the first time I have ever thought you were wrong about a science or statistics question, but I disagree with your second statement here.

What I mean is this:

- Suppose there are less than 4 retained pairs after shuffling. Obviously this doesn't prove that they have been shuffled well. Pran gives an example of how bad shuffling could lead to zero retained pairs.

- Suppose there are much more than 4 retained pairs. Say 51 retained pairs, i.e. the order is exactly the same as before shuffling. For all practical purposes this proves that the cards have not been shuffled well.

Do you really disagree with that?

Seems like I somehow didn't make myself clear, since Aquahombre says the same as I say and then says he agrees with your disagree

[jdonn]

Yes it is weird he said he agreed with me then made an argument that agreed with you.

Anyway, not to answer a question with a question, but isn't what you are claiming here the exact same logic the director in this thread used to make an adjustment with no legal justification, that of there being 'too big' of a coincidence?

[aguahombre]

helene_t, on Dec 10 2009, 10:02 AM, said:

Seems like I somehow didn't make myself clear, since Aquahombre says the same as I say and then says he agrees with your disagree

Perhaps it was I who didn't make myself clear. Subtle difference between not shuffling well and not really shuffling at all. My example was the second case --and in effect what the computer did in the OP.

If there was a feeble attempt at shuffling, I doubt anyone could really spot that without artificial help (computer analysis, e.g.).

[helene_t]

jdonn, on Dec 10 2009, 06:12 PM, said:

Yes it is weird he said he agreed with me then made an argument that agreed with you.

Anyway, not to answer a question with a question, but isn't what you are claiming here the exact same logic the director in this thread used to make an adjustment with no legal justification, that of there being 'too big' of a coincidence?

Ah right, yes indeed it is the same.

And it is a valid argument, I think.

It depends how much you trust the dealing process. Say some random sucker deal the cards manually, and there are 4 retained pairs. I would say that suggests that he didn't deal them perfectly but I could easily be wrong. Now suppose there are 8 retained pairs. I would then say that it proves beyond reasonable doubt that he didn't shuffle them perfectly.

OTOH, if Big Deal produces a shuffle with 4 retained pairs (OK, that may be nonsense, as we there may be no such thing as an original order of the cards, but you get the idea), it wouldn't raise my suspicion at all. 8 retained pairs may warrant an investigation but I would still find it more likely that it was a coincidence. 51 retained pairs? Now I would be pretty sure it was a bug.

When I say you may be able to establish that the cards were not shuffled well I mean that you can prove it beyond reasonable doubt. Not that you can prove in the mathematical sense. Statistical analysis can't prove anything in the mathematical sense.

[jdonn]

I guess you are right, although we all know any particular order of the cards after 'perfect' shuffling is as likely as any other, including the original order they started in.

Ok I'll shut up now.

[helene_t]

jdonn, on Dec 10 2009, 06:50 PM, said:

we all know any particular order of the cards after 'perfect' shuffling is as likely as any other, including the original order they started in.

Yes but you have to apply Bayes' theorem:

Say that a priori there is a probability of 90% that I shuffle well.

Given that I shuffle well, there is a probability of 1.8% that I will produce 4+ retained pairs.

Given that I shuffle badly, the probability of 4+ retained pairs is higher. Say 50%.

What is the probability that I shuffle badly, given that there are 4+ retained pairs?

Shuffle well: 0.018 * 90% = 0.0162
Shuffle badly: 0.50 * 10% = 0.05

So the probability that I shuffle badly is now 0.05 / ( 0.05+0.0162) or some 76%.