Really wish op would at least just say he's trolling

He is not trolling, and to his credit he did not cave into the "LOL I was trolling all along" or start to behave like a troll to imply that. The guy is true to his beliefs, weird as those may be, and that is partially why this thread had the life it did. 10 days and 750+ posts is several times the usual length of these types of discussions with a single person like him.

Nah, the only reason why this thread got to where it was was due to several people going through the entire exercise with that person, from trying to clarify whatever he was talking about in his incoherent talk early on, to actually working out a way to test his silly theory (beyond watching and manually recording hands dealt in final table replay broadcasts and play money home game simulations), to then fully testing his theory and showing it is (to nobody's surprise) a figment of his imagination.

Nobody will do a significant amount of additional work for him at this point, regardless of which meaningless patterns he discovers when using hindsight cherry picking. Nobody will try that hard to make him understand why he has no idea what he is talking about. If he continues to rumble (without being genuinely freshly entertaining) then people will put him on ignore/block him and perhaps he will eventually be banned (which as he has said - is a constant end to his debates over the past 10 years).

Hey, I'll be the first one to hope he brings some new and amusing crazy to the mix, but his arguments and even his language (ostensible, ostensible , ostensible) are exceedingly repetitive, so I am not holding out hope.

I'm going to run 10 batches of 50k results. I will post them later as I just started the run and am otw out the door to play.

I will make the bold prediction that you will not find anything of concern. Still, all of this data will potentially give him a new life on a different forum after he looks it all over and cherry picks out the new patterns that concern him, and modifies his theory to incorporate that. You should feel ?/3 guilt enabling him in such a manner

I thank this forum for its patience in dealing with something rather strange. Please allow it to continue until the final test results have been done allowing me to communicate for a final time.

I have had an eureka moment and one new test would confirm something. But you all have done enough so with due respect, I will step out the thread once I have said my final on the numbers of the last tests of the numbers.

I wonder curiously , how many Ad would be simultaneously occupying the top card position of 10,000 shuffled decks.

I think it would be more than 1.

I think we just did the wrong tests ..........we needed to see how many aces occupied the top card spot after the 10,000 deck ,


How many aces are there as a top card of 10,000 top cards.

We did it wrong.

All the tests were wrong up to this point, and now you require new tests that will test what you want to test, until after it is tested and shown to not be significant, which at that point another test will be needed to test another test given the testing of the last tests did not pass your test test. What an unexpected twist!

Everyone - back to work!

?/?

All the best.

Just sauntering on by and noticed that we have a thread here where the ratio of very numerous replies to views is gigantic. I haven't had time to read it yet but the original thesis must certainly be of exquisite brilliance to engender such a response.

Yes, we need to see a consistency of 192 , in several runs .

Sorry, I clicked on a while ago, the test was not showing how many {a}/y

Oh you're going to love it.

Almost certainly more than 1.

The chances of there being 1 is so small that when I try to type the formula into google to get a result, it switches to using scientific notation. I think the chances of there being 1 Ad is 9.1359656e-87.

Same thing with the chances of there being 0. When I type that formula in, I get the chances of there being 0 Ad as 4.6593425e-85.

And when I try to put in the formula to calculate the chances of there being more than 1 Ad, google just rounds the answer to 1 (aka 100%).



The "expected number" of Ad would be approximately 192 (slightly more than that actually).

So do you agree the odds are not 1/52?

Not really. The chance of several runs in a row, producing exactly 192 Ad is very low. You need to learn how to use standard deviation and confidence intervals, which has been posted about a few times in this thread.

Actually I have an idea. Since these tests were all wrong and the algorithm could be rigged as well, why don't we turn a few sweat shops in India into card dealing factories? We'll have kids deal cards 18/7 to get a true and fair sample with real cards.

There was nothing up with the test we was just testing for the wrong thing.

I would expect some deviance of course, there is a chance although small of 0 ad. There is also a chance of all of them being ad.

No, I do not agree.


EDIT:

Just to skip ahead:

192/10,000 = 1/52 (give or take due to rounding)

A 192 out of 10,000 chance is the same thing as a 1 out of 52 chance.





Maybe a better series of tests would be to have two bags of marbles.

Bag 1 has 12 blue marbles and 1 red marble.

Bag 2 has 48 blue marbles and 4 red marbles.

Then we just draw marbles from each bag 1,000,000 times (after each draw replacing the drawn marble before the next draw). See how many times we draw a red marble from Bag 1 and how many times from Bag 2.

The point being to try to illustrate that the chance to draw a red marble from Bag 1 is exactly the same as the chance to draw a red marble from Bag 2 because 1/13 = 4/52.

I'm not gonna do it though.

Why do you think taking elements away is logical in an equation?

1/12 is not the same physically a 4/48


Try it this way,

I have a target that is 1cm² that is in the centre of a 12cm² board,


I have a second target that is 4cm² that is in the centre of a 48cm² board.




Which one is the easiest to shoot?

That is not a probability question.

The size of the target and the size of the board do not make up the numerator and denominator of a fraction; they do not have any relationship to one another. The size of the board isn't even relevant information to the question. You just presented it to try to make the question look similar.

A larger target is easier to hit than a smaller target. A 4cm² target is easier to hit than a 1cm² target.



Go try doing the marbles test yourself and see whether or not you draw roughly the same amount of red marbles from each bag.:


Bag 1 has 12 blue marbles and 1 red marble.

Bag 2 has 48 blue marbles and 4 red marbles.

Then we just draw marbles from each bag 1,000,000 times (after each draw replacing the drawn marble before the next draw). See how many times we draw a red marble from Bag 1 and how many times from Bag 2.

The point being to try to illustrate that the chance to draw a red marble from Bag 1 is exactly the same as the chance to draw a red marble from Bag 2 because 1/13 = 4/52.









And/or you could try watching the below video. Note that I only watched the first 30 seconds of it. It seemed like it was going to be right and a simple instructional video.











I'm gonna go take a shower and then do something else.






EDIT:

I would guess that the marbles test would be relatively easy to program. So maybe someone will be kind enough to write a program to do it for you.

I think you meant ?/2 showers (depending on the cm² of the shower). Just wait until time travel gets into the mix.

It appears that some people in this thread have no clue about probability and statistics (especially when it comes to testing or understanding randomness).

So I will repeat a post I made in the Internet Poker "Rigged" thread since it seems relevant once again and forever.

Then I am going to find the nearest wall and bang my head against it until I forget about this thread.

When this all began, I think most of us could have predicted two things:

1) the results of the test would show that there is absolutely nothing fundamentally wrong with the way cards are distributed; and
2) no matter what the test results were, pkdk would insist that the wrong test was being done and something had to be changed.

Congratulations to all who would have made those predictions.

I have looked at four tests that were done, two with 10,000 and two with 1 million. I have no idea what differences there were between the tests, because I did not conduct them, or read the source code. I just looked at the results. For some reason, the ace of diamonds seems to be of interest. I would expect that 1/52, or 1.9231% of the cards would be the ace of diamond. In any particular test, I would expect the result to be close to that. The results in tests 1,2,3 and 4 respectively were 2.04%, 2.07%, 1.923% and 1.92%. I have all the confidence in the world that you can do this 52 more times, or 100 more times, or 1000 more times, and the ace of diamonds will be selected very close to once for every 52 opportunities.

I don't know if my next sentence applies to this thread or not but there is a mathematical fact that I am not sure everyone knows.

The EV and/or the probability of an experiment that can be conducted in a various ways is the EV of the EVs or the probability of the probabilities. If I am 30% to be 20% and 70% to be 80% I am a 62% shot. I need not know any details.

Wow. Every time I think you've reached a maximum level of ridiculousness with your posts, you find a new one. I don't think I've ever seen someone so able to misapply knowledge again, and again, and again. And to do it with such arrogance, such a continual ability to look down on others who don't "understand" is...incredible.

The thought that you've been at this for a decade is sad. Really, really sad.

FYI the 10 batches have been posted to my drive in a new subfolder called "10 batches"